# Lakhmir Singh Class 10 Physics 5th Chapter Refraction of Light Solution

## Lakhmir Singh Class 10 Physics 5th Chapter Refraction of Light Solution

Lakhmir Singh Manjit Kaur Physics Solution: Refraction of Light Chapter 5. Here you get easy Solutions of Lakhmir Singh Class 10 Physics Solution Chapter 5. Here we have given Chapter 5 all Solution of Class 10. Its help you to complete your homework.

• Board – CBSE
• Text Book – Physics
• Class – 10
• Chapter – 05

### Lakhmir Singh Class 10 Physics 5th Chapter Solution

#### Very Short Answer Type Questions

1) If a ray of light goes from a rarer medium to a denser medium, will it bend towards the normal or away from it?

Ans: If a ray of light goes from a rarer medium to a denser medium then it bends towards the normal.

2) If a ray of light goes from a denser medium to a rarer medium, will it bend towards the normal or away from the normal?

Ans: If a ray of light goes from a denser medium to a rarer medium then the ray bends away from the normal.

3) A beam of light travelling in a rectangular glass slab emerges into air) Draw a ray-diagram indicating the change in its path.

Ans: The following figure shows the change in path of the light when a beam of light is traveling in a rectangular glass slab emerges into air. Fig. Refraction of light through rectangular glass slab

4) A beam of light travelling in air is incident on water. Draw a ray-diagram indicating the change in its path in water.

Ans: The following diagram shows the path of ray in water when a beam of light is traveling in air is incident on water. 5) A ray of light travelling in water emerges into air. Draw a ray-diagram indicating the change in its path.

Ans:

The following diagram shows the change in path of the ray of light which is traveling in water emerges into air. 6) A ray of light travelling in air is incident on a parallel-sided glass slab (or rectangular glass slab). Draw a ray-diagram indicating the change in its path in glass.

Ans:

The following figure shows the change in the path of light ray in glass when a ray of light travelling in air is incident on a parallel sided glass slab. Fig. Refraction of light through rectangular glass slab

7.) A ray of light travelling in glass emerges into air. State whether it will bend towards the normal or away from the normal.

Ans:

• When a ray of light travelling in glass emerge into air then it bends away from the normal.
• Because, we know that when a ray of light is travelling from denser medium to rarer medium it bends away from the normal.
• And here, the ray of light is traveling from glass to air hence it bends away from the normal when emerge in the air. Since glass is denser than air medium.

8.) A ray of light travelling in air enters obliquely into water. Does the ray of light bend towards the normal or away from the normal? Why?

Ans:

• We know that, when a ray of light is traveling from rarer medium to denser medium then it bends towards the normal.
• Here, the ray of light is traveling from air medium into water and hence it bends towards the normal. Since water is denser than air medium.

9.) A ray of light goes from water into air. Will it bend towards the normal or away from the normal?

Ans:

• We know that, the ray of light traveling from denser medium into rarer medium bends away from the normal.
• Here, the ray of light is traveling from water into air hence it bends away from the normal.
• Since, water is denser than air medium.

10.) State two effects caused by the refraction of light.

Ans:

• When we kept the lemon inside the glass tumbler filled with full of water then if we see that lemon from outside it appears bigger than its actual size and this is due the refraction of light.
• When we kept the thick glass slab over the printed paper having some letters and if we see that letter from top side of the glass slab then the letters appear to be raised and this is also due to the refraction of light.

11.) Name the phenomenon due to which a swimming pool appears less deep than it really is.

Ans:

Due to the phenomenon of refraction of light a swimming pool appears less deep than it really is.

12. When a ray of light passes from air into glass, is the angle of refraction greater than or less than the angle of incidence?

Ans:-

When a ray of light passes from air into glass then the angle of refraction is less than the angle of incidence. Since when a ray of light travels from air into glass it bends towards the normal.

13.) A ray of light passes from air into a block of glass. Does it bend towards the normal or away from it?

Ans:

A ray of light passes from air into a block of glass bends towards the normal.
Because, when a ray of light is traveling from air medium to denser medium it bends towards the normal. Since glass is denser than air medium.

14.) As light rays pass from water into glass, are they refracted towards the normal or away from the normal?

Ans:

When a ray of light pass from water into glass it bends towards the normal.
Because, when a ray of light travels from rarer medium into denser medium it bends towards the normal. Since, glass is denser than water medium.

15.) In which material do you think light rays travel faster–glass or air?

Ans:

The medium which is denser has less speed of light in it. Here glass is denser than air and hence the speed of light in air medium is faster than in the glass medium.

16.) Which phenomenon of light makes the water to appear shallower than it really is?

Ans:

The phenomenon of refraction of light makes the water to appear shallower than it really is.

17.) State whether the following statement is true or false: Refraction occurs because light slows down in denser materials.

Ans:

The above statement is true.

18.) Why does a ray of light bend when it travels from one medium to another?

Ans:

When ray of light travels from one medium to another it bends because the speed of light changes with respect the density of the medium.
In rarer medium the velocity of light is more than in denser medium.

19.) Fill in the following blanks with suitable words:

Ans:

a) Light travelling along a normal is not refracted.
b) Light bends when it passes from water into air. We say that it is refracted.

20) What is meant by ‘refraction of light’? Draw a labelled ray diagram to show the refraction of light.

Ans:

• When a ray of light travels from one medium to another medium then its path changes in another medium which is called as refraction of light.
• The refraction of light is due to the change in the speed of the light in different media with respect to their optical density.
• The following diagram shows the refraction of light. Fig. Refraction of light through rectangular glass slab

21) A ray of light travelling in air is incident on a rectangular glass block and emerges out into the air from the opposite face. Draw a labelled ray diagram to show the complete path of this ray of light. Mark the two points where the refraction of light takes place. What can you say about the final direction of ray of light?

Ans:

• When a ray of light travelling in air is incident on the rectangular glass block then it bends towards the normal as shown in figure below.
• After that it again emerges into air from glass block then it bends away from the normal.
• In figure, O and O’ are points where refraction of light takes place.
• The ray finally emerge out from the glass block into air is the emergent ray which is parallel to the incident ray as shown in figure, 22) Draw a labelled ray diagram to show how a ray of light is refracted when it passes:

(a) from air into an optically denser medium.

(b) from an optically denser medium into air.

Ans:

• The following figure shows the refraction of light when it is passes from air into an optically denser medium. • The following figure shows the refraction of light when it passes from an optically denser medium into air. 23) The diagram given alongside shows a ray of light entering a rectangular block of glass.

(a) Copy the diagram and draw the normal at the point of entry.

(b) Draw the approximate path of the ray of light through the glass block and out of the other side.

Ans:

a) b) Fig. Refraction of light through rectangular glass slab

The above figure shows the approximate path of ray of light through the glass block and out of the other side.

24) What is meant by the ‘angle of incidence’ and the ‘angle of refraction’ for a ray of light? Draw a labelled ray diagram to show the angle of incidence and the angle of refraction for a refracted ray of light.

Ans:

• The angle made by the incident ray with the normal at the point of incidence is called as angle of incidence. It is denoted by i.
• The angle made by the refracted ray with the normal at the point of incidence is called as the angle of refraction. It is denoted by r.
• The following figure shows the angle of incidence i and the angle of refraction r. 25) Light travels more quickly through water than through glass.

(a) Which is optically denser: water or glass?

(b) If a ray of light passes from glass into water, which way will it bend: towards the normal or away from the normal?

Ans:

a) The medium through which the speed of light is less it is the optically denser medium. As light quickly pass through the water than glass that means glass is optically denser than water.

b) If a ray of light passes from glass into water it bends away from the normal. Since when a ray of light travels from denser medium to rarer medium it bends away from the normal.

26) Draw a labelled ray diagram to show how a ray of light passes through a parallel sided glass block:

(a) if it hits the glass block at 90° (that is, perpendicular to the glass block)

(b) if it hits the glass block at an angle other than 90° (that is, obliquely to the glass block).

Ans:

a) b) 27) When a light ray passes from air into glass, what happens to its speed? Draw a diagram to show which way the ray of light bends.

Ans:

• When a ray of light travels from air medium into the glass medium it bends towards the normal. Because glass is more optical denser than the air medium. And we know that the speed of light in more optically denser medium is less.
• Hence the speed of light ray decreases when it enters the glass medium from the air medium.
• The following figure shows the bending of ray of light from air medium into glass medium towards the normal. 28) (a) Explain why, a stick half immersed in water appears to be bent at the surface. Draw a labelled diagram to illustrate your answer.

(b) A coin in a glass tumbler appears to rise as the glass tumbler is slowly filled with water. Name the phenomenon responsible for this effect.

Ans:

• A stick half immersed in water appears to be bent at the surface because of the refraction of light when light passes from water into air.
• In following figure, a straight stick AO whose lower part BO is immersed in the water as shown in figure below. Although the stick is already straight but when it is immersed in water appears bent at point B in the direction BI.
• When the light ray OC comes from the lower end of the stick O from water to air at point C there is refraction takes place and it bends away from the normal as CX.
• Similarly, the ray OD get refracted in air as DY as shown in figure below.
• These two refracted rays CX and DY when we extend them in backward direction then they meet at point I and it is the virtual image of the end point O of the stick AO, which is formed due to the refraction of light when it enters from water into the air.
• Thus, the person seeing from position E sees the end O of the stick at point I which is nearer to the water surface.
• Thus, we can say that the virtual image of the lower end BO of the stick is formed as BI. Hence, when we see stick inside the water actually it is the virtual image BI of the portion BO.
• And we observe that, the upper end of the stick which in air AB and virtual image BI are not in the same direction or straight line and thus the stick AO appears like bent at the point B in water as BI.
• Thus, the actual stick ABO appears like ABI but it not means that the stick is bent really. But only the light coming from water into the air get bent.
• Thus, when the pencil is partially immersed in water such that it held obliquely to the surface then we see that the pencil appears bent at the surface of water. • When a coin in glass tumbler is immersed then the coin will be under water and due to refraction of light a virtual image of the coin is formed nearer to surface of water. And we see only that virtual image which appears to the water surface hence the coin appears to rise after adding the water in glass tumbler.
• Thus, the phenomenon of refraction is responsible for the rise of coin which is under glass tumbler filled water slowly.

29) (a) With the help of a labelled diagram, explain why a tank full of water appears less deep than it actually is.

(b) Name the phenomenon due to which a pencil partly immersed in water and held obliquely appears to be bent at the water surface.

Ans:

a)

• See the following figure which shows a tank of water. We consider any point O at the bottom of the tank and this point is under the surface of water. We can see that point due to the light rays coming from that point. Let us consider the ray of light OA is coming from the point O from water into the air at point A. As it is traveling from water into air get bends away from the normal in the direction as AX as shown in figure.
• Similarly, another ray of light OB which is coming from point O travels from water into air get refracted at point B and hence get bend away from the normal in the direction as BY.
• When we extend these refracted rays in backward direction then we see that they meet at point I under the water. And when we see these refracted rays AX and BY then they appear like coming from point I.And point I is the virtual image of point O. And the virtual image I of the point O is nearer to the water surface than point O. Thus, we can see that the point O which is at the bottom of the water tank appears much nearer to position I.
• And we can apply the same phenomenon to all the points which forms the base PQ of the tank. And hence due to refraction of light the bottom PQ of the tank appears more nearer to water surface as P’Q’ and hence this is the reason due to which the tank full of water appears less deep than actually it is.
• Thus, when we see in the water of tank then we observe the bottom of tank is raised to some height but this is the virtual image of the bottom of tank which is formed due to the refraction of light coming from water into air. b)

• when the pencil is partially immersed in water such that it held obliquely to the surface then we see that the pencil appears bent at the surface of water.
• This is due to the phenomenon of refraction of light coming from water into the air.

30) (a) With the help of a diagram, show how when light falls obliquely on the side of a rectangular glass slab, the emergent ray is parallel to the incident ray.

(b) Show the lateral displacement of the ray on the diagram.

(c) State two factors on which the lateral displacement of the emergent ray depends.

Ans:

a)

• The following diagram shows the when light falls obliquely on the side of a rectangular glass slab the emergent ray is parallel to the incident ray.
• In figure, the incident ray AO bends towards the normal at point O whereas the refracted ray OB bends away from the normal at point B by an equal amount. Due to which, the light emerges from the parallel sides glass slab in a direction parallel with that ray of light which enters the glass slab.
• In figure emergent ray BC is parallel to the incident ray AO. b)
In figure, the emergent ray BC is laterally displaced from the original path of the incident ray AO by the perpendicular distance CD which is called as the lateral displacement.

The perpendicular distance between the original path of the incident ray and the emergent ray coming out of the glass slab is called as lateral displacement of the emergent ray as shown in figure.

c)
Lateral displacement depends on the three factors mainly:

• Angle of incidence, thickness of glass slab and refractive index of the glass slab.
• And also, the lateral displacement of the emergent ray is directly proportional to the angle of incidence, thickness of glass and refractive index of the glass lab.

31.) Explain with the help of a labelled ray diagram, why a pencil partly immersed in water appears to be bent at the water surface. State whether the bending of pencil will increase or decrease if water is replaced by another liquid which is optically denser than water. Give reason for your answer.

Ans:

• A pencil half immersed in water appears to be bent at the surface because of the refraction of light when light passes from water into air.
• In following figure, a straight pencil AO whose lower part BO is immersed in the water as shown in figure below. Although the pencil is already straight but when it is immersed in water appears bent at point B in the direction BI.
• When the light ray OC comes from the lower end of the pencil O from water to air at point C there is refraction takes place and it bends away from the normal as CX.
• Similarly, the ray OD get refracted in air as DY as shown in figure below.
• These two refracted rays CX and DY when we extend them in backward direction then they meet at point I and it is the virtual image of the end point O of the pencil AO, which is formed due to the refraction of light when it enters from water into the air.
• Thus, the person seeing from position E sees the end O of the pencil at point I which is nearer to the water surface.
• Thus, we can say that the virtual image of the lower end BO of the pencil is formed as BI. Hence, when we see pencil inside the water actually it is the virtual image BI of the portion BO.
• And we observe that, the upper end of the pencil which in air AB and virtual image BI are not in the same direction or straight line and thus the pencil AO appears like bent at the point B in water as BI.
• Thus, the actual pencil ABO appears like ABI but it not means that the pencil is bent really. But only the light coming from water into the air get bent.
• Thus, when the pencil is partially immersed in water such that it held obliquely to the surface then we see that the pencil appears bent at the surface of water. And if we replaced water by another optically denser medium than water then the bending of pencil increases because in more optically denser medium the refraction of light takes place is more and hence bending of light is also more.

### Laws of Refraction:

#### Very Short Answer Type Questions

1.) What name is given to the ratio of sine of angle of incidence to the sine of angle of refraction?

Ans:

The ratio of sine of angle of incidence to the sine of angle of refraction is constant for given pair of media and it is called as the refractive index of the second medium in which light enters.
Thus, sini/sinr = refractive index of the second medium

2.) Write the relation between the angle of incidence and the angle of refraction for a medium.

Ans:

The relation between angle of incidence and angle of refraction for a medium is given by
Refractive index of the medium = Sin(i)/ Sin(r)

3.) What is the unit of refractive index?

Ans:

Refractive index has no unit since it is the ratio of two similar quantities.

4.) Which has higher refractive index: water or glass?

Ans:

The medium which is more optically denser has the more or higher refractive index. Here glass is more optically denser than the water and hence glass has higher refractive index than the water.

5.) Refractive indices of carbon disulphide and ethyl alcohol are 1.63 and 1.36 respectively. Which is optically denser?

Ans:

The material medium having higher refractive index is the more optically denser medium.
Here, the refractive index of carbon disulphide is more than ethyl alcohol hence carbon disulphide is more optically denser than ethyl alcohol.

6.) The refractive index of diamond is 2.42. What is the meaning of this statement in relation to the speed of light?

Ans:

The refractive index of diamond is 2.42. It means that the ratio of speed of light in air to the speed of light in diamond is 2.42.
Thus, speed of light in air/ speed of light in diamond = 2.42

7.) If the refractive index for light going from air to diamond be 2.42, what will be the refractive index for light going from diamond to air?

Ans:

We know that, the refractive index of diamond is 2.42 means the ratio of speed of light in air to the speed of light in diamond only when light travels from air to diamond.

Hence, 2.42= speed of light in air / speed of light in diamond
Thus, when light goes from diamond into air then refractive index will be
Speed of light in diamond/ speed of light in air= 1/(2.42)= 0.41

Thus, the refractive index will be 0.41

8.) How is the refractive index of a material related to the speed of light in it?

Ans:

When a ray of light travels from air into the another medium then the refractive index of the medium is given by

Refractive index of the medium = speed of light in air/ speed of light in medium.

9.) Fill in the following blank with a suitable word:

Ans:

When a ray of light goes from air into a clear material, you see the ray bend. How much the ray bends is determined by the refractive index of the material.

10. Give three examples of materials that refract light rays. What happens to the speed of light rays when they enter these materials?

Ans:

• The materials which refract the light rays are water, air, glass, diamond etc.
• When the ray of light traveling from one medium to other changes its speed and hence there is change in the direction of path of the light ray.
• The optical density of the medium decides the how much light will be refracted through it
• More is the refractive index more is the optical density of the medium and hence refraction of light also more.

11.) Define Snell’s law of refraction. A ray of light is incident on a glass slab at an angle of incidence of 60°. If the angle of refraction be 32.7°, calculate the refractive index of glass. (Given: sin 60° = 0.866, and sin 32.7° = 0.540).

Ans:

The ratio of sine of angle of incidence to the sine of angle of refraction is constant for a pair of media and it is called as the refractive index of the medium. This law is called as Snell’s law.

Thus, Sin(i) / Sin(r) = constant= refractive index of the medium

Given that,
Angle of incidence = 60°
Angle of refraction = 32.7°

We know that,
Sin (60°)/ Sin (32.7°) = refractive index of the glass
Thus, refractive index of the glass = 0.866/ 0.540 = 1.6
Thus, the refractive index of the glass is 1.6

12.) The speed of light in vacuum and in two different glasses is given in the table below: Medium Speed of light Vacuum 3.00 × 108 m/s Flint glass 1.86 × 108 m/s Crown glass 1.97 × 108 m/s

(a) Calculate the absolute refractive indexes of flint glass and crown glass.

(b) Calculate the relative refractive index for light going from crown glass to flint glass.

Ans:

a)

• The absolute refractive index of the flint glass is given by
• Absolute refractive index of the flint glass = speed of light in vacuum/ speed of light in flint glass
• Absolute refractive index of flint glass = 3*108/ 1.86*108 = 1.61
• Thus, absolute refractive index of the flint glass is 1.61
• The absolute refractive index of the crown glass is given by,
• Absolute refractive index of crown glass = speed of light in vacuum/ speed of light in crown glass
• Thus, absolute refractive index of crown glass= 3*108/ 1.97*108 = 1.52
• Thus, the absolute refractive index of the crown glass is 1.52

b)

• The relative refractive index for light going from crown glass to flint glass is given by,
• Relative refractive index for light going from crown glass to flint glass= speed of light in crown glass/ speed of light in flint glass
• Thus, relative refractive index = 1.97*108/ 1.86*108 = 1.059
• Thus, the relative refractive index for light going from crown glass to flint glass is 1.059

13.) The speed of light in air is 3 × 108 m/s. In medium X its speed is 2 × 108 m/s and in medium Y the speed of light is 2.5 × 108 m/s.

Calculate: (a) air nX (b) air nY (c) xnY

Ans:

Given that,
Speed of light in air = 3*108 m/s
Speed of light in medium X = 2*108 m/s
Speed of light in medium Y = 2.5*108 m/s

a)
air n X = speed of light in air/ speed of light in medium X= 3*108/ 2*108= 1.5

b)
air n Y = speed of light in air / speed of light in medium Y = 3*108/ 2.5*108 = 1.2

c)
X n Y = speed of light in medium X/ speed of light in medium Y = 2*108/ 2.5*108 = 0.8

14.) What is the speed of light in a medium of refractive index 6/5 if its speed in air is 3,00,000 km/s?

Ans:

Given that,
Speed of light in air = 3,00,000
Refractive index of the medium = 6/5

We know that,
Refractive index of the medium = speed of light in air/ speed of light in medium
Thus, speed of light in medium = speed of light in air/ refractive index of medium
= 3,00,000*5/6 = 2,50,000 km/s
Thus, the speed of light in medium is 2,50,000 km/s

15.) The refractive index of glass is 1.5. Calculate the speed of light in glass. The speed of light in air is 3.0 × 108 ms–1.

Ans:

Given that,
Refractive index of glass = 1.5
Speed of light in air = 3*108m/s

We know that,
Refractive index of the glass = speed of light in air/ speed of light in glass
Thus, speed of light in glass = speed of light in air/ refractive index of the glass
= 3*108/ 1.5 = 2*108 m/s

Thus, the speed of light in glass is 2*108 m/s.

16.) The speed of light in water is 2.25 × 108 m/s. If the speed of light in vacuum be 3 × 108 m/s, calculate the refractive index of water.
Ans:
Given that,
Speed of light in water = 2.25*108 m/s
Speed of light in vacuum = 3*108 m/s
We know that,
Refractive index of water = speed of light in air/ speed of light in water = 3*108/ 2.25*108 = 1.33
Thus, the refractive index of water is 1.33

17.) Light enters from air into diamond which has a refractive index of 2.42. Calculate the speed of light in diamond. The speed of light in air is 3.0 × 108 ms–1.

Ans:

Given that,
Refractive index of diamond = 2.42
Speed of light in air = 3*108 m/s

We know that,
Refractive index of diamond = speed of light in air/ speed of light in diamond
Thus, speed of light in diamond = speed of light in air/ refractive index of diamond
= 3*108/ 2.42= 1.24*108 m/s
Thus, the speed of light in diamond is 1.24*108 m/s

18.) (a) State and explain the laws of refraction of light with the help of a labelled diagram.

(b) What is meant by the refractive index of a substance?

(c) Light travels through air at 300 million ms–1. On entering water, it slows down to 225 million ms–1. Calculate the refractive index of water.

Ans:

a)
When a ray of light travels from one optical medium to another optical medium then it changes its path of motion due to change in speed indirectly due to change in refractive index of the medium.
The laws of refraction are stated as follows:

1) First law of refraction states that, the incident ray, refracted ray and normal at the point of incidence all lie in the same plane.

In figure below, the incident ray AO, refracted ray OB and normal ON all lie in the same plane. • The second law of refraction states that, the ratio of sine of angle of incidence to the sine of angle of refraction is constant for a given pair of media. And this constant is called as the refractive index of the medium. This law is also called as Snell’s law.

Thus, Sin(i) / Sin(r) = constant = refractive index of medium

b)

• The refractive index of the medium 2 with respect to medium 1 is equal to the ratio of speed of light in medium 1 to the speed of light in medium 2.

Thus, we can write as

1 n 2 = speed of light in medium 1 / speed of light in medium 2

• Also, the refractive index of medium or substance with respect to the air medium is called as absolute refractive index.
• And the ratio of speed of light in air to the speed of light in a medium is called as refractive index of the medium.

Thus, refractive index of medium = speed of light in air/ speed of light in vacuum

c)

Speed of light in air = 3*108 m/s

Speed of light in water = 2.25*108 m/s

We know that,

Refractive index of the water = speed of light in air/ speed of light in water

Refractive index of water= 3*108/ 2.25*108 = 1.33

Thus, the refractive index of water is 1.33

### Refraction of Light by Spherical Lenses

#### Very Short Answer Type Questions –

1.) Name the lens which can concentrate sun’s rays to a point and burn a hole in a piece of paper.

Ans:

A concave lens is used to concentrate sun’s rays to a point and burn a hole in a piece of paper.

2.) Give the usual name for the following: A point inside a lens through which the light passes undeviated .

Ans:

A point inside a lens through which the light passes undercoated is the optical centre.

3.) A 1 cm high object is placed at a distance of 2f from a convex lens. What is the height of the image formed?

Ans:

• Since, here the object is placed at a distance of 2f from a convex lens then the height of object is equal to the height of the image formed.
• Hence, here the height of the image will be also 1cm.

4.) If the image formed by a convex lens is of the same size as that of the object, what is the position of the image with respect to the lens?

Ans:

If the image formed by a convex lens is of the same size as that of the object, then the position of the image with respect to the lens is always at 2F that means at a distance of twice the focal length.

5.) If an object is placed at the focus of a convex lens, where is the image formed?

Ans:

If the object is placed at the focus of a convex lens then the image will be formed at very large distance i.e. at infinity.

6.) Where should an object be placed in order to use a convex lens as a magnifying glass?

Ans:

To use the convex lens as a magnifying glass we have to place the object at a distance which is less than the focal length.

7.) Where should an object be placed in front of a convex lens so as to obtain its virtual, erect and magnified image?
Ans:
We have to place the object within focus of a convex lens so as to obtain it’s virtual, erect and magnified image.

8.) Where should an object be placed in front of a convex lens so as to obtain its real, inverted and magnified image?

Ans:

We have to place the object between F and 2f of a convex lens to obtain it’s virtual, inverted and magnified image.

9.) For what position of an object a real, diminished image is formed by a convex lens?

Ans:

When the object is placed beyond 2f then only the image formed is real, diminished by the convex lens.

10.) If an object is at a considerable distance (or infinity) in front of a convex lens, where is the image formed ?

Ans:

If an object is at a considerable distance or infinity in front of a convex lens then the image will be formed at focus F of convex lens.

11. Draw the given diagram in your answer book and complete it for the path of a ray of light after passing through the lens.

Ans: 12.) What type of lens would you use as a magnifying glass? How close must the object be to the lens?

Ans:

We will use convex lens as the magnifying glass. And to get magnified image of the object we have to place it in between optical centre C and the focus F of the convex lens.

13.) Name two factors on which the focal length of a lens depends.

Ans:

The following are the two factors on which focal length of the lens depends:

1) The curvature of the lens

2) And the material medium from which the lens is made.

14.) State any two uses of convex lenses.

Ans:

Following are the uses of convex lens:

• To correct the vision defect called as hypermetropia convex lens are used.
• Convex lens is also used in making simple cameras
• They are also used as magnifying glass.

15. Fill in the following blanks with suitable words:

Ans:

• Parallel rays of light are refracted by a convex lens to a point called focus.
• The image in a convex lens depends upon the distance of the object from the lens.

16) What is a lens? Distinguish between a convex lens and a concave lens. Which of the two is a converging lens: convex lens or concave lens?

Ans:

Lens is the transparent glass which is made up of two spherical surfaces. There are two types of spherical lens that are concave lens and convex lens which are explained as follows.

Concave lens:

• Concave lens is also called as diverging lens because when parallel beam of light rays is incident on it, it diverges them.
• Concave lens is thin at centre and it is bulged at its edges.
• It forms the images which are always diminished and virtual.
• The focus of the concave lens is virtual. Fig. Concave lens

Convex lens:

• Convex lens is also called as converging lens because when parallel beam of light rays is incident on it, it converges them.
• Convex lens is thin at its edges and it is bulged at its centre.
• It forms the image which is real or virtual and diminished or magnified.
• The focus of the convex lens is real. Fig. Convex lens

17) (a) Explain with the help of a diagram, why the convex lens is also called a converging lens.

(b) Define principal axis, principal focus and focal length of a convex lens.

Ans:

a)

The following diagram shows why convex lens is called as converging lens. Fig. Converging action of convex lens

b)

Principal axis of convex lens:

It is the line which is passing through optical centre of the lens and also perpendicular to the both the faces of the lens.

Principal focus of convex lens:

It is the point on the principal axis and when the parallel beam of rays is incident parallel to principal axis then after refraction all the rays meet in a single point on the principal axis which is called as principal focus of the convex lens.

Convex lens has real focus.

Focal length of convex lens:

The distance between optical centre and the principal focus of the convex lens is called as focal length of convex lens.

18) (a) Explain with the help of a diagram, why the concave lens is also called a diverging lens.

(b) Define the principal focus of a concave lens.

Ans:

a)

The following figure shows that why the concave lens is also called as diverging lens. Fig. Diverging action of concave lens

b)

Principal focus of the concave lens:

When parallel beam of rays is incident on the concave lens which is also parallel to principal axis of concave lens then after refraction all the rays appears like diverging from the point on the principal axis and that point is called as principal focus of concave lens.

Concave lens has virtual focus.

19) Draw a ray diagram to show the formation of a real magnified image by a convex lens. (In your sketch the position of object and image with respect to the principal focus of lens should be shown clearly).

Ans:

The following diagram shows the formation of real and magnified image by a convex lens.

When the object is placed between F1 and C1 then the image will be formed beyond C2 which is enlarged, real and inverted as shown in figure below. When the object is placed at F2 then image will be formed at infinity which is highly enlarged, real and inverted as shown in following figure. 20) Describe with the help of a ray-diagram, the formation of image of a finite object placed in front of a convex lens between f and 2f. Give two characteristics of the image so formed.

Ans:

The following diagram shows the formation of image of a finite object placed in front of a convex lens between F and 2f. When the object is placed between F1 and C1(2F) then the image will be formed beyond C2 which is enlarged, real and inverted as shown in figure above.

21) Describe with the help of a ray diagram the nature, size and position of the image formed when an object is placed in front of a convex lens between focus and optical centre. State three characteristics of the image formed.

Ans:

The following diagram shows the formation of image when an object is placed in front of a convex lens between focus and optical centre. When the object is placed between focus F and optical centre O of convex lens then the image will be formed on the same side of the lens as the object. And the image formed is enlarged, virtual and erect as shown in figure above.

22) An object is placed at a distance equal to 2f in front of a convex lens. Draw a labelled ray diagram to show the formation of image. State two characteristics of the image formed.

Ans:

The following diagram shows the formation of image when an object is placed at a distance equal to 2f in front of a convex lens. When the object is placed at a distance equal to 2f i.e. at C1 then the image formed will be at C2 which is of the same size, real and inverted as shown in figure above.

23) Describe with the help of a ray-diagram, the size, nature and position of the image formed by a convex lens when an object is placed beyond 2f in front of the lens.

Ans:

The following diagram shows the formation of image when an object is placed beyond 2f in front of the lens. When the object is placed beyond 2f i.e. beyond C1 then the image will be formed between F2 and C2 which is diminished, real and inverted as shown in figure above.

24) Describe with the help of a ray diagram the nature, size and position of the image formed when an object is placed at infinity (considerable distance) in front of a convex lens. State three characteristics of the image so formed.

Ans:

The following diagram shows the formation of image when an object is placed at infinity in front of convex lens. When the object is placed at infinity then the image will be formed at focus F2 which is highly diminished, point sized, real and inverted as shown in figure above.

25) (a) What type of lens is shown in the diagram on the right? What will happen to the parallel rays of light? Show by completing the ray diagram.

(b) Your eye contains a convex lens. Why is it unwise to look at the sun?

Ans:

a) The lens shown in the diagram is the convex lens. When the parallel beam of light is incident on the convex lens b)

In case of convex lens which in our eye, when beam of sunlight incident on it it converges all the sun rays to a single point i.e. focus due to which our eye may get damaged. Hence, it is unwise to look at the sun.

26) Where must the object be placed for the image formed by a converging lens to be:

(a) real, inverted and smaller than the object?

(b) real, inverted and same size as the object?

(c) real, inverted and larger than the object?

(d) virtual, upright and larger than the object?

Ans:

a) To get the image of the object as real, inverted and of the smaller size as that of the object we have to place the object beyond 2f in case of convex lens.

b) In case of convex lens, to get the real, inverted and of the same size as that of the object we have to place the object at 2f.

c) In case of convex lens, to get real, inverted and larger image of the object we have to place the object between F and 2f of the convex lens

d) In case of convex lens, to get virtual, upright and larger image of the object we have to place the object between F and optical centre C of the convex lens.

27) Draw a diagram to show how a converging lens held close to the eye acts as a magnifying glass. Why is it usual to choose a lens of short focal length for this purpose rather than one of long focal length?

Ans:

• When the object is placed within the focus of the convex lens then the image will be formed behind the object which is virtual, erect and larger than the object as shown in figure below. • The above figure shows the use of convex lens as a magnifying glass. To see the object as magnified it should be placed within the focus of the convex lens die to which the image formed will be erect and magnified.
• That means we have to place object at a distance less than the focal length of the convex lens.
• If we are using the convex lens of focal length 6cm then we have to place the object at a distance less than 6cm to get magnified image. If such lens we are using to see magnified image of letters on paper, then we have to hold the paper within 6cm so that we will get magnified image in the convex lens.
• Also, if the focal length of the lens is smaller its magnifying power is greater. Hence usually convex lens of short focal length is sued for the magnifying purpose rather than one of long focal length.

28) How could you find the focal length of a convex lens rapidly but approximately?

Ans:

• When the object is placed at infinity then only the image distance and focal length of the lens are equal. And this fact is used to determine the focal length of the convex lens rapidly but approximately.
• For this we kept the convex lens in a holder and placed it distant apart in front of the window and due to which the rays coming from windows pass through it. We have kept some cardboard like screen behind the lens and we are changing the distance of the screen from the lens so that we get the clear inverted image of the window on the screen.
• Now, after forming the clear image on the screen we have measured the distance of the screen from the lens with the help of scale. And this distance is nothing but the focal length of the convex lens.
• In this way, we can find the focal length of convex lens rapidly but approximately.

29. (a) With the help of a labelled diagram explain how a convex lens converges a beam of parallel light rays. Mark the principal axis, optical centre, principal focus and focal length of the convex lens on the diagram.

(b) State whether convex lens has a real focus or a virtual focus.

(c) List some things that convex lens and concave mirror have in common.

Ans:

a)

The following diagram shows the how convex lens converges the beam of parallel light rays. b)

A convex lens has real focus.

c)

• When the object is placed between pole P and focus F of a convex lens then the image formed will be virtual, erect and magnified.
• When the object is placed between centre C and focus F of a concave mirror then the image formed will be virtual, erect and magnified.

30) (a) With the help of a labelled diagram, explain how a concave lens diverges a beam of parallel light rays. Mark the principal axis, optical centre, principal focus and focal length of the concave lens on the diagram.

(b) State whether concave lens has a real focus or a virtual focus.

(c) List some things that concave lens and convex mirror have in common.

Ans:

a)

The following diagram shows the how concave lens diverges a beam of parallel light rays. b)

A concave lens has virtual focus.

c)

• When the object is placed between pole P and focus F of a convex mirror then the image formed will be virtual, erect and magnified.
• When the object is placed between centre C and focus F of the concave lens then the image formed will be virtual, erect and magnified.

31.) Draw ray diagrams to represent the nature, position and relative size of the image formed by a convex lens for the object placed:

(a) at 2F1,

(b) between F1 and the optical centre O of the lens. Which of the above two cases shows the use of convex lens as a magnifying glass? Give reasons for your choice.

Ans:

a)

The following diagram shows that, when the object is placed at 2f1 of a convex lens then the image formed will be at C2 which is of same size, real and inverted as shown in figure below. b)

The following diagram shows that, when the object is placed between F1 and optical centre O of the convex lens then the image formed will be on the same side of the lens as object which is enlarged, virtual and erect as shown in figure below. Also, when object is placed within the focus of the convex lens then only the image formed will be erect and magnified as shown in figure below. And in this case only convex lens is used as a magnifying glass. 32) (a) An object is placed well outside the principal focus of a convex lens. Draw a ray diagram to show how the image is formed, and say whether the image is real or virtual.

(b) What is the effect on the size and position of the image of moving the object (i) towards the lens, and (ii) away from the lens?

Ans:

a)

The following diagram shows, when the object is placed well outside the principal focus i.e. between F and 2F of the convex lens then the image formed will be real and inverted as shown in figure below. b)

• As the object is moved towards the convex lens then the size of the image goes in increasing and the position of the image moves away from the lens towards the infinity till the object is placed at the focus of the lens.
• And the image formed here is virtual, erect and magnified.
• If the object is moved away from the lens,then image also moves towards the lens and finally it get diminished so that it becomes point size.

33) (a) Explain what is meant by a virtual, magnified image.

(b) Draw a ray diagram to show the formation of a virtual magnified image of an object by a convex lens. In your diagram, the position of object and image with respect to the principal focus should be shown clearly.

(c) Three convex lenses are available having focal lengths of 4 cm, 40 cm and 4 m respectively. Which one would you choose as a magnifying glass and why?

Ans:

a)

Virtual image:

• Virtual image is the image which cannot be taken on the screen.

Magnified image:

• Magnified image is the image which larger in size than the object size.

b)

The following diagram shows the formation of virtual and magnified image of an object by a convex lens. c)

• As we already know that, the magnifying power of the convex lens is more when the focal length of the lens is smaller.
• Here, the convex lens with lowest focal length is the convex lens with focal length 4cm and hence it can be used as a magnifying glass.

34.) (a) Explain why, a real image can be projected on a screen but a virtual image cannot.

(b) Draw a ray diagram to show the formation of a real diminished image of an object by a convex lens. In your diagram, the position of object and image with respect to the principal focus should be shown clearly.

(c) Name one simple optical instrument in which the above arrangement of convex lens is used.

Ans:

a)
Following is the explanation which gives that real image is projected on the screen while virtual image cannot:

• We know that, when the actual rays meet at a point then that image will be taken on screen which is called as real image.
• While virtual image is not formed by the meeting of actual rays at a single point. Hence, we cannot project the virtual image on the screen.

b)

The following diagram shows the formation of real and diminished image of an object by a convex lens. c)

In simple camera the above arrangement of convex lens is used.

### Sign Conventions for Spherical Lenses

#### Very Short Answer Type Questions

1.) Write the formula for a lens connecting image distance (v), object distance (u) and the focal length (f). How does the lens formula differ from the mirror formula?

Ans:

• The formula which gives the relationship between focal length f, image distance v and the object distance u in case of lens is called as lens formula.
• It is given by,. 1/f = 1/v – 1/u
•  The lens formula differs from the mirror’s formula only in the sign between 1/v and 1/u.

2.) Write down the magnification formula for a lens in terms of object distance and image distance. How does this magnification formula for a lens differ from the corresponding formula for a mirror?

Ans:

• The magnification formula for a lens in terms of object distance and image distance is given by,
• Magnification m = v/u
• And magnification formula for mirrors has negative sign but magnification formula for lens has no minus sign.

3.) What is the nature of the image formed by a convex lens if the magnification produced by the lens is +3?

Ans:

If the magnification produced by the lens is +3 then the nature of the image will be virtual and erect.

4.) What is the nature of the image formed by a convex lens if the magnification produced by the lens is, – 0.5?

Ans:

If the magnification produced by the lens is -0.5 then the nature of the image will be real and inverted.

5.) What is the position of image when an object is placed at a distance of 10 cm from a convex lens of focal length 10 cm?

Ans:

When the object is placed at a distance of 10cm from a convex lens of focal length 10cm then the image will be formed at infinity.

6) Describe the nature of image formed when an object is placed at a distance of 30 cm from a convex lens of focal length 15 cm.

Ans-

When the object is placed at a distance of 30cm from a convex lens of focal length 15cm then the image formed will be real and inverted.:

7. At what distance from a converging lens of focal length 12 cm must an object be placed in order that an image of magnification 1 will be produced?

Ans:

Given that,
Focal length f = 12cm
Magnification m = 1
Here, magnification produced is 1 which means that image distance and object distance both are same. And such case happens only when the object is placed at 2f.

Hence, object distance = 2f = 2*12= 24cm

8.) State and explain the New Cartesian Sign Convention for spherical lenses.

Ans:

The following are the new Cartesian sign conventions for spherical lenses:

• The sign conventions for spherical lenses are similar to the sign conventions for spherical mirror as given below.
• All the distances are measured from the optical centre O of the lens.
• All the distances measured to the right of optical centre are taken as positive while the distances measured to the left of the optical centre are taken as negative.
• Distances measured above & normal to principal axis are taken as positive.
• Distances measured below & normal to principal axis are taken as negative.
• If the image is real then image distance is taken as negative.
• If the image is virtual then the image distance is taken as positive.
• Here all the distances are taken from optical centre O.
• The focal length of convex lens is positive and that of concave lens is negative

9.) An object 4 cm high is placed at a distance of 10 cm from a convex lens of focal length 20 cm. Find the position, nature and size of the image.

Ans:

Given that,
Object height h1 = 4cm
Object distance u = -10cm
Focal length f = 20cm

We know that,
1/f = 1/v – 1/u
Thus, 1/v = 1/f + 1/u = 1/20 -1/10 = -1/20
Thus, v = -20cm
Thus, the image is formed at a distance of 20cm in front of the convex lens i.e. On left side.
And magnification m = v/u = -20/-10= 2
Thus, the image formed will be virtual, erect and magnified.

Now, magnification m = h2/h1
Thus, h2 = m*h1 = 2*4= 8cm
Thus, the height of the image is 8cm.

10.) A small object is so placed in front of a convex lens of 5 cm focal length that a virtual image is formed at a distance of 25 cm. Find the magnification.

Ans:

Given that,
Focal length f = 5cm
Image distance v =- 25cm
We know that, 1/f = 1/v – 1/u
Thus, 1/u = 1/v – 1/f = -1/5 -1/25 = -6/25
Thus, u = -25/6 cm

Now, magnification m = v/u = -25/(-25/6) = +6

11.) Find the position and nature of the image of an object 5 cm high and 10 cm in front of a convex lens of focal length 6 cm.

Ans:

Given that,
Object height h1= 5cm
Object distance u = -10cm
Focal length f = 6cm

We know that,
1/f = 1/v – 1/u
Thus, 1/v = 1/f + 1/u = 1/6 – 1/10
2/v = 1/3 – 1/5
2/v = 2/15
Thus, v/2 = 15/2
Thus, v = +15cm
Thus, the image will be formed at a distance of 15cm behind the convex lens i.e. on the right side of convex lens.
Magnification m = v/u = 15/-10= -3/2= -1.5

Thus, the image formed will be real and inverted.

12.) Calculate the focal length of a convex lens which produces a virtual image at a distance of 50 cm of an object placed 20 cm in front of it.

Ans:

Given that,
Image distance v = -50cm
Object distance u= -20cm
We know that,
1/f = 1/v – 1/u = -1/50 +1/20
10/f = -1/5 + ½ = 3/10
Thus, f/10 = 10/3
Thus, f = 100/3 = 33.33cm
Thus, the focal length of convex lens is 33.33 cm

13. An object is placed at a distance of 100 cm from a converging lens of focal length 40 cm. (i) What is the nature of image? (ii) What is the position of image?

Ans:

Given that,
Object distance u= -100cm
Focal length f = 40cm

We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u = 1/40 – 1/100
10/v = ¼ – 1/10
20/v = ½ – 1/5 = 3/10
Thus, v/20= 10/3
Thus, v = 200/3= 66.66cm
Thus, the image is formed behind the convex lens at a distance of 66.66cm.

Magnification m = v/u = 200/(-100*3)= -2/3= – 0.66

1) As the magnification m produced is negative hence the image formed is real and inverted.

2) As v = 66.66cm , hence the image formed is at a distance of 66.66cm behind the convex lens.

14.) A convex lens produces an inverted image magnified three times of an object placed at a distance of 15 cm from it. Calculate focal length of the lens.

Ans:

Given that,
Magnification m = -3
Object distance u= -15cm
We know that, magnification m = v/u
Thus, -3 = v/-15
Thus, v = 45cm
And 1/f = 1/v – 1/u = 1/45 + 1/15 = 4/45
Thus, f = 45/4 = 11.25cm
Thus, the focal length of lens is 11.25cm

15.) A converging lens of focal length 5 cm is placed at a distance of 20 cm from a screen. How far from the lens should an object be placed so as to form its real image on the screen?

Ans:

Given that,
Focal length f = 5cm
Image distance v = +20cm
We know that,
1/f = 1/v – 1/u
Thus, 1/u = 1/v – 1/f = 1/20 – 1/5 = -3/20
Thus, u = -20/3 = -6.6cm
Thus, to form the real image of the object it should be placed at a distance of 6.6cm from the lens.

16. An object 5 cm high is held 25 cm away from a converging lens of focal length 10 cm. Find the position, size and nature of the image formed. Also draw the ray diagram.

Ans:

Given that,
Object height h1 = 5cm
Focal length f = 10cm
Object distance u= -25cm

We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u = 1/10 – 1/25
5/v = ½ – 1/5 = 3/10
Thus, v/5 = 10/3
Thus, v = 50/3 = 16.66cm
Thus, the image is formed at a distance of 16.66cm behind the convex lens.
We know that, magnification m = v/u = 50/(-3*25) = -2/3
And also, m = h2/h1
Thus, h2 = m*h1 = -2/3*5= -10/3= -3.33cm
Thus, the height of the image is 3.33cm.
And the image formed will be real and inverted.

17. At what distance should an object be placed from a convex lens of focal length 18 cm to obtain an image at 24 cm from it on the other side? What will be the magnification produced in this case?

Ans:

Given that,
Focal length f= 18cm
Image distance v = 24cm
We know that,
1/f = 1/v – 1/u
1/u = 1/v – 1/f = 1/24 – 1/18
3/u = 1/8 – 1/6
6/u = ¼ – 1/3 = -1/12
Thus, u = -12*6= -72cm
Thus, the object should be placed at a distance of 72cm from the lens
Magnification m = v/u = 24/-72= 2/-6= -1/3

18. An object 2 cm tall is placed on the axis of a convex lens of focal length 5 cm at a distance of 10 m from the optical centre of the lens. Find the nature, position and size of the image formed. Which case of image formation by convex lenses is illustrated by this example ?

Ans:

Given that,
Object height h1 = 2cm
Focal length f = 5cm
Object distance u= -10m = -1000cm

We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u
1/v = 1/5 -1/1000= 199/1000
Thus, v = 1000/199= 5cm
Thus, the image formed will be at a distance of 5cm behind the lens.
We have, magnification m = h2/h1 = v/u
Thus, h2 = v/u*h1 = 5*2/-1000= -1/100= – 0.01cm
Thus, the image height is 0.01cm which is real, inverted and highly diminished.
From this example, we illustrate that when the object is placed beyond 2f then the image will be formed between F and 2f which is real, inverted and highly diminished.

19.) The filament of a lamp is 80 cm from a screen and a converging lens forms an image of it on a screen, magnified three times. Find the distance of the lens from the filament and the focal length of the lens.

Ans:

Given that,
Image distance + object distance = 80cm
Hence, v + u = 80cm
Magnification m = v/u =3
Thus, v = 3u
Put v = 3u in equation v + u = 80
Thus, 3u + u = 4u = 80
Hence, u = 20cm
Object distance u = -20
Hence, v = 3u = 60cm

We know that,
1/f = 1/v -1/u = 1/60 + 1/20= 4/60= 2/30 = 1/15
Thus, f = 15cm
Thus, the focal length of the convex lens is 15cm.

20. An erect image 2.0 cm high is formed 12 cm from a lens, the object being 0.5 cm high. Find the focal length of the lens.

Ans:

Given that,
Image height h2 = 2cm
Image distance v= -12cm
Object height h1 = 0.5cm

We know that,
Magnification m = v/u = h2/h1
Thus, u= v*h1/h2 = -12*0.5/2= -12/4= -3cm
And we know that,
1/f = 1/v – 1/u = -1/12 + 1/3= 3/12= ¼
Thus, f = 4cm
The focal length of the lens is 4cm.

21.) A convex lens of focal length 0.10 m is used to form a magnified image of an object of height 5 mm placed at a distance of 0.08 m from the lens. Calculate the position, nature and size of the image.

Ans:
Given that,
Focal length f = 0.10m= 10cm
Object height h1= 5mm = 0.5cm
Object distance u= -0.08m = -8cm

We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u = 1/10 – 1/8 = -2/80= -1/40
Thus, v = -40cm = -0.4m
Thus, the position of the image is 0.4m from the lens on the same side of the object.
Magnification m = v/u = h2/h1
Thus, h2 = v/u*h1 = -40*0.5/-8= 2.5cm= 25mm
Thus, the image height is 25mm and it is virtual and erect.

22.) A convex lens of focal length 6 cm is held 4 cm from a newspaper which has print 0.5 cm high. By calculation, determine the size and nature of the image produced.

Ans:

Given that,
Focal length f= 6cm
Object distance u = -4cm
Object height h1= 0.5cm

We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u = 1/6 -1/4 = -2/24= -1/12
Thus, v = -12cm
Also, magnification m= v/u = h2/ h1
Thus, m = -12/-4= h2/0.5
Thus, m = 3= h2/0.5
Thus, h2 = 1.5cm and m= 3
The size of the image produced is 1.5cm and it is virtual, erect and magnified.

23.) Determine how far an object must be placed in front of a converging lens of focal length 10 cm in order to produce an erect (upright) image of linear magnification 4.

Ans:

Given that,
Magnification m = +4
Focal length f= 10cm
We have, magnification m = v/u
Thus, 4= v/u
And V = 4u
Now, 1/f = 1/v – 1/u
1/f = 1/4u – 1/u = -3/4u
Thus, f = -4u/3
And u= -10*3/4= -30/4= -7.5cm
Thus, the object should be placed at a distance of 7.5cm in front of the lens.

24.) A lens of focal length 20 cm is used to produce a ten times magnified image of a film slide on a screen. How far must the slide be placed from the lens?

Ans:

Given that,
Focal length f= 20cm
Magnification m = -10
We know that, object distance is always negative.
Thus, object distance u= -u

We know that,
Magnification m = v/u = v/-u
Thus, v = -mu = 10u
Now, we know that,
1/f = 1/v – 1/u = 1/10u +1/u = 11/10u
Thus, f = 10u/11
And u = 20*11/10= 22cm
Hence object distance= -u = -22cm
Thus, the object should be placed at a distance of 22cm in front of the lens.

25.) An object placed 4 cm in front of a converging lens produces a real image 12 cm from the lens.

(a) What is the magnification of the image?

(b) What is the focal length of the lens?

(c) Draw a ray diagram to show the formation of image. Mark clearly F and 2F in the diagram.

Ans:

Here the lens is the converging lens which means that the convex lens.
Given that,
Object distance u= -4cm
Image distance v= 12cm

a)

We know that,
Magnification m = v/u = 12/-4= -3
Thus, the magnification produced is -3.

b)

We know that,
1/f = 1/v -1/u = 1/12 +1/4= 4/12= 1/3
Thus, f= 3cm
Thus, the focal length of the lens is 3cm.

c)

The following diagram shows the formation of image in above case. 26.) (a) An object 2 cm tall stands on the principal axis of a converging lens of focal length 8 cm. Find the position, nature and size of the image formed if the object is:

(i) 12 cm from the lens

(ii) 6 cm from the lens

(b) State one practical application each of the use of such a lens with the object in position (i) and (ii).

Ans:

a)

Object height h1= 2cm
Focal length f= 8cm
1) Object distance u= -12cm
We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u = 1/8 -1/12 = 4/96= 1/24
Thus, v = 24cm
Thus, the image formed is at a distance of 24cm from the lens.

And magnification m = v/u = 24/-12= -2

Thus, the image formed is real and inverted.

2) Object distance u= -6cm
We know that,
1/f = 1/v -1/u
1/v = 1/f + 1/u = 1/8 – 1/6 = -2/48= -1/24
Thus, v= -24cm
Thus, the image formed will be at a distance of 24cm in front of the lens.
Now, magnification m = v/u = -24/-6= 4

Now, magnification m = h2/h1 = v/u
Thus, h2 = h1*m = 2*4= 8cm
Thus, the size of the image is 8cm.
Thus, the image formed is virtual and erect.

b)
The lens in first case is used for correcting the vision defect.
And the lens in second case is used as magnifying lens to see big letters while reading.

27. (a) An object 3 cm high is placed 24 cm away from a convex lens of focal length 8 cm. Find by calculations, the position, height and nature of the image.

(b) If the object is moved to a point only 3 cm away from the lens, what is the new position, height and nature of the image?

(c) Which of the above two cases illustrates the working of a magnifying glass?

Ans:

a)
Object height h1= 3cm
Object distance u= -24cm
Focal length f = 8cm

We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u = 1/8 -1/24= 2/24= 1/12
Thus, v = 12cm
Thus, the image is formed at a distance of 12cm behind the lens.
Now, magnification m= v/u = 12/-24= -1/2
Thus, the magnification is less than one and negative hence the image formed is diminished.
And m = h2/h1 = -1/2
Thus, h2= -1/2*h1= -1/2*3= -3/2= -1.5cm
Thus, the height of the image is 1.5cm and it is real , inverted.

b)
Given that,
Object distance u = -3cm
We know that,
1/f = 1/v – 1/u
1/v = 1/f +1/u = 1/8 -1/3 = -5/24
Thus, v = -24/5= -4.8cm
Thus, the image is formed at a distance of 4.8cm in front of the lens.
Magnification m= v/u = -4.8/-3= 1.6
And m = h2/h1= 1.6
Thus, h2= 1.6*h1= 1.6*3= 4.8cm
Thus, the image height is 4.8cm and which is virtual and erect.

c)
Case b of the above mentioned illustrates the working of magnifying glass.

28.) (a) Find the nature, position and magnification of the images formed by a convex lens of focal length 0.20 m if the object is placed at a distance of:

(i) 0.50 m (ii) 0.25 m (iii) 0.15 m

(b) Which of the above cases represents the use of convex lens in a film projector, in a camera, and as a magnifying glass?

Ans:

a)

1) Given that,

Focal length f= 0.2m = 20cm
Object distance u= -0.5m= -50cm

We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u
1/v = 1/20 – 1/50
10/v = ½ – 1/5 = 3/10
Thus, v/10= 10/3
Thus, v= 10/3*10= 100/3= 33.3cm= 0.33m
Thus, the image is formed at a distance of 0.33m behind the lens.
Magnification m= v/u =- 100/3*50= -2/3= -0.66

Thus, the image formed is real and inverted.

2) Given that,
Focal length f= 0.2m = 20cm
Object distance u= -0.25m= -25cm

We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u = 1/20 – 1/25
5/v = ¼ – 1/5 = 1/20
Thus, v/5 = 20
And V = 100cm = 1m

Thus, the image is formed at a distance of 1m behind the lens.
Magnification m= v/u = 100/-25= -4

Thus, the image formed is real and inverted.

3) Object distance u= 0.15m = -15cm
Focal length f= 0.2m = 20cm

We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u = 1/20 -1/15
5/v = ¼ – 1/3= -1/12
Thus, v/5 = -12
And V= -12*5= -60cm = -0.6m
Thus, the image is formed at a distance of 0.6m in front of the lens.
Magnification m= v/u = -60/-15= 4

Thus, the image formed will be virtual and erect.

b)

• The case 1 represent the use of convex lens in a camera.
• The case 2 represent the use of convex lens in film projector.
• The case 3 represent the use of convex lens in magnifying glass.

#### RULES FOR OBTAINING IMAGES FORMED BY CONCAVE LENSES

1) If the image formed by a lens is always diminished and erect, what is the nature of the lens?

Ans:- If the image formed by a lens is always diminished and erect then the nature of the lens is concave lens.

2) Copy and complete the diagram below to show what happens to the rays of light when they pass through the concave lens: Principal axis F 252

Ans: 3. Which type of lenses are:

(a) thinner in the middle than at the edges?

(b) thicker in the middle than at the edges?

Ans:

a) Concave lenses are thinner in the middle than at the edges.

b) Convex lenses are thicker in the middle than at the edges.

4. A ray of light is going towards the focus of a concave lens. Draw a ray diagram to show the path of this ray of light after refraction through the lens.

Ans: A ray of light passing through focus of the concave lens is parallel to the principal axis of the concave lens after refraction as shown in figure below.

5. (a) What type of images can a convex lens make?

(b) What type of image is always made by a concave lens?

Ans:

a) Convex lens makes real and virtual images.
b) Virtual images are always made by a concave lens.

6. Take down this figure into your answer book and complete the path of the ray.

Ans: 7. Fill in the following blanks with suitable words:

Ans:

a) A convex lens converges rays of light, whereas a concave lens diverges rays of light.

b) Lenses refract light to form images: a converging lens can form both real and virtual images, but a diverging lens forms only virtual images.

8. Things always look small on viewing through a lens. What is the nature of the lens ?

Ans:

Things always look small on viewing through a lens. That lens is called as concave lens.

9.) An object lies at a distance of 2f from a concave lens of focal length f. Draw a ray-diagram to illustrate the image formation.

Ans:

When the object is placed at a distance of 2f from the concave lens of focal length f then the image formed will be between optical centre C and focus F of the lens and which is virtual, erect diminished as shown in figure below. 10) Show by drawing a ray-diagram that the image of an object formed by a concave lens is virtual, erect and diminished.

Ans:

When the object is placed beyond the focus F of the concave lens then the image formed will be between C and F which is virtual, upright and diminished as shown in figure below. 11.) Give the position, size and nature of image formed by a concave lens when the object is placed:

(a) anywhere between optical centre and infinity.

(b) at infinity.

Ans:

• When the object is placed anywhere between optical centre and infinity of a concave lens then the image formed will be between optical centre C and focus F which is virtual, erect and diminished as shown in figure below. • When the object is placed at infinity in case of a concave lens then the image will be formed at the focus which is virtual, erect and highly diminished.

12. Which type of lens is:

(a) a converging lens, and which is

Ans:

a) Convex lens is called as converging lens.

When a parallel beam of light rays which are also parallel to the axis of convex lens are passed through the convex lens they get refracted according to laws of refraction.
And after refracting through the convex lens they meet or converges at a single point F on the other side of lens and that point is called as principal focus of the convex lens.
The following figure shows that, how convex lens acts as a converging lens.
Hence, convex lens has real focus. b) Concave lens is called as diverging lens.

When a parallel beam of light rays which are also parallel to principal axis of the concave lens are passed through the concave lens then they get refracted and after refraction the rays will spreads as shown in figure.

As they are diverging from each other hence they can’t meet at a actual single point. When we produce them in backward direction they appear to meet at a point F on the left side of lens and it is called as the focus of the concave lens as shown in figure.

Hence, concave lens has virtual focus. 13.) With the help of a diagram, explain why the image of an object viewed through a concave lens appears smaller and closer than the object.

Ans:

• The following figure shows the concave lens with optical center C and F is the focus of the lens. Let pointing arrow in upward AB is the object. To see the image of object AB and to find the nature of the image we follow the steps as discussed below.
• From starting point A of AB we draw a line which is parallel to principal axis and by rule this ray appear to be coming from focus F of the concave lens after refraction.
• And hence we join DF by dotted lines as shown in figure and we extend FD in upward direction as DX by solid line as shown in. Now, we drawn the second ray of light which also starts from point A of AB and which passes through the optical centre C.
• And hence we joined the points A and C by line AC. And this ray AC also passing through optical centre C goes straight and hence we extend as CY which is the second refracted ray.
• These two diverging rays appears to meet at point A’ on the left side of the lens when we produce them in backward direction as shown in figure.
• So when we see from the right side of the concave lens the top of the arrow AB appears A’ and hence A’ is the virtual image of the point A of the object.
• And AB’ is the complete image of the object AB which is virtual, erect and smaller than the object. 15) (a) An object is placed 10 cm from a lens of focal length 5 cm. Draw the ray diagrams to show the formation of image if the lens is (i) converging, and (ii) diverging.

(b) State one practical use each of convex mirror, concave mirror, convex lens and concave lens.

Ans:

a)

1)

Given that, object is placed 10cm from converging lens i.e. convex lens of focal length 5cm that means the object is placed at 2f then the image formed will be at 2F on the other side of the lens which is real, inverted and of the same size as shown in figure.  b)

Practical uses of concave and convex mirrors:

• Concave mirror are used as shaving mirrors to see large images of the faces.
• Concave mirrors are also used by dentist to see the large images of the teeth of patients
• Convex mirrors are used as rear view mirrors in vehicles to see traffic from the backside of the vehicle.
• Big convex mirrors are used as shop security mirrors.

Practical uses of concave lens and convex lens:

• Convex lenses are used for making a simple camera.
• Convex lenses are used as magnifying glass.
• Concave lenses are used as eye lens in Galilean telescope.
• Also, concave lenses are used in wide angle spyhole in doors.

16. (a) Construct ray diagrams to illustrate the formation of a virtual image using

(i) a converging lens, and (ii) a diverging lens.

(b) What is the difference between the two images formed above?

Ans:

a)

1) The following diagram shows the formation of a virtual image using a converging lens i.e. convex lens. 2) The following diagram shows the formation of a virtual image using a diverging lens i.e. concave lens. b)

The virtual image formed by a converging lens is magnified whereas that formed by a diverging lens is diminished.

#### LENS FORMULA AND MAGNIFICATION FORMULAE FOR A CONCAVE LENS

1.) The lens A produces a magnification of, – 0.6 whereas lens B produces a magnification of + 0.6.

(a) What is the nature of lens A?

(b) What is the nature of lens B?

Ans:

a)
Lens A has magnification of -0.6 hence it is the convex lens.

b)
Lens B has magnification of +0.6 hence it is the concave lens.

2.) A 50 cm tall object is at a very large distance from a diverging lens. A virtual, erect and diminished image of the object is formed at a distance of 20 cm in front of the lens. How much is the focal length of the lens?

Ans:

Given that,
Object height h1= 50cm
Image distance v= 20cm

• As the object is at larger distance that means at infinity from the diverging lens I.e. Concave lens then the image formed will be only virtual, erect and diminished image of the object which is at a distance of 20cm from the lens.
• And in this case the image distance is equal to the focal length of the concave lens.
• Hence, the focal length of the lens is 20cm.

3. An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image.

Ans:

Given that,
Object distance u= -4cm
Focal length f= -12cm
We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u = 1/-12 – ¼ = -4/12 = -1/3
Thus, v = -3cm
The image formed is at a distance of 3cm in front of the concave lens.
Thus, the image formed is virtual and erect.

4. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray-diagram.

Ans:

Given that,
Focal length f= -15m
Image distance v= -10cm
We know that,
1/f = 1v – 1/u
1/u = 1/v – 1/f
1/u = -1/10 + 1/15= -5/150= -1/30
Thus, u = -30cm
Thus, the object is placed in front of the concave lens at a distance of 30cm
The following diagram shows the above condition. 5.) An object 60 cm from a lens gives a virtual image at a distance of 20 cm in front of the lens. What is the focal length of the lens? Is the lens converging or diverging? Give reasons for your answer.

Ans:

Object distance u = -60cm
Image distance v = -20cm
Focal length f = ?

We know that,
Magnification m = v/u = -20/-60= 1/3
As the magnification factor is less than 1, the image formed is diminished.
And given that, image formed is virtual so the nature of lens is diverging lens and which is concave lens.

Also, we have
1/f = 1/v – 1/u
1/f = -1/20 +1/60= -2/60= -1/30
Thus, f = -30cm

As the focal length of the lens is negative it means that the lens is concave lens which is diverging in nature.

6.) A concave lens of 20 cm focal length forms an image 15 cm from the lens. Compute the object distance.

Ans:

Given that,
Focal length f= -20cm
Image distance v= -15cm
We know that,
1/f = 1/v – 1/u
1/u = 1/v – 1/f
1/u = -1/15 + 1/20 = -5/300= -1/60
Thus, u = -60cm
Thus, the object distance will be 60cm and it is placed in the left side of the concave lens.

7. A concave lens has focal length 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also find the magnification produced by the lens.

Ans:

Focal length f = -15cm
Image distance v= -10cm

We know that,
1/f = 1/v – 1/u
1/u = 1/v – 1/f
1/u = -1/10 + 1/15= -5/150= -1/30
Thus, u = -30cm

Hence, we can say that the object is placed at a distance of 30cm from the concave lens on its left side.
Now, we know that
Magnification m = v/u = -10/-30= 1/3
Since, the magnification produced by the lens is less than 1 and hence the image formed is diminished.

8. Calculate the image distance for an object of height 12 mm at a distance of 0.20 m from a concave lens of focal length 0.30 m, and state the nature and size of the image.
Ans:

Given that,
Object height h1= 12mm = 0.012m
Object distance u= -0.2m
Focal length f= -0.3m

We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u = -1/0.3 -1/0.2
1/v = -10/3 – 10/2 = -10/3 -5= -25/3
Thus, v/10= -3/25
And v = -3/25= -0.12m = -12cm

Thus, the image formed is on the left side of the concave lens which is at a distance of 12cm.

We know that,
Magnification m = v/u = -0.12/-0.2 = 1.2/2= 0.6
And m = h2/h1
Here, 1.2/2= h2/0.012
Thus, h2 = 1.2/2*0.012= 1.2*0.006= 0.0072m= 0.72cm = 7.2mm

Thus, the image formed is virtual and erect with size of 7.2mm

9. A concave lens has a focal length of 20 cm. At what distance from the lens a 5 cm tall object be placed so that it forms an image at 15 cm from the lens? Also calculate the size of the image formed.

Ans:

Given that,
Focal length f = -20cm
Object height h1= 5cm
Image distance v = -15cm

We know that,
1/f = 1/v – 1/u
1/u = 1/v – 1/f
1/u = -1/15 + 1/20 = -5/300= -1/60
Thus, u = -60cm
Thus, the object is placed at a distance of 60cm from the lens.

We know that,
Magnification m = v/u = h2/h1
Thus, -15/-60= h2/5
And ¼*5= h2
Thus, h2= 5/4= 1.25cm

Thus, the size of the image formed is 1.25cm

10. An object is placed 20 cm from (a) a converging lens, and (b) a diverging lens, of focal length 15 cm. Calculate the image position and magnification in each case.

Ans:

a) For a converging lens:

Given that,
Focal length f= 15cm
Object distance u= -20cm
We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u = 1/15 -1/20 = 5/300= 1/60
Thus, v = 60cm
Thus, the image formed is at a distance of 60cm from the converging lens.
And magnification m = v/u = 60/-20= -3

b) For diverging lens:

Given that,
Focal length f = -15cm
Object distance u= -20cm
We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u = -1/15 -1/20 = -35/300= -7/60
Thus, v = -60/7 = -8.5cm
Thus, the image formed is at a distance of 8.5cm from the diverging lens.

We know that,
Magnification m = v/u = -8.5/-20= 0.85/2= 0.42

11. A 2.0 cm tall object is placed 40 cm from a diverging lens of focal length 15 cm. Find the position and size of the image.

Ans:

Focal length f = -15cm
Object distance u= -40cm
Object height h1= 2cm

We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u = -1/15 -1/40
5/v = -1/3 -1/8 = -11/24
Thus, v/5 = -24/11
And v = -24*5/11= -120/11= -10.9cm

Thus, the image is formed at a distance of 10.9cm on the left of the diverging lens.

Now, magnification of the lens is given by,
Magnification m = v/u = h2/h1
Hence, -10.9/-40= h2/2
Thus, h2= 10.9/20= 1.09/2 = 0.54cm
The size of the image formed is 0.54cma

12. (a) Find the position and size of the virtual image formed when an object 2 cm tall is placed 20 cm from: (i) a diverging lens of focal length 40 cm. (ii) a converging lens of focal length 40 cm.

(b) Draw labelled ray diagrams to show the formation of images in cases

(i) and (ii) above (The diagrams may not be according to scale).

Ans:

a)
1)
Given that,
Object height h1= 2cm
Object distance u= -20cm
Focal length f = -40cm

We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u
1/v = -1/40 -1/20
10/v = -1/4 – ½ = -3/4
Thus, v/10= -4/3
And v = -40/3 = -13.33cm
Thus, the image formed is virtual and erect and at a distance of 13.33cm from the diverging lens.

Now, magnification m = v/u = h2/h1
Thus, -13.33/-20= h2/2
Hence, h2 = 13.33/10= 1.33cm

Thus, the height of the image formed is 1.33cm

2)
Given that,
Object height h1= 2cm
Object distance u= -20cm
Focal length f = 40cm

We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u
1/v = 1/40 –1/20 = -1/40
Thus, v = -40cm
Thus, the image formed is at a distance of 40cm from the converging lens.
Magnification produced is m = v/u = h2/h1
Thus, -40/-20= h2/2
Thus, h2= 40/10= 4cm

Thus, the height of the image formed is 4cm.  13. (a) A small object is placed 150 mm away from a diverging lens of focal length 100 mm.

(i) Copy the figure below and draw rays to show how an image is formed by the lens. Object F F

(ii) Calculate the distance of the image from the lens by using the lens formula.

(b) The diverging lens in part (a) is replaced by a converging lens also of focal length 100 mm. The object remains in the same position and an image is formed by the converging lens. Compare two properties of this image with those of the image formed by the diverging lens in part (a).

Ans:

a)

Given that,
Object distance u= -150mm = -15cm
Focal length f= -100mm = -10cm

1)

The image formed is as shown in figure which is virtual, erect and diminished. 2)

We know that,

1/f = 1/v – 1/u

1/v = 1/f + 1/u

1/v = -1/10 -1/15

5/v = -1/2 – 1/3 = -5/6

Thus, v/5 = -6/5

And v= -6cm

Thus, the image formed is at a distance of 6cm from the lens.

b)

For converging lens:

Focal length f= 100mm = 10cm

Object distance u= -150mm = -15cm

We know that,

1/f = 1/v – 1/u

1/v = 1/f + 1/u = 1/10 – 1/15 = 5/150= 1/30

Thus, v = 30cm

Thus, the image is formed at a distance of 30cm from the converging lens.

Thus, the image formed is behind the converging lens which is real, inverted and magnified.

But, in case of diverging lens the image formed is in front of the diverging lens which is virtual, erect and diminished.

#### POWER OF A LENS

1. The lens A has a focal length of 25 cm whereas another lens B has a focal length of 60 cm. Giving reason state, which lens has more power: A or B.

Ans:

We know that, the power of a lens is given by
P = 1/f
Thus, the power of lens is inversely proportional to the focal length of the lens.
If the focal length is more then power of the lens is less and vice versa.
Hence, here the power of lens A with focal length 25cm is more.

2. Which causes more bending (or more refraction) of light rays passing through it: a convex lens of long focal length or a convex lens of short focal length?

Ans:

The convex lens of short focal length causes more bending of light rays passing through it.

3. Name the physical quantity whose unit is dioptre.

Ans:

The power of the lens is measured in dioptre.

4. Define 1 dioptre power of a lens.

Ans:

The power of the lens is said to be 1 dioptre whose focal length is 1 meter.

5. Which type of lens has (a) a positive power, and (b) a negative power?

Ans:

a)
The power of a convex lens is positive.

b)
The power of a concave lens is negative.

6. Which of the two has a greater power: a lens of short focal length or a lens of large focal length?

Ans:

• We know that, the power of lens is inversely proportional to the focal length of the lens.
• Hence the lens with short focal length has the greater power than the lens of large length.

7. How is the power of a lens related to its focal length?

Ans:

The power of the lens is related to its focal length by the following relation as P = 1/f.

Thus, the power of the lens is the reciprocal of the focal length of the lens.

8. Which has more power: a thick convex lens or a thin convex lens, made of the same glass? Give reason for your choice.

Ans:

The thick convex lens has more power as it has shorter focal length.

9. The focal length of a convex lens is 25 cm. What is its power?

Ans:

Given that,
Focal length f = 25cm= 0.25m
Then power of the convex lens is given by,
P= 1/f = 1/0.25 = +4D

10. What is the power of a convex lens of focal length 0.5 m?

Ans:

The power of the convex lens with focal length 0.5m is
P= 1/f = 1/0.5= +2D

11. A converging lens has a focal length of 50 mm. What is the power of the lens?

Ans:

Given that, focal length f= 50mm= 5cm = 0.05m
The power of the converging lens is given by,
P= 1/f = 1/0.05 = +20D

12. What is the power of a convex lens whose focal length is 80 cm?

Ans:

Given that, focal length f = 80cm= 0.8m
The power of the convex lens is given by, P= 1/f = 1/0.8= 10/8= 1.25D
Thus, power of the lens is +1.25D

13. A diverging lens has a focal length of 3 cm. Calculate the power.

Ans:
Given that, focal length f = -3cm = -0.03m
The power of the lens is given by,
P= 1/f = 1/-0.03= 100/-3= -33.3D

14. The power of a lens is + 0.2 D. Calculate its focal length.

Ans:

Given that, power of lens = +0.2D
The focal length of the lens is given by, f = 1/P = 1/0.2= 10/2= +5m
Thus, the focal length of the lens is +5cm.

15. The power of a lens is, – 2 D. What is its focal length?

Ans:

Given that, power of the lens = -2D
The focal length of the lens is given by, f = 1/P = 1/-2= -0.5m = -50cm
Thus, the focal length of the lens is -50cm.

16. What is the nature of a lens having a power of + 0.5 D?

Ans:

The nature of the lens having power of +0.5D is convex lens.

17. What is the nature of a lens whose power is, – 4 D?

Ans:

The nature of the lens having power -4D is concave lens.

18. The optician’s prescription for a spectacle lens is marked + 0.5 D. What is the:

(a) nature of spectacle lens?

(b) focal length of spectacle lens?

Ans:

a)
As the power of the lens is 0.5D, hence it is the convex lens.

b)
Given that,
Power of the lens= +0.5D
The focal length of the lens is given by, f = 1/P = 1/0.5= 10/5= 2m

19. A doctor has prescribed a corrective lens of power, –1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Ans:

Given that, power of the lens = -1.5D
The focal length of the lens is given by, f = 1/P = 1/-1.5= -10/15= -0.666m = -66.6cm
Thus, the focal length of the lens is -66.66cm

As the focal length of the lens is -66.66cm hence it is the diverging lens.

20. A lens has a focal length of –10 cm. What is the power of the lens and what is its nature?

Ans:

Given that, focal length f = -10cm= -0.1m

The power of the lens is given by, P = 1/f = 1/-0.1= -10D

As the focal length of the lens is -10D, it is the concave lens.

21. The focal length of a lens is +150 mm. What kind of lens is it and what is its power?

Ans:

Given that, focal length f = +150mm= 15cm= 0.15m
The power of the lens is given by, P= 1/f = 1/0.15= 100/15= +6.6D
Thus, the power of the lens is +6.6D
And as the focal length of the lens is +150mm hence it is the convex lens.

22. Fill in the following blanks with suitable words:

Ans:
a) The reciprocal of the focal length in meters gives you the power of the lens, which is measured in dioptres.

b) For converging lenses, the power is positive while for diverging lenses, the power is negative.

23.) An object of height 4 cm is placed at a distance of 15 cm in front of a concave lens of power, –10 dioptres. Find the size of the image.

Ans:

Given that,
Object height h1= 4cm
Object distance u= – 15cm
Power of lens P= -10D

We know that,
P= 1/f
And f= 1/P = -1/10= -0.1m= -10cm
Hence, the focal length of the concave lens is -10cm.

We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u = -1/10 – 1/15 = -25/150= -1/6
Thus, v = -6cm
Thus, the image formed is at a distance of 6cm.
We have, magnification m = v/u = h2/h1
And -6/-15= h2/4
Thus, h2= 24/15= 8/5= 1.6cm
Thus, the height of the object is 1.6cm

24. An object of height 4.25 mm is placed at a distance of 10 cm from a convex lens of power +5 D. Find (i) focal length of the lens, and (ii) size of the image.

Ans:

Given that,
Object height h1= 4.25mm = 0.425cm
Object distance u= -10cm
Power of lens P= +5D

We know that,
P= 1/f
Thus, f = 1/P = 1/5= 0.2m = 20cm
Thus, the focal length of the convex lens is 20cm

We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u = 1/20 -1/10 = -1/20
Thus, v = -20cm
Thus, the image formed is at a distance of 20cm from the convex lens.

Magnification m = v/u = h2/h1
Thus, -20/-10= h2/0.425
Hence, h2 = 2*0.425= 0.85cm = 8.5mm

Thus, the size of the image formed is 8.5mm which is virtual and erect.

25. A convex lens of power 5 D and a concave lens of power 7.5 D are placed in contact with each other. What is the:

(a) power of this combination of lenses?

(b) focal length of this combination of lenses?

Ans:

Given that,
Power of convex lens P’= 5D
Power of concave lens P” = -7.5D

a)
The power of this combination of lenses is given by,
P= P’ + P” = 5 – 7.5 = -2.5 D
Thus, the power of combination of given lenses is -2.5D

b)
Now, the focal length of this combination of lenses is
P= 1/f
Thus, f = 1/P = 1/-2.5= -10/25= -2/5= -0.4m = -40cm
Thus, the focal length of the combination of lenses is 40cm and here negative sign indicates that the lens formed due to combination of lenses is also like the concave lens.

26. A convex lens of focal length 25 cm and a concave lens of focal length 10 cm are placed in close contact with one another.

(a) What is the power of this combination?
(b) What is the focal length of this combination?
(c) Is this combination converging or diverging?

Ans:

Given that,
Focal length of convex lens = 25cm = 0.25m
Focal length of concave lens = -10cm = -0.1m

a)
The power of convex lens is P’ = 1/0.25= 100/25= 4D
The power of the concave lens is P” = 1/-0.1= -10/1= -10D
Hence, the power of combination of this lenses is given by,
P= P’ + P” = 4 – 10= -6D
Thus, the power of the combination of this lenses is -6D.

b)
The focal length of this combination of lenses is given by,
F= 1/P = 1/-6= -1/6= -0.166m = -16.66cm
Thus, the focal length of the combination of this lenses is -16.66cm

c)
As the focal length of this combination of lenses is -16.66cm and it is negative which shows that the combination of lenses acts like a concave lens.
Hence, the combination of this lenses acts as a diverging lens.

27. The power of a combination of two lenses X and Y is 5 D. If the focal length of lens X be 15 cm:

(a) calculate the focal length of lens Y.

(b) state the nature of lens Y.

Ans:

Given that,
Power of combination of lenses X and Y is P = 5D
Focal length of lens X = f’ = 15cm

a)

Here, power of lens X = P’ = 1/f’ = 1/0.15= 100/15= 6.67D
We know that,
Power of combination of lenses X and Y = Power of lens X + Power of lens Y
P = P’ + P”
Thus, 5 = 6.67 + P”
Hence, P” = 5 – 6.67= -1.67D
Thus, the power of the lens Y is -1.67D
And focal length of lens Y is f” = 1/P” = -1/1.67= -0.60m = -60cm
Thus, the focal length of the lens Y is -60cm.

b)

As the focal length of lens Y is -60cm and it is negative hence it is the concave lens and diverging in nature.

28. Two lenses A and B have focal lengths of + 20 cm and, –10 cm, respectively.

(a) What is the nature of lens A and lens B?

(b) What is the power of lens A and lens B?

(c) What is the power of combination if lenses A and B are held close together?

Ans:
Given that,
Focal length of lens A = 20cm
Focal length of lens B = -10cm

a)
As the focal length of lens A is 20cm and which is positive and hence it is the convex lens and converging in nature.
As the focal length of lens B is -10cm and which is negative and hence it is the concave lens and diverging in nature.

b)
The power of the lens A is = 1/f = 1/0.2= 10/2= +5D
The power of the lens B is = 1/f = 1/-0.1= -10D

c)
The power of the combination of lenses A and B is given by,
P = 5 – 10= -5D

29. (a) What do you understand by the power of a lens? Name one factor on which the power of a lens depends.

(b) What is the unit of power of a lens? Define the unit of power of a lens.

(c) A combination of lenses for a camera contains two converging lenses of focal lengths 20 cm and 40 cm and a diverging lens of focal length 50 cm. Find the power and focal length of the combination.

Ans:
a)

• The power of lens is the reciprocal of the focal length of the lens and it is given by
P= 1/f
• And hence the power of the lens depends on the focal length of the lens.
• Thus, shorter is the focal length then more is the power of the lens and if longer is the wavelength less is the power of the lens.

b)

• Power of the lens is dioptre and it is denoted by D.
• The power of the lens is 1 dioptre when the focal length of the lens is 1meter.

c)

Given that,

Focal length of lens 1 is f1= 20cm= 0.2m
Focal length of lens 2 is f2= 40cm= 0.4m
Focal length of lens 3 is f3= -50cm = -0.5m
Power of lens 1 is P1= 1/0.2= 10/2= +5D
Power of lens 2 is P2= 1/0.4= 10/4= 2.5D
Power of lens 3 is P3 = 1/-0.5= -10/5= -2D

Thus, the power of combination of lenses 1, 2 and 3 is given by
P= P1 + P2 + P3 = 5 + 2.5 – 2= 7.5 – 2 = 5.5 D
Hence, the power of the combination of lenses is 5.5D.

And focal length of the combination of lenses is f = 1/5.5 = 0.1818m = 18.18cm
Thus, the focal length of the combination of lenses is 18.18cm

30. (a) Two lenses A and B have power of (i) + 2 D and (ii) – 4 D respectively. What is the nature and focal length of each lens?
(b) An object is placed at a distance of 100 cm from each of the above lens’s A and B. Calculate (i) image distance, and (ii) magnification, in each of the two cases.

Ans:

a)

Power of lens A = +2D
Power of lens B = -4D
The focal length of lens A = 1/P = ½= 0.5m = 50cm
The focal length of lens B = 1/P = 1/-4= -1/4= -0.25m = -25cm
As the focal length of the lens A is positive hence it is the convex lens and as the focal length of the lens B is negative hence it is the concave lens.

b)

Given that,
Object distance u = -100cm
For convex lens A:
Focal length f = 50cm

We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u = 1/50 – 1/100 = 1/100
Thus, v = 100cm
Thus, the image formed is at a distance of 100cm in case of lens A.
And magnification produced is m = v/u = 100/-100= -1

For lens B:
Object distance u = -100cm
Focal length f= -25cm

We know that,
1/f = 1/v – 1/u
1/v = 1/f + 1/u
1/v = -1/25 – 1/100 = -5/100= -1/20
Thus, v = -20cm
Hence, the image formed is at a distance of 20cm in case of lens B.
And magnification produced is, m = v/u = -20/-100= 1/5
Thus, m= 0.2

Here is your solution of Lakhmir Singh Class 10 Physics 5th Chapter Refraction of Light

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Updated: May 28, 2022 — 4:16 pm