# Selina Concise Class 10 Physics Chapter 5 Refraction Through A Lens Solution

## Selina Concise Class 10 Physics Solution Chapter No. 5 – ‘Refraction through a Lens’ For ICSE Board Students.

Selina Concise Class 10 Physics Chapter 5 Refraction through a Lens Exercise All Questions and Answers by Physics Teacher here in this post.

#### Exercise 5 (A)

Question: 1

What is a lens?

Solution: A transparent refracting medium bounded by either two spherical surface or one spherical surface and the other surface plane is called as lens.

Question: 2

Name the two kinds of lens? Draw diagrams to illustrate them.

Solution: lenses are of two kinds,

Convex or converging lens

Concave or diverging lens. Question: 3

State difference between a convex and a concave lens in there

(a.) appearance, and

(b.) action on the incident light.

Solution:

 Convex lens Concave lens It is thick in the middle and thin at the ends. It is thin in the middle and thick on the ends. It converges the incident ray towards the principal axis. It diverges the light ray from the principal axis.

Question: 4

Which lens is converging:

(i.) an equiconcave lens or an equiconvex lens?

(ii.) a concavo-convex lens or a convexo-concave lens?

Solution:

i.) An equiconvex lens is converging lens.

ii.) A concavo-convex lens is a converging lens.

Question: 5

Out of the two lenses, one concave and the other convex, state which one will show the divergent action on a light beam. Draw diagram to illustrate your answer.

Solution: Concave lens acts as diverging lens for a light beam. Question: 6

Show by a diagram the refraction of two light rays incident parallel to the principal axis on a convex lens by treating it as a combination of a glass slab and two triangular glass prisms.

Solution: Question: 7

Show by a diagram the refraction of two light rays incident parallel to the principal axis on a concave lens by treating it as a combination of a glass slab and two triangular glass prisms.

Solution: Question: 8

How does the action of the convex lens differ from that of a concave lens on a parallel beam of light incident on them? Draw diagrams to illustrate your answer.

Solution: The convex lens converges the light at its focus. The concave lens diverges the light. Question: 9

Define the term principal axis of a lens.

Solution: Line passing through centres of curvature of the two surfaces of lens is called principal axis.

Question: 10

Explain the optical centre of a lens with the help of proper diagram(s).

Solution: The point on the principle axis inside the lens through which the ray of light undeviated is called as optical centre. Question: 11

A ray of light incident at a point on the principal axis of a convex lens passes undeviated through the lens.

(a.) What special name is given to this point on the principal axis?

(b.) Draw a labelled diagram to support the answer in part (a).

Solution: (a) The point on the principle axis inside the lens through which the ray of light undeviated is called as optical centre.

(b) Question: 12

State the condition when a lens is called an equiconvex or equi-concave.

Solution: an equiconvex or equiconcave lens has equal radii of the both the curved surfaces of the lens.

Question: 13

Define the term principal foci of a convex lens and illustrate your answer with the aid of proper diagrams.

Solution: The point on which the convex lens converges all the incident ray is called as principal foci of the lens. It is present on the either side of principle axis.  Question: 14

Define the term principal foci of a concave lens and show them with the help of proper diagrams.

Solution: The point on the opposite side of diffracting ray where they appear to meet is called as principal foci of concave lens. It is present on either side.  Question: 15

Draw a diagram to represent the second focus of a concave lens.

Solution: Question: 16

Draw a diagram to represent the second focus of a convex lens.

Solution: Question: 17

A ray of light, after refraction through a concave lens emerges parallel to the principal axis.

(a.) Draw a ray diagram to show the incident ray and its corresponding emergent ray.

(b.) The incident ray when produced meets the principal axis at a point F. Name the point F.

Solution: Point F is called first focus.

Question: 18

A ray of light, after refraction through a convex lens emerges parallel to the principal axis.

(a.) Draw a ray diagram to show it.

(b.) The incident ray passes through a point F on the principal axis. Name the point F.

Solution: The point is first focus.

Question: 19

A beam of light incident on a convex lens parallel to its principal axis converges at a point F on the principal axis. Name the point F. Draw a ray diagram to show it.

Solution: Question: 20

A beam of light incident on a thin concave lens parallel to its principal axis diverges and appears to come from a point F on the principal axis. Name the point F. Draw a ray diagram to show it.

Solution: Question: 21

Define the term focal length of a lens.

Solution: A distance from the optical centre of lens to the point where the light ray is focused is called as focal length.

Question: 22

What do you mean by the focal plane of a lens?

Solution: A plane perpendicular to the principal ais passing through focus is called as focal plane.

Question: 23

State the condition for each of the following:

(i.) a lens has both its focal lengths equal.

(ii.) a ray passes undeviated through the lens.

Solution:

i.) If the focal lengths are equal then the medium must be same on the either side of lens.

ii.) If a ray passes through optical centre of lens, it passes undeviated.

Question: 24

A parallel oblique beam of light falls on a (i) convex lens, (ii) concave lens. Draw a diagram in each case to show the refraction of light through the lens.

Solution:  Question: 25

The diagram below shows a lens as a combination of a glass block and two prisms.

(i.) Name the lens formed by the combination.

(ii.) What is the line XX’ called?

(iii.) Complete the ray diagram and show the path of the incident ray AB after passing through the lens.

(iv.) The final emergent ray will either meet XX’ at a point or appear to come from a point on XX’. Label the point as F. What is this point called?

Solution:

i.) It is a combination of convex lens.

ii.) XX’ is called as principal axis.

iii.) iv.) focus.

Question: 26

The diagram below shows a lens as a combination of a glass slab and two prisms.

(i.) Name the lens formed by the combination.

(ii.) What is the line XX’ called?

(iii.) Complete the path of the incident ray AB after passing through the lens.

(iv.) The final emergent ray either meets XX’ at a point or appears to come from a point on XX’. Label it as F. What is this point called?

Solution:

i.) It is a combination of concave lens.

ii.) XX’ is called as principal axis.

iii.) iv.) Focus.

Question: 27

In Fig. (a) and (b), F1 and F2 are the positions of the two foci of the thin lenses. Draw the path taken by the light ray AB after it emerges from each lens.

Solution:

a.) b.) Question: 29

Complete the following sentences:

(a.) If half part of a convex lens is covered, the focal length __________ change, but the intensity of image _________.

(b.) A convex lens is placed in water. Its focal length will ________.

(c.) The focal length of a thin convex lens is _________ than that of a thick convex lens.

Solution:

a.) Does not, decreases

b.) Increase

c.) More

MULTIPLE CHOICE TYPE

Question: 1

A ray of light after refraction through a lens emerges parallel to the principal axis of the lens. The incident ray either passes through:

(a) its optical centre

(b) its first focus

(c) its second focus

(d) it’s centre of curvature of the first surface

Solution: (b) its first focus

Question: 2

A ray of light incident on a lens parallel to its principal axis, after refraction passes through or appears to come from:

(a) Its first focus

(b) Its optical entre

(c) Its second focus

(d)The centre of curvature of its second surface

Solution: (c) Its second focus

Exercise 5 (B)

Question: 1

What are the three principal rays that are drawn to construct the ray diagram for the image formed by a lens? Draw diagram to support your answer.

Solution: One ray is which that passes undeviated. Other one which passes through focus. And the last one is parallel to principal axis. Question: 2

In the diagrams below, XX’ represents the principal axis, O the optical centre and F the focus of the lens. Complete the path of the rays A and B as they emerge out of the lens.

Solution:

a.)  Question: 3

Where must a point source of light be placed in front of a convex lens so as to obtain a parallel beam of light?

Solution: To obtain parallel beam of light, source would placed at the firsts focal point on the left of the convex lens.

Question: 4

Distinguish between a real and a virtual image.

Solution:

 Real image Virtual image A real image is formed due to actual intersection of the rays refracted by the lens. A virtual image is formed when the rays refracted by the lens appears to meet. It can be obtained on the screen. It can not be obtained on the screen. It is inverted. It is straight.

Question: 5

Study the diagram given below.

(a.) Name the lens LL’.

(b.) What are the points O and O’ called?

(c.) Complete the diagram to form the image of the object AB.

(d.) State the three characteristics of the image.

(e.) Name a device in which this action of lens is used.

Solution:

a.) Lens LL’ is a convex lens.

b.) The first and second focal points are O and O’.

c.) Question: 6

Study the diagram below.

(i.) Name the lens LL’.

(ii.) What are the points O and O’ called?

(iii.) Complete the diagram to form the image of the object AB.

(iv.) State three characteristics of the image.

Solution:

i.) Lens LL’ is a concave lens.

ii.) The second and first focal points are O and O’.

iii.) Question: 7

The following diagram in Fig. shows an object AB and a converging lens L with foci F1 and F2.

(a.) Draw two rays from the object AB and complete the diagram to locate the position of the image CD. Also mark on the diagram the position of eye from where the image can be viewed.

(b.) State three characteristics of the image in relation to the object.

Solution: Image is magnified, virtual and upright.

Question: 8

The diagram given below in fig. shows the position of an object OA in relation to a converging lens L whose foci are at F1 and F2.

(i.) Draw two rays to locate the position of the image.

(ii.) State the position of the image with reference to the lens.

(iii.) Describe the three characteristics of the image.

(iv.) Describe how the distance of the image from the lens and the size of the image change as the object move towards F1.

Solution: (i) (ii.) image will be beyond the 2F2

(iii.) image will be magnified, real and inverted.

(iv.) When object moves towards F1 the image will shift away from F2 and get magnified.

Question: 9

A converging lens forms the image of an object placed in front of it, beyond 2F2 of the lens.

(a) Where is the object placed? (b) Draw a ray diagram to show the formation of an image. (c) State its three characteristics of the image.

Solution: (a) object is placed beyond 2F1.

(b) (c.) image is real, inverted and diminished.

Question: 10

A convex lens forms an image of an object equal to the size of the object. (a) Where is the object placed in front of the lens? (b) Draw a diagram to illustrate it. (c) State two more characteristics of the image.

Solution: (a.) object placed at 2F1.

(b.) (c.) Image is real and inverted and of same size.

Question: 11

A lens forms an erect, magnified and virtual image of an object.

(a.) Name the kind of lens.

(b.) Where is the object placed in relation to the lens?

(c.) Draw a ray diagram to show the formation of the image.

(d.) Name the device which uses this principle.

Solution: (a) convex lens

(b) object id placed between optical centre and focus.

(c) d.) Magnifying lens

Question: 12

A lens always forms an image between the object and the lens. (a) name the lens. (b) Draw a ray diagram to shown the formation of such an image. (c) state three characteristics of the image.

Solution: (a.) concave lens.

(b.) d.) image is virtual, erect and diminished.

Question: 13

Classify as real or virtual, the image of a candle flame formed on a screen by a convex lens. Draw a ray diagram to illustrate how the image is formed.

Solution: If the object is placed beyond 2F1 the required image will form between F2 and 2F2. Question: 14

Show by a ray diagram that a diverging lens cannot form a real image of an object placed anywhere on its principal axis.

Solution: Question: 15

Draw a ray diagram to show how a converging lens can form a real and enlarged image of an object.

Solution: Question: 16

A lens forms an upright and diminished image of an object placed at its focal point. Name the lens and draw a ray diagram to show the formation of an image.

Solution: The lens is concave lens. Question: 17

Draw a ray diagram to show how a converging lens is used as a magnifying glass to observe a small object. Mark on your diagram the foci of the lens and the position of the eye.

Solution: Question: 18

Draw a ray diagram to show how a converging lens can form an image of the sun. Hence give a reason for the term ‘burning glass’ for a converging lens used in this manner.

Solution: Question: 19

A lens forms an inverted image of an object.

(a.) Name the kind of lens.

(b.) State the nature of the image whether real or virtual?

Solution:

a.) Convex lens invert image.

b.) The convex lens produces real image.

Question: 20

A lens forms an upright and magnified image of an object.

(a.) Name the lens.

(b.) Draw a labelled ray diagram to show the image formation.

Solution:

a.) Convex lens is used.

b.) Question: 21

(a.) Name the lens which always forms an erect and virtual image.

(b.) State whether the image in part (a) is magnified or diminished?

Solution:

a.) Concave lens produces erect and virtual image.

b.) The concave lens produces diminished image.

Question: 22

Can a concave lens form an image of size two times that of the object? Give reason?

Solution: Concave lens only produces diminished images because it diverges rays. Hence, it can’t form an image of size two times that of object.

Question: 23

Give two characteristics of the image formed by a concave lens.

Solution: Concave lens produces virtual and diminished image.

Question: 24

Give two characteristics of the virtual image formed by a convex lens.

Solution: The convex lens produces magnified and erect virtual image.

Question: 25

In each of the following cases, where must an object be placed in front of a convex lens so that the image formed is

(a.) at infinity,

(b.) of same size as the object,

(c.) inverted and enlarged,

(d.) upright and enlarged?

Solution:

a.) To form image at infinity object must be put at focus.

b.) To form image of same size object must be put at 2F.

C.) To form image inverted and enlarged object must be put between the focus and 2F.

d.) To form image upright and enlarged object must be put between optical centre and focus.

Question: 26

Complete the following table:

 Type of lens Position of object Nature of image Size of image Convex Between the optical centre and focus Convex At focus Concave At infinity Concave At any distance

Solution:

 Type of lens Position of object Nature of image Size of image Convex Between the optical centre and focus Virtual and upright Magnified Convex At focus Real and inverted Very much magnified Concave At infinity Virtual and upright Highly diminished Concave At any distance Virtual and upright Diminished

Question: 27

State the changes in the position, size and nature of the image when the object is brought from infinity up to the convex lens. Illustrate your answer by drawing the ray diagrams.

Solution: when the object is at infinity the image will form at principal focus. When the object is just beyond twice of focus the image will form between focus and twice of focus on the other side. When the object is placed at twice of focus the image will form at twice of focus on the other side. When the object is placed between twice of focus and focus the image will form beyond the twice of focus on the other side. Question: 29

Complete the following sentence

(a.) An object is placed at a distance of more than 40 cm from a convex lens of focal length 20 cm. The image formed is real, inverted and…………….

(b.) An object is placed at a distance 2f from a convex lens of focal length f. The image formed is…………….that of the object

(c.) An object is placed at a distance 5 cm from a convex lens of focal length 10 cm. The image formed is virtual, upright and…………..

Solution:

(a) diminished

(b) equal to

(c) magnified

Question: 30

State whether the following statements are ‘true’ or ‘false’ by writing T/F against them.

(a.) A convex lens has a divergent action and a concave lens has a convergent action.

(b.) A concave lens, if kept at a proper distance from an object, can form its real image

(c.) A ray of light incident parallel to the principal axis of a lens, passes undeviated after refraction

(d.) A ray of light incident at the optical centre of the lens, passes undeviated after refraction

(e.) A concave lens forms a magnified or diminished image depending on the distance of the object from it.

Solution:

(a.) False

(b.) False

(c.) False

(d.) True

(e.) False

MULTIPLE CHOICE TYPE

Question: 1

For an object placed at a distance 20 cm in front of a convex lens, the image is at a distance 20 cm behind the lens. The focal length of the convex lens is:

(a.) 20 cm

(b.) 10 cm

(c.) 15 cm

(d.) 40 cm

Solution: (b) 10 cm

Question: 2

For the object placed between the optical centre and focus of a convex lens, the image is:

(a) Real and enlarged

(b) Real and diminished

(c) Virtual and enlarged

(d) Virtual and diminished.

Solution: (c) Virtual and enlarged

Question: 3

A concave lens forms the image of an object which is:

(a) Virtual, inverted and diminished

(b) Virtual, upright and diminished

(c) Virtual, inverted and enlarged

(d) Virtual, upright and enlarged

Solution: (b) Virtual, upright and diminished

#### Exercise 5(c)

Question: 1

State the sign convention to measure the distances for a lens.

Solution:

(a.) The optical centre of the lens is chosen as the origin of the coordinate system.

(b.) The object is considered to be placed onthe principal axis to the left of the lens.

(c.) All distances are measured along the principal axis from the optical centre of the lens. The distance of an object from the lens is denoted by u, the distance of image by v and the distance of second focus by f

(d.) Distances measured in the direction of the incident ray are taken positive, while distances opposite to the direction of the incident ray are taken negative.

(e.) The length above the principal axis is taken positive, while the length below the principal axis is taken negative.

Question: 2

The focal length of a lens is (i) positive, (ii) negative.

In each case, state the kind of lens.

Solution:

i.) Convex lens has the positive focal length.

ii.) Concave lens gas the negative focal length.

Question: 3

Write the lens formula explaining the meaning of the symbols used.

Solution: lens formula is

1/v -1/u = 1/f

v = distance of image from optical centre.

u = distance of object from optical centre.

f = focal length.

Question: 4

What do you understand by the term magnification? Write expression for it for a lens, explaining the meaning of the symbols used.

Solution: The ratio of the size of the image and size of object is called as magnification.

m = v/u

where, m is linear magnification

v = length of image

u = length of object.

Question: 5

What information about the nature of the image (i) real or virtual, (ii) erect or inverted, do you get from the sign of magnification + or -?

Solution:

i.) Real image shown by negative sign and virtual image shown by positive magnification.

ii.) Erect image is shown by positive magnification and inverted image is indicated by negative sign.

Question: 6

Define the term power of a lens. In what unit is it expressed?

Solution: Power of a lens is a reciprocal of its focal length. It is the deviation of incident light ray produced by a lens on refraction through it. The unit of power is dioptre.

Question: 7

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

Solution: Power of a lens is a reciprocal of its focal length.

Question: 8

How does the power of a lens change if its focal length is doubled?

Solution: power will get halved as the power is reciprocal of focal length.

Question: 9

How is the sign (+ or -) of power of a lens related to its divergent or convergent action?

Solution: If a lens shows convergent action its power is positive and if a lens shows divergent action its power is negative.

Question: 10

The power of a lens is negative. State whether it is convex or concave?

Solution: As power of lens is negative it is a concave lens.

Question: 11

Which lens has more power: a thick lens or a thin lens?

Solution: thick lens has more power as power is reciprocal of focal length thick lens has less focal length implies to more power.

MULTIPLE CHOICE TYPE

Question: 1

If the magnification produced by a lens is – 0.5, the correct statement is :

(a.) The lens is concave

(b.) The image is virtual

(c.) The image is magnified

(d.) The images are real and diminished formed by a convex

Solution: (d) The images are real and diminished formed by a convex

Question: 2

The correct lens formula is

(a.) 1 / u + 1 / v = 1 / f

(b.) 1 / u – 1 / v = 1 / f

(c.) f = uv/ (u – v)

(d.) f = (u + v) / uv

Solution: (c) f = uv/ (u – v)

Question: 3

On reducing the focal length of a lens, its power:

(a.) Decreases

(b.) Increases

(c.) Does not change

(d.) First increases then decreases.

Solution: (b) Increases

Question: 4

The lens of power + 1.0 D is :

(a) convex of focal length 1.0 cm

(b) convex of focal length 1.0 m

(c) concave of focal length 1.0 cm

(d) concave of focal length 1.0 m

Solution:(b) convex of focal length 1.0 m

NUMERICAL

Question: 1

(a.) At what position a candle of length 3 cm be placed in front of a convex lens so that its image of length 6 cm be obtained on a screen placed at distance 30 cm behind the lens?

(b.) What is the focal length of the lens in part (a)?

Solution: Given

Height of a candle = 3cm

Height of image = 6cm

Lens used = convex

Image distance, v= 30 cm

For object distance we can use magnification formula as,

M = height of image / height of object

M = 6/3 =2

And

M = image distance / object distance

M = 30 / object distance

Object distance, u= 30/2 = 15cm u is negative

Focal length,

1/f = 1/v – 1/u

1/f = 1/30 – 1/(-15)

1/f = 1/30 + 1/15

1/f = 1/30 + 2 /30

1/f = 3/30

1/f = 1/10

f =10 cm

As the lens is convex focal length is + 10 cm.

Question: 2

A concave lens forms the image of an object kept at a distance 20 cm in front of it, at a distance 10 cm on the side of the object.

(a.) What is the nature of the image?

(b.) Find the focal length of the lens.

Solution:

a.) The image will virtual because it is on same side of the object. Because of the concave lens image will be erect and diminished.

b.) Given

Distance of image, v = 10 cm

Distance of object, u = 20 cm

Focal length

1/f = 1/v – 1/u

1/f = 1/10 – 1/20

1/f = 2-1/20

1/f = 1/20

f = 20 cm

Question: 3

The focal length of a convex lens is 25 cm. At what distance from the optical centre of the lens an object be placed to obtain a virtual image of twice the size?

Solution: Given

Focal length, f = 25 cm

Magnification is twice to the object

Therefore

v = 2u

therefore,

1/f = 1/v – 1/u

1/f = 1/2u – 1/u

1/f = 1-2 /2u

1/25 = -1/2u

2u = -25

u = -12.5 cm

Question: 4

Where should an object be placed in front of a convex lens of focal length 0.12 m to obtain a real image of size three times the size of the object, on the screen?

Solution: Given

Focal length, f = + 0.12 m

Magnification is three times to the object but image is real

Therefore

v = -3u

therefore,

1/f = 1/v – 1/u

1/f = 1/(-3)u – 1/u

1/f = -1 -3 /3u

1/0.12 = -4/3u

3u/4 = -0.12

3u = – 0.48

u = – 0.48 /3

u = – 0.16 m

Question: 5

An illuminated object lies at a distance 1.0 m from a screen. A convex lens is used to form the image of object on a screen placed at distance 75 cm from the lens. Find: (i) the focal length of lens, and (ii) the magnification.

Solution: Given

Distance of image, v = 75 cm

Distance of object, u = -25 cm

focal length,

1/f = 1/v – 1/u

1/f = 1/75 – 1/(-25)

1/f = 1 + 3 /75

1/f = 4/75

f = 18.75 cm

magnification

m = v/ u

m = 75 / -25

m = -3

Question: 6

A lens forms the image of an object placed at a distance 15 cm from it, at a distance 60 cm in front of it. Find: (i) the focal length, (ii) the magnification, and (iii) the nature of image.

Solution: Given

Distance of image, v = -60 cm

Distance of object, u = -15 cm

i.) focal length,

1/f = 1/v – 1/u

1/f = 1/(-60) – 1/(-15)

1/f = -1/60 + 1 /15

1/f = -1 + 4 / 60

f = 20 cm

ii.) magnification

m = v/ u

m = -60 / -15

m = +4

iii.) nature of the image is virtual, erect and 4 times magnified.

Question: 7

A lens forms the image of an object placed at a distance of 45 cm from it on a screen placed at a distance 90 cm on the other side of it. (a) name the kind of lens. (b) find: (i) the focal length of lens, (ii) the magnification of the image.

Solution: Given

Distance of image, v = 90 cm

Distance of object, u = -45 cm

a.) as image is real lens used is convex lens.

b.) (i) focal length,

1/f = 1/v – 1/u

1/f = 1/ (90) – 1/ (-45)

1/f = 1/90 + 1/45

1/f = 1 + 2/ 90

f = 30 cm

ii.) magnification

m = v/ u

m = 90 / -45

m = -2

Question: 8

An object is placed at a distance of 20 cm in front of a concave lens of focal length 20 cm. find: (a) the position of the image, and (b) the magnification of the image

Solution: Given

Focal length, f = -20 cm

Distance of object, u = -20 cm

focal length,

1/f = 1/v – 1/u

1/ (-20) = 1/v – 1/ (-20)

1/v = -1/20 – 1/20

1/v = -2/20

v = -10 cm

magnification

m = v / u

m = -10 / -20

m = +1/2

Question: 9

A convex lens forms an inverted image of size same as that of the object which is placed at a distance 60 cm in front of the lens. Find:

(a) The position of image, and

(b) The focal length of the lens

Solution: A convex lens forms an inverted image of same size when object is placed at 2f

As object is placed at 60 cm.

2f = 60 cm

f = 60 /2

f = 30 cm.

in this scenario the image will be formed at 2f on the other side of the object.

Question: 10

A concave lens forms an erect image of 1/3rdsize of the object which is placed at a distance 30 cm in front of the lens. Find:

(a.) The position of image, and

(b.) The focal length of the lens.

Solution: Given

Magnification = 1/3

Distance of object, u = -30 cm

For position of object

m = v /u

1/3 = v / -30

V = -10 cm

For focal length

1/f = 1/v – 1/u

1/f = 1/ (-10) -1/ (-30)

1/f = -2/30

f = -15 cm

Question: 11

The power of a lens is +2.0 D. Find its focal length and state what kind of lens it is?

Solution: Given

Power = +2.0 D

Power = 1/ f

f = 1/P

f = ½ m = 0.5 m.

this lens is a convex lens.

Question: 12

Express the power (with sign) of a concave lens of focal length 20 cm.

Solution: Given

Focal length, f = -20cm = -0.2 m

As the lens is concave

Power = 1/f

P = 1/ -0.2

Power = – 5 D

Question: 13

The focal length of a convex lens is 25 cm. Express its power with sign.

Solution: Given

Focal length, f = +25cm = 0.25 m

As the lens is convex,

Power = 1/f

P = 1/ 0.25

Power = 4 D

Question: 14

The power of a lens is -2.0 D. Find its focal length and its kind.

Solution: Given

Power = – 2.0 D

Power = 1/ f

f = 1/P

f = -½ m = -0.5 m.

this lens is a concave lens.

Question: 15

The magnification by a lens is -3. Name the lens and state how are u and v related?

Solution: Given

Magnification is – 3

As magnification is -3 image is enlarged then lens should be convex lens.

Magnification = v/-u

v/-u = -3

v = 3u

Question: 16

The magnification by a lens is +0.5. Name the lens and state how are u and v related?

Solution: Given

Magnification is +0.5

As magnification is +0.5 image is diminished hen lens should be concave lens.

Magnification = v/u

v/u = 0.5

v = u/2

u = 2v

Question: 17

A concave lens is a focal length of 30 cm. Find the position and magnification (m) of image for an object placed in front of it at distance 30 cm. State whether the image is real on virtual?

Solution: Given

Focal length, f = -30 cm

Distance of object, u = -30 cm

focal length,

1/f = 1/v – 1/u

1/ (-30) = 1/v – 1/ (-30)

1/v = -1/30 – 1/30

1/v = -2/30

v = -15 cm

magnification = v/u

m = -15 / -30

m = ½ = 0.5

the image formed will be virtual and erect.

Question: 18

Find the position and magnification of the image of an object placed at distance of 8.0 cm in front of a convex lens of focal length 10.0 cm. Is the image erect or inverted?

Solution: Given

Focal length, f = 10 cm

Distance of object, u = -8 cm

focal length,

1/f = 1/v – 1/u

1/ (10) = 1/v – 1/ (-8)

1/v = 1/10 – 1/8

1/v = 4 – 5 / 40

v = -40cm

magnification,

m = v/u

m = -40 / -8

m = + 5

Image will be erect.

#### Exercise 5(d)

Question: 1

What is magnifying glass? State its two uses.

Solution: Magnifying lens is a convex lens of short focal length. It used in simple and compound microscope, telescope camera etc.

Question: 2

Draw a neat labelled ray diagram to locate the image formed by a magnifying glass. State three characteristics of the image.

Solution: Image is virtual, magnified and erect.

Question: 3

Where is the object placed in reference to the principal focus of a magnifying glass, so as to see its enlarged image? Where is the image obtained?

Solution: The object should be placed in between lens and focal distance of magnifying lens to see enlarged image and image is obtained between lens and principal focus.

Question: 4

Write an expression for the magnifying power of a simple microscope. How can it be increased?

Solution: Magnifying power of the microscope is given by

M = 1 + (D/f)

D = least distance of distinct vision for normal eye.

f = focal length of the lens used.

Question: 5

State two applications each of a convex lens and concave lens.

Solution: Our eye is also a convex lens with changing focal lengths. Convex lens also used in simple and compound microscope, telescope camera etc.

Person suffering from myopia uses concave lens in its spectacles. Concave lens is used in combination with convex lens to overcome chromatic aberration.

Question: 7

The diagram in Fig. shows the experimental set up for the determination of the focal length of a lens using a plane mirror.

(i.) Draw two rays from the point O of the object pin to show the formation of image I at O itself.

(ii.) What is the size of the image I?

(iii.) State two more characteristics of the image I.

(iv.) Name the distance of the object O from the optical centre of the lens.

(v.) To what point will the rays return if the mirror is moved away from the lens by a distance equal to the focal length of the lens?

Solution:

(ii.) size of image will be the same.

(iii.) image will be real and inverted.

(iv.) Distance will be the focal length.

(v.) The distance of mirror will not affect image till the rays are perpendicular to the mirror.

Question: 9

How will you differentiate between a convex and a concave lens by looking at (i) a distant object and (ii) a printed page?

Solution: (i) Convex lens will form inverted image of distant object on the other hand concave lens will form upright image of the distant object.

(ii) The convex lens will the enlarge the letters on the printed page on the other side concave lens will diminish it.

MULTIPLE CHOICE TYPE

Question: 1

A magnifying glass form:

(a.) A real and diminished image

(b.) A real and magnified image

(c.) A virtual and magnified image

(d.) A virtual and diminished image

Solution:(c.) A virtual and magnified image

Question: 2

The maximum magnifying power of a convex lens of focal length 5 cm can be:

(a.) 25

(b.) 10

(c.) 1

(d.) 6

Solution:(d) 6

Updated: March 28, 2023 — 10:24 am