Class 10th light reflection and refraction notes
NCERT Solutions Class 10 Science Chapter 10 Light Reflection and Refraction
NCERT Solutions Class 10 Science Chapter 10 Light Reflection and Refraction: National Council of Educational Research and Training Class 10 Science Chapter 10 Solutions – Light Reflection and Refraction. NCERT Solutions Class 10 Science Chapter 10 PDF Download.
NCERT Solutions Class 10 Science Chapter 10: Overview
Board |
NCERT |
Class |
10 |
Subject |
Science |
Chapter |
10 |
Chapter Name |
Light Reflection and Refraction |
Topic |
Exercise Solutions |
NCERT Solutions Class 10 Science Chapter 10 – Light Reflection and Refraction
Page No. 168
1.) Define the principal focus of a concave mirror.
Ans:
- When a parallel beam of light is incident on a concave mirror as shown in figure then after reflection all the rays meet at a single point on the principal axis and that point is called as the principal focus of the concave mirror.
- It is denoted by F.
2.) The radius of curvature of spherical mirror is 20cm. What is its focal length.
Ans:
We already know that, the radius of curvature of spherical mirror is twice the focal length of the spherical mirror.
Hence, R= 2f
Given that, focal length f= 20cm
Hence, R= 2*20= 40cm
Thus, the radius of curvature of spherical mirror is found to be 40cm.
3.) Name a mirror that can give an erect and enlarged image of an object.
Ans:
Concave mirror is the mirror which gives enlarged and erect image of the object which is formed behind the mirror when the object is placed between pole P and focus F of the mirror.
4.) Why do we prefer a convex mirror as a rear view mirror in vehicles.
Ans:
We prefer mostly a convex mirror as a rear view mirror in vehicles because convex mirror always gives an erect and diminished image.
Also convex mirror has wider field of view due to which the driver can see the large portion of traffic or vehicles coming from backside in his mirror and thereby it become safe while driving.
Page No. 171
1.) Find the focal length of a convex mirror whose radius of curvature is 32cm.
Ans:
We all know that, the radius of curvature of convex mirror or any spherical mirror is twice the focal length of that mirror.
Given that, focal length of convex mirror f= 32cm.
Hence, the radius of curvature is given by,
R=2*f = 2*32= 64cm
Thus, the radius of curvature of a given convex lens is 64 cm.
2.) A concave mirror produces three times magnified i.e. enlarged real image of an object placed at 10cm in front of it. Where is the image located?
Ans:
Given that,
For concave mirror:
Magnification m= -3
Object distance u= -10cm
We know that, magnification of concave mirror is given by,
m= -v/u
-3= -v/-10
Thus, v= -30 cm
Thus, the image formed will be at a distance of 30 cm in front of the mirror which is real image.
Page No. 176
1.) A ray of light traveling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Ans:
- We know that, when a ray of light travels from one transparent medium to another transparent medium then it changes its direction of path of traveling and the phenomenon is called as refraction.
- But, according to laws of refraction when a ray of light traveling from rarer medium to denser medium then it bends towards the normal.
- While if it is traveling from denser medium to rarer medium then it bends away from the normal.
- Here, ray of light traveling from air to water i.e.rarer medium to denser medium hence it bends towards the normal.
2.) Light enters from air to glass having refractive index 1.5. what is the speed of light in the glass? The speed of light in vacuum is 3*108 m/s.
Ans:
- Given that, refractive index of glass is 1.5
- And light speed in vacuum is 3*108 m/s.
- We know that, the refractive index of the medium is given by,
- Refractive index of medium= speed of light in air or vacuum/ speed of light in medium
- Here, glass is the medium hence we can write,
Refractive index of glass= speed of light in vacuum/ speed of light in glass
1.5= 3*108/ speed of light in glass
Thus, speed of light in glass= 3*108/1.5
Speed of light in glass = 2*108 m/s.
3.) Find out, from table 10.3 the medium having highest optical density. Also find the medium with lowest optical density.
Ans:
- We know that, if the refractive index of the medium is larger then it’s optical density is also larger and it behaves as optically denser medium.
- While if the refractive index of the medium is lesser it means that it’s optical density is also less and it behaves like a optically rarer medium.
- From table 10.3, the medium with lowest refractive index is water having refractive index 1.0003 and hence it is the medium with lowest optical density.
- While the medium having highest refractive index is the diamond with refractive index 2.42 and hence it is the medium with highest optical density.
4.) You are given kerosene, turpentine and water. In which of this does the light travel fastest? Use the information given in the table 10.3
Ans:
From table,
Refractive index of kerosene= 1.44
Refractive index of turpentine= 1.47
Refractive index of water= 1.33
- As we already know that more is the refractive index more is the optical density and lower is the refractive index lower is the optical density of the medium.
- But, the speed of light in optically denser medium is low while in optically rarer medium it is larger.
- Here, the refractive index of water is lower than water and turpentine which means it’s optical density is lower than kerosene and turpentine.
- Hence, water is the rarer medium than kerosene and turpentine and hence the speed of light in water is faster.
5.) The refractive index of diamond is 2.42. what is the meaning of this statement.
Ans:
- We know that, refractive index of the medium gives its ability to pass the light through it in short it gives the optical density of the medium.
- As the refractive index of diamond is highest hence it is highly optically denser medium due to which light travels through it with lowest speed.
Page No. 184
1.) Define 1Diopter of lens.
Ans:
- The SI unit of power of lens is the dioptre and it is denoted by D.
- Hence, 1 dioptre is the power of that lens which has its focal length equal to one meter.
- Hence, we can write as 1D = 1m-1.
2.) A convex lens forms a real and inverted image of a needle at a distance of 50cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also find the power of the lens.
Ans:
Given that,
For convex lens:
Image distance v= 50cm
Also, m= -1
We know that, magnification is given by
m= -v/u
-1= -50/u
Thus, u= 50cm
Thus, we have to place the needle at a distance of 50cm in front of the convex lens.
By lens formula,
1/f = 1/v – 1/u
1/f = 1/50 +1/50
1/f = 2/50= 1/25
Thus, f= 25cm
And we know that, power of lens is the reciprocal of its focal length.
Hence, P= 1/f= 1/0.25= 4D
Thus, the power of given convex lens will be 4D.
3.) Find the power of concave lens of focal length 2m.
Ans:
Given that,
For concave lens:
Focal length f= -2m
We know that, power of lens is the reciprocal of it’s focal length.
Thus, P= 1/f = 1/2= 0.5D
Hence, the power of given concave lens will be 0.5D.
Exercises Solution
1.) Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Ans: (d) Clay
Explanation: Because, glass and water are more transparent while plastics are transparent to some extent and hence light can be passed through them.
While clay is the totally opaque material and hence we cannot transmit the light through it.
2.) The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.
Ans:(d) Between the pole of the mirror and its principal focus.
Explanation: Because, when the object is placed between focus F and pole P of the mirror then only the image will be formed virtual, erect and enlarged as shown in figure.
3.) Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.
Ans:(b) At twice the focal length
Explanation: Because, when the object is placed at twice its focal length in front of convex lens then only the image formed will be of the same size and real.
4.) A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be
(a) both concave
(b) both convex
(c) the mirror is concave and the lens is convex
(d) the mirror is convex, but the lens is concave.
Ans:(a) both concave
Explanation: Because, the concave mirror and concave lens both has negative focal length.
5.) No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) only plane
(b) only concave
(c) only convex
(d) either plane or convex.
Ans:(d) either plane or convex
Explanation: Because, no matter how far we stand from either plane mirror or convex your image appears erect.
6.) Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm
(b) A concave lens of focal length 50 cm
(c) A convex lens of focal length 5 cm
(d) A concave lens of focal length 5 cm
Ans:(c) A convex lens of focal length 5 cm
Explanation: Because, convex lens produces magnified image when the object is placed between focus and radius of curvature.
And this is possible only when the focal length lens is shorter and hence we can easily read the small letters found in dictionary.
7.) We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Ans:
- We know that, the concave mirror produces an erect image only when the object is placed between pole P and principal focus F of the mirror.
- Given that, the concave mirror is a focal length 15 cm which is nothing but the distance between pole P and principal focus F of the concave mirror. Hence, we can place the object anywhere in between pole P and principal focus F i.e. between 0 cm to 15 cm in between P and F.
- The image formed here will be erect, virtual and magnified i.e. larger than the object. The following ray diagram shows the formation of image in case of concave mirror of focal length 15 cm.
Alternative Answer –
We know that concave mirror gives an erect image when we place an object between its principle focus (F) and pole(P) of the concave mirror. Therefore the Range of distance of the object must be less than 15cm from the mirror. The formed image will be virtual and erect in nature and this image will be larger than the object.
8.) Name the type of mirror used in the following situations.
(a) Headlights of a car
(b) Side/rear-view mirror of a vehicle
(c) Solar furnace
Support your answer with reason.
Ans:
a) In headlights of car concave mirror is used because when object is placed at the principal focus of the concave mirror then it gives the more intense parallel beam of light due to which we can see the coming vehicles from front side of car easily.
The following figure explains the above points in detail.
b) In side view or rear view mirror of a vehicle convex mirrors are used because convex mirrors are having larger filed of view only.
c) In solar furnaces concave mirrors are used because when parallel beam of light is incident on the concave mirror then after reflection it will be concentrated at the focus and hence large amount of heat will be generated.
The following figure explains in detail.
Alternative Answer –
- We can use Concave mirror for headlights of the car. Because when light is falls on concave mirror, then the concave mirror convert light into strong beam of light so we can see vehicle and the road clearly.
- We can use Convex mirror for side/ rear view mirror of a vehicle. Because when light falls on convex mirror it produces virtual and erect image of the object which helps to driver to see vehicle,road and traffic clearly.
- We can use Concave mirror for solar furnace. Because when light falls on concave mirror it converge all light rays into at a point. So in this way solar furnace produces high temperature from sunlight.
In case you are missed :- Previous Chapter Solution
9.) One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Ans:
- When one half of the convex lens is covered with a black paper then also the lens produces complete image of the object. This can be verified by performing following experiment.
- Take a convex lens and cover its half portion with black paper and fix it on a stand.
- Now take a burning candle and place it in front of convex mirror and take a screen place it on the opposite side of the candle that means on other side of the convex lens.
- Now, by adjusting the burning candle in a such way that the complete clear image of the burning candle will be formed on the screen. Here, you may adjust the screen also to get clear image of the candle.
- Now, we finally observe that the complete image of burning candle will be formed on the screen although the one half portion of the convex lens is covered with black paper.
- This concludes that, the size of the convex lens doesn’t decide the size of the image. If we have taken a small part of convex lens then it also produces the whole or complete image of the object.
- But, as the some portion of convex lens is covered with black paper, the some of the light rays get blocked at that portion and hence we cannot get the clear or bright image.
- Thus, the size of the image of the object does not depends on the size of convex mirror. But, the brightness of the image will be depends on the size of the convex mirror.
Alternative Answer –
Even if half of the convex lens is covered with black paper,it shows the full image or complete image of the object. Here is an experiment to prove it.
Light a candle in front of a convex lens and place a white screen behind the lens and adjust the distance of the object so that the image of the candle is project into the screen. Then when half of the lens is covered with black paper we notice that even now the full image comes on the screen, there is a slight problem in clarity due to the partial cover. This is because not all rays of light fall on lens,yet from this we know that the image of an object does not depend on the size of the lens.
10.) An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Ans:
Given that,
For concave lens:
Object height h1= 5cm
Object distance u= -25cm
Focal length f= 10cm
According to lens formula,
1/f = 1/v – 1/u
1/v = 1/f + 1/u = 1/10 -1/25 = 0.1 – 0.04 = 0.06
Hence, v = 1/0.06 = 16.66cm
Thus, the image will be formed at a distance of 16.66 cm on the other side of the lens.
We know that, the magnification of the lens is given by,
Magnification m= h2/h1 = -v/u
Thus, h2/h1 = – 16.66/25 = 0.6664
Hence, h2/h1 =- 0.6664
And h2/5 = -0.6664
Thus, h2= -3.33 cm
Thus, the image formed will be inverted, real and diminished as shown in the following ray diagram.
Alternative Answer –
Given that
Height of the object, h_{O} = 5cm
Position of the object, u = -25cm
The focal length of the lens, f = 10cm
To find the position of the image
w.k.t the lens formula, (1/v) – (1/u) = 1/f
Substituting the above values in the equation then,
(1/v) + 1/25 = 1/10
1/v = (1/10) – (1/25)
1/v = ( 25 – 10)/ 250
= 15/ 250
= 3/50
v = 50/3
v = 16.66 cm
The position of the image is 16.66 cm from the lens on the other side.
Next to find the size and the nature of the image
We use the formula, Magnification = v/u
m = (16.66)/-25
= -0.66
But m = ( h_{i} / h_{o} )
( Magnification = height of the image / height of the object )
-0.66 = h_{i }/ 5 cm
h = -0.66*5 cm
Height of the image, h = -3.3 cm
The size or height of the image is 3.3 cm.
The nature of the image formed is inverted and real.
11.) Concave lens of focal length 15cm forms an image 10cm from the lens. How far is the object placed from the lens? Draw a ray diagram.
Ans:
Given that,
For concave lens:
Focal length f = 15cm
Image distance v= -10cm
According to lens formula,
1/f = 1/v – 1/u
Hence, 1/u = 1/v – 1/f
1/u = 1/-10 +1/15
1/u = -0.1+ 0.066
1/u = -0.034
Hence, u= -29.41 cm
Thus the image will be formed in front of lens which is at a distance of ~30 cm.
The following ray diagram shows in detail.
Alternative Answer –
Given that,
Focal length of the lens, f = -15 cm
Image distance from the lens, v = -10 cm
( Focal length and image distance of the concave lens is always nagetive )
w.k.t lens formula (1/v) – (1/u) = 1/f
1/u = (1/v) – (1/f)
Substituting the above values in the equation.
1/u = (1/-10)+(1/15)
= (15-10)/ -150
= 5/ -150
= 1/ -30
u = -30 cm
The nagetive sign indicates that the object is placed at the left side of the lens.
Therefore the object distance from the lens is 30 cm.
12.) An object placed at a distance of 10cm from a convex mirror of focal length 15cm. Find position and the nature of the image?
Ans:
Given that,
For convex mirror,
Object distance u= -10cm
Focal length f= 15cm
By mirror formula,
1/f = 1/v + 1/u
Hence,1/v = 1/f – 1/u
1/v = 1/15 + 1/10
1/v = 25/150= 5/30= 1/6
Thus, v= 6cm
Hence, we can say that the image formed will be beyond the mirror at a distance of 6cm.
We know that, the magnification formula is given by
m = -v/u = -6/-10= 0.6
Thus, the image formed will be erect and diminished.
Alternative Answer –
Object distance,u = -10 cm
The focal length of the mirror,f = 15 cm
w.k.t the mirror formula, (1/v) + (1/u) = 1/f
1/v = (1/f) – (1/u)
Put the above values in the equation, 1/v = (1/15) – (1/-10)
= (1/15) + (1/10)
= (10+15) / 150
= 25/ 150
= 1/ 6
v = 6 cm
The image is formed 6 cm behind the mirror which produces an virtual and erect image.
13.) The magnification produced by the plane mirror is +1. What does this mean?
Ans:
- We know that, the image formed by the plane mirror is virtual and erect.
- Given that, the magnification produced by the plane mirror is +1, which means that the image formed will be of the same size of the object.
Alternative Answer –
We know that Magnification, m = Image height / Object height
= h_{i} / h_{o}
= v / u
m = +1
The magnification produced by plane mirror is +1 indicates that the image size or height is same as the object size or height and the positive sign of magnification indicates that the image produced by the lens is virtual and erect in nature.
14.) An object 5cm in length is placed at a distance of 20cm in front of convex mirror of radius of curvature 30cm. Find the position of the image, nature and its size.
Ans:
Given that,
For convex mirror,
Object height h1= 5cm
Object distance u= -20cm
R= 30cm
We know that,
R=2f
Thus, 30=2f
Hence, f= 15cm
By mirror formula,
1/f = 1/v + 1/u
Hence, 1/v = 1/f – 1/u
1/v = 1/15 +1/20
1/v = (20+15)/300= 35/300= 0.1166
Thus, v= 8.57~8.6cm
Thus, the image formed will be behind the mirror at a distance of 8.6cm from the mirror.
By magnification formula,
Thus, m= h2/h1 = -v/u
Hence, h2/5= -8.6/20= 0.43
Thus, h2 = 0.43*5= 2.15 cm
Hence, the height of image will be 2.15cm.
Thus, the image formed will be virtual and erect.
Alternative Answer –
Given that,
Height or length of the object, h_{o} = 5.0 cm
Distance of the object from mirror, u = -20 cm
Radius of curvature of the given convex mirror, R = 30 cm
Then Focal length, f = R/2
= 30/2 = 15 cm
To find the position or distance of the image from the mirror,using the mirror formula,
(1/v) + (1/u) = 1/f
1/v = (1/f) – ( 1/u)
1/v = (1/15) – (1/-20)
= (20+15)/300
= 35/300
1/v = 0.1166
v = 8.571 cm
To find size or height of the image
We use the formula Magnification, m = h_{i} / h_{o }= -v / u
-v / u = h_{i} / h_{o}
8.87/20 = h_{i}/ 5
h_{i} = (8.57*5)
h_{i} = 2.142 cm
The position of image is behind the mirror and it produce virtual and erect image as it’s nature and size of the image will be reduced. The nature of the image is virtual and erect.
15.) An object of size 7cm is placed at 27cm in front of a concave mirror of focal length 18cm. At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained? Find the size and nature of the images.
Ans:
Given that,
For concave mirror:
Object height h1= 7cm
Object distance u= -27cm
Focal length f= -18cm
By mirror formula,
1/f= 1/v + 1/u
Thus, 1/v = 1/f – 1/u
1/v = 1/-18 + 1/27
9/v = -1/2 + 1/3
Hence, 9/v = (-3+2)/6 =-1/6
9/v = -1/6
hence, v/9 = -6
Thus, v= -54cm
Hence, the image will be formed at a distance of 54 cm from the mirror.
Now, magnification is given by,
m= h2/h1 = -v/u
h2/7= -54/27= -2
Thus, h2 = -14cm
Hence, we conclude that, the image formed will be real, inverted and having size 14cm.
Alternative Answer –
Given that,
The object size or height, h_{o} = 7.0 cm
Object distance, u = -27 cm
(Object distance taken as always nagetive)
The focal length of the mirror,f = -18 cm
(In concave mirror focal length is negative)
To find image distance we can use mirror formula,
(1/ v) + (1/u) = 1/f
1/v = (1/f) – (1/u)
1/ v = (1/-18) – (1/-27)
= (-1/18) + (1/27)
= ( -27+18 ) / 486
= -9 / 486
= -1/ 54.00
v = -54 cm
The image distance is 54 cm on the object side.
To find size or height of the image we use the formula, Magnification m = v / u = h / h
-54/-27 = h_{i} / 7
h_{i} = (-54*7) / 27
h_{i} = -14 cm
The size or height of the image is 14 cm and the nature of the image is real and inverted.
16.) Find the focal length of the lens of power -2.0 D. What type of lens this.
Ans:
Given that,
Power of lens P= -2D
And we know that, the power of lens is nothing but the reciprocal of its focal length.
Hence, P=1/f
Thus, f=1/P = 1/-2 = -0.5m = -50cm
Hence, focal length of lens will be -50cm.
Here, the focal length and power of lens both are negative hence the given lens is the concave lens.
Alternative Answer –
The power of the lens, P = (-2.0)D
w.k.t power of the lens P = 1/ f
( Where f is focal length)
f = 1 / P
= 1 / -2.0
f = -0.5 m
The negative sign indicates that the lens is Concave lens.
17.) A doctor has prescribed a corrective lens of power +1.5D. Find the focal length of the lens. Is the prescribed lens is diverging or converging?
Ans:
Given that,
Power of lens P= +1.5D
We know that, power of lens P is nothing but the reciprocal of its focal length f.
Hence, P= 1/f
Thus, f= 1/P = 1/1.5= 0.66m = 66cm
Thus, the focal length of lens will be 66cm.
As the focal length and power of given lens is positive hence the given lens is the convex lens i.e. converging lens.
Alternative Answer –
Given that,
The power of lens, P = 1.5 D
w.k.t the power of lens, P = 1 / f
Where f is a focal length of the lens.
f = 1/ P
= 1 / 1.5
= 0.666 m
f = 0.666*10^{2} cm
f = 66.6 cm
The positive sign indicates that the lens is convex lens which is converging.
Therefore the prescribed lens is converging.
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