NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

These NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Light Reflection and Refraction NCERT Solutions for Class 10 Science Chapter 10

Class 10 Science Chapter 10 Light Reflection and Refraction InText Questions and Answers

In-text Questions (Page 168)

Question 1.
Define the principal focus of a concave mirror.
Answer:
A point on the principal axis at which all parallel rays to the principal axis meets is called the principal focus of the concave mirror.

Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer:
Focal length (l)

Question 3.
Name a mirror that can give an erect and enlarged image of an object.
Answer:
Concave mirror.

Question 4.
Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
Convex mirrors are preferred because they always give an erect, through diminished, image. Also, they have a wider field of view as they are curved outwards.

In-text Questions (Page 171)

Question 1.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
R = 2f
∴ f = \(\frac{R}{2}=\frac{32 \mathrm{~cm}}{2}\) = 16 cm

Question 2.
A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Answer:
Magnification (m) = \(\frac{h^{1}}{h}=-\frac{v}{u}\)
3 = – = \(-\frac{v}{(+4)}=-\frac{v}{4}\)
u = (-3) × 4 = – 12 cm
The image is located in front of the mirror at 12 cm from the pole of the concave mirror.

In-text Questions (Page 176)

Question 1.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
The light ray bend towards the normal. When light travels from rarer (air) to denser (water) medium it bends towards the normal.

Question 2.
Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 10s m/sec.
Answer:
The refractive index of the medium ‘nm’ is given by

Question 3.
Find out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.

Answer:
Higher will be the refractive index, higher will be the optical density of the material and lower refractive index means lower will be the optical density.

Diamond has the highest optical density (Refractive index 2.42) and air has the lowest optical density, (refractive index 1.0003)

Question 4.
You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 10.3.
Answer:
The refractive index of kerosene, turpentine and water as follows :
Kerosene – 1.44
Water – 1.33
Turpentine – 1.47
The light travel fastest in water because it has lowest refractive index.

Question 5.
The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
Refractive index of the medium (or material) is the ratio of the speed of light in air and speed of light in medium S.e.

The refractive index of diamond of means that the ratio of the speed of light in air and the speed of light in diamond is 2.42.

In-text Questions (Page 184)

Question 1.
Define 1 dioptre of power of a lens.
Answer:
1 dioptre is the power of a lens whose focal length is 1 metre.

Question 2.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm 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.
Answer:
We know that

Distance of the image from the lens = + 50 cm.
Now by lens formula

Question 3.
Find the power of a concave lens of focal length 2 m.
Answer:

Class 10 Science Chapter 10 Light Reflection and Refraction Textbook Questions and Answers

Page no. 185

Question 1.
Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer:
(d) Clay

Question 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.
Answer:
(d) Between the pole of the mirror and its principal focus.

Question 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.
Answer:
(b) At twice the focal length

Question 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) convex
(c) The mirror is concave and the lens is convex.
(d) The mirror is convex, but the lens is concave.
Answer:
(a) Both are concave.

Question 5.
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) plane.
(b) concave.
(c) convex.
(d) either plane or convex.
Answer:
(c) Convex only

Question 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.
Answer:
(d) A concave lens of focal length 5 cm.

Question 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.
Answer:
To obtain an erect image of an object it should be placed between ‘P’ and ‘F’ i.e., between pole and pricipal focus. So the range of the distance of the object from the mirror will be greater than zero and less than fifteen.

The image will be virtual and formed behind the mirror. The image is larger than the object.

Question 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.
Answer:
(a) Headlight of a car: Concave mirror is used in head lights of a car to get a powerful parallel beams of light.
(b) Side/rear-view mirror of a vehicle: Convex mirros are commonly used because they always given an erect, through diminished image and they also have a wider field of view.
(c) Solar furnace : Large concave mirrors are used in the solar furnace to concentrate sunlight to produce heat in the furnace.

Question 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.
Answer:
When one half of a convex lens is covered by a black paper it may or may not be form a complete image. The image formation is based on the position of the object placed in front of the lens.
Experimently we observed that.

• When object is placed at infinity highly diminished and point-sized real and inverted image is formed.
• When object is placed beyond 2F₁ at 2F₁ between F₁ and 2F₁, and at focus F₁ not a complete image is formed.
• When the object is placed between focus F₁ and optical centre ‘O’ and enlarged, virtual and erect image is formed.

Question 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.
Answer:
Given h = 5 cm
u = -25 cm
f = 10 cm
To find v = ?
h₁ = ?
From the lens formula

The positive sign of ‘v’ means, image is formed at a distance of 50/3 cm to the right of the optical centre of the lens.
now m= \(\frac{h^{1}}{h}=\frac{v}{u}\)

The negative sign of h’ suggests that the image is real, inverted and diminished.

Question 11.
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.
Answer:
Given v = -10 cm
f = -15 cm
u = ?


\(\frac{1}{u}=\frac{-3+2}{30 \mathrm{~cm}}=-\frac{1}{30 \mathrm{~cm}}\)
or, u = -30 cm

Question 12.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer:
Given u = -10 cm
f = -15 cm
v = ?
By mirror formula

Because v is positive, so image will form behind the mirror. The image will be erect.

So the size of image will be 0.6 times the object size.

Question 13.
The magnification produced by a plane mirror is +1. What does this mean?
Answer:
Magnification (m)

So, Height of the image (h’) = Height of object (h)
‘+1′ magnification means the height of the object is equal to the height of the image and the image is virtual and erect.

Question 14.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
Given h = 5.0 cm
R = 30 cm
u = – 20 cm
v = ?
h’ = ?
We know that R = 2f

So the image will form behind the mirror and it will be virtual and erect.

Question 15.
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
Answer:
Given h = 7.0 cm
u = -27 cm
f = -18 cm
v = ?
h’ = ?
From the mirror fourmla :

= \(\frac{(-54 \mathrm{~cm}) \times 7 \mathrm{~cm}}{(-27 \mathrm{~cm})}\)
= -14 cm
So the image is inverted, real and of 14 cm height and lies in the direction opposite to the incident light ray.

Question 16.
Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer:
Given p = -2.0 D
f = ?
We know that p = \(\frac{1}{f}\)
f = \(\frac{1}{P}=\frac{1}{(-2.0 D)}\) = -0.5 m
This is a concave lens.

Question 17.
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?
Answer:

Class 10 Science Chapter 10 Light Reflection and Refraction Textbook Activities

Activity 10.1 (Page 161)

• Take a large shining spoon. Try to view your face in its curved surface,
• Do you get the image? Is it smaller or larger?
• Move the spoon slowly away from your face. Observe the image. How does it change?
• Reverse the spoon and repeat the Activity. How does the image look like now?
•  Compare the characteristics of the image on the two surfaces.

Observation:

• Yes we get the image. It is a virtual image. It is smaller when spoon is moved slowly away from the face/the size of the image decreases.
• In reverse spoon. We observe an inverted image i.e., real image. When spoon is moved slowly from the face the size of the image decreases.

Real Image	Virtual Image
1. It is inverted with respect to the object.	1. It is erect with respect to the object.
2. It can be obtained on a screen.	2. It cannot be obtained on a screen.

Activity 10.2 (Page 162)

CAUTION : Do not look at the Sun directly or even into a mirror reflecting sunlight. It may damage your eyes.

• Hold a concave mirror in your hand and direct its reflecting surface towards the Sun.
• Direct the light reflected by the mirror on to a sheet of paper held close to the mirror.
• Move the sheet of paper back and forth gradually until you find on the paper sheet a bright, sharp spot of light

Question 1.
Hold the mirror and the paper m the same position for a few minutes. What do you observe? Why?
Answer:
Observation : The paper at first begins to burn producing smoke. The light rays from the sun is converged at a point, as a sharp, bright spot by the mirror. The heat produced due to the concentration of sunlight ignites the paper.

Activity 10.3 (Page 163)

You have already learnt a way of determining the focal length of a concave mirror. In Activity 10.2, you have seen that the sharp bright spot of light you got on the paper is, in fact, the image of the Sun., it was a tiny, real, inverted image. You got the approximate focal length of the concave mirror by measuring the distance of the image from the mirror.

• Take a concave mirror. Find out its approximate focal length in the way described above. Note down the value of focal length. (You can also find it out by obtaining image of a distant object on a sheet of paper.)
• Mark a line on a Table with a chalk. Place the concave mirror on a stand. Place the stand over the line such that its pole lies over the line.
• Draw with a chalk two more lines parallel to the previous line such that the distance between any two successive lines is equal to the focal length of the mirror. These lines will now correspond to the positions of the points P» F and C, respectively. Remember – For a spherical mirror of small aperture, the principal focus Flies mid-way between the pole P and the centre of curvature C.
• Keep a bright object, say a burning candle, at a position far beyond C. Place a paper screen and move it in front of the mirror till you obtain a sharp bright image of the candle flame on it.
• Observe the image carefully. Note down its nature, position and relative size with respect to the object size.
• Repeat the activity by placing the candle – (a) just beyond C, (b) at C, (c) between F and C, (d) at F, and (e) between P and F.
• In one of the cases, you may not get the image on the screen. Identify the position of the object in such a case. Then, look for its virtual image in the mirror itself.
• Note down and tabulate your observations.

Observation:

Activity 10.4 (Page 166)

• Draw neat ray diagrams for each position of the object shown in Table 10.1.
• You may take any two of the rays mentioned in the previous section for locating the image.
• Compare your diagram with those given in Fig. 10.7.
• Describe the nature, position and relative size of the image formed in each case.
• Tabulate the results in a convenient format.

Observation : The result of activity 10.3 in table 10.1.
The rays diagram obtained are similar to the rays diagram given in Fig,

Activity 10.5 (Page 167)

• Take a convex mirror. Hold it in one hand
• Hold a pencil in the upright position in the other hand.

Question 1.
Observe the image of the pencil in the mirror. Is the image erect or inverted? Is it diminished or enlarged?
Answer:
Image is erect. Image is diminished.

Question 2.
Move the pencil away from the mirror slowly. Does the image become smaller or larger?
Answer:
Image become smaller; when the pencil is moved away from the mirror slowly.

Question 3.
Repeat this Activity carefully. State whether the image will move closer to or farther away from the focus as the object is moved away from the mirror?
Answer:
If object is moved away from the mirror, image moves closer to the focus.

Activity 10.6 (Page 167)

• Observe the image of a distant object, say a distant tree, in a plane mirror.

Question 1.
Could you see a full-length image?
Answer:
We see full-length image of a tall building or tree in a small convex mirror.

Question 2.
Try with plane mirrors of different sizes. Did you see the entire object in the image?
Answer:
A full-length image is observed in plane mirror (in different type of mirror).

Question 3.
Repeat this Activity with a concave mirror. Did the mirror show full length image of the object?
Answer:
A diminished image is observed which is virtual and erect by using a convex mirror.

Question 4.
Now try using a convex mirror. Did you succeed? Exp1air your observations with reason.
Answer:
In concave mirror, high!y diminished point sized. real and in.erted image s formed when object is placed at infinity, but in convex mirror highly diminished point-sized virtual and erect image is formed when object is placed at infinity, this image is formed at the focus F, behind the mirror.

Activity 10.7 (Page 172)

• Place a coin at the bottom of a bucket filled with water.
• With your eye to a side above water, try to pick up the coin in one go. Did you succeed in picking up the coin?
• Repeat the Activity. Why did you not succeed in doing it in one go?
• Ask your friends to do this. Compare your experience with theirs.

Observation: No we cannot succeed in picking up the coin. This is because of the fact the rays of light traveling upward from the coin of the buket changes their direction due to refraction as they reach the surface of the water

Activity 10.8 (Page 172)

• Place a large shallow bowl on a table and put a coin in it.
• Move away slowly from the bowl. Stop when the coin just disappears from your sight.
• Ask a friend to pour water gently into the bowl without disturbing the coin.

Question 1.
Keep looking for the coin from your position. Does the coin becomes visible again from your position? How could this happen?
Answer:
Observation: The coin becomes visible again on pouring water into the bowl. The coin appears slightly raised above its actual position due to refraction of light.

Activity 10.9 (Page 172)

• Draw a thick straight line in ink, over a sheet of white paper placed on a table.
• Place a glass slab over the line in such a way that one of its edges makes an angle with the line.

Question 1.
Look at the portion of the line under the slab from the sides. What do you observe? Does the line under the glass slab appear to be bent at the edges?
Answer:
We observe that the line inside the slab appears to be bent at the edges.

Question 2.
Next, place the glass slab such that it is normal to the line. What do you observe now? Does the part of the line under the glass slab appear bent?
Answer:
We find that there is no bending of the line.

Question 3.
Look at the line from the top of the glass slab. Does the part of the line, beneath the slab, appear to be raised? Why does this happen?
Answer:
We find the portion of tire line between the edges of the slab appears to be raised.

Activity 10.10 (Page 173)

• Fix a sheet of white paper on a drawing board using drawing pins,
• Place a rectangular glass slab over the sheet in the middle.
• Draw the outline of the slab with a pencil. Let us name the outline as ABCD.
• Take four identical pins.
• Fix two pins, say E and F, vertically such that the line joining the pins is inclined to the edge AB.
• Look for the images of the pins E and F through the opposite edge. Fix two other pins, say G and H, such that these pins and the images of E and F lie on a straight line.
• Remove the pins and the slab.
• Join the positions of tip of the pins E and F and produce the line up to AB. Let EF meet AB at O. Similarly, join the positions of tip of the pins G and H and produce it up to the edge CD. Let HG meet CD at O’, in Fig.

In this Activity, you will note, the light ray has changed its direction at points O and O’. Note that both the points O and O’ lie on surfaces separating two transparent media. Draw a perpendicular NN’ to AB at O and another perpendicular MM’ to CD at O’. The light ray at point O has entered from a rarer medium to a denser medium, that is, from air to glass. Note that the light ray has bent towards the normal. At O’, the light ray has entered from glass to air, that is, from a denser medium to a rarer medium The light here has bent away from the normal. Compare the angle of incidence with the angle of refraction at both refracting surfaces AB and CD.

In Fig., EO is the incident ray. OO’ is the refracted ray and O’ H is the emergent ray. We observe that the emergent ray is parallel to the direction of the incident ray. The extent of bending of the ray of light at the opposite parallel faces AB (air-glass interface) and CD (glass-air interface) of the rectangular glass slab is equal and opposite. This is why the ray emerges parallel to the incident ray. However, the light ray is shifted sideward slightly.

Activity 10.11 (Page 177)

CAUTION: Do not look at the Sun directly or through a lens while doing this Activity or otherwise. You may damage your eyes if you do

• Hold a convex lens in your hand. Direct it towards the Sun.
• Focus the light from the Sun on a sheet of paper. Obtain a sharp bright image of the Sun.

Question 1.
Hold the paper and the lens in the same position for a while. Keep observing the paper. What happened? Why? Recall your experience in Activity 10.2.
Answer:
Observation : The paper begins to burn producing smoke. It may even catch fire after a while. Why does this happen? The light from the Sun constitutes parallel rays of light. These rays were converged by the lens at the sharp bright spot formed on the paper. The bright spot you got on the paper is a real image of the Sun.

Activity 10.12 (Page 178)

• Take a convex lens. Find its approximate focal length in a way described in Activity 10.11.
• Draw five parallel straight lines, using chalk, on a long Table such that the distance between the successive lines is equal to the focal length of the lens.
• Place the lens on a lens stand. Place it on the central line such that the optical centre of the lens lies just over the line.
• The two lines on either side of the lens correspond to F and 2F of the lens respectively. Mark them with appropriate letters such as 2F₁, F₁, F2 and 2F₂, respectively.
• Place a burning candle, far beyond 2F₁ to the left. Obtain a clear sharp image on a screen on the opposite side of the lens.
• Note down the nature, position and relative size of the image.
• Repeat this Activity by placing object just behind 2F₁ between F₁ and 2F₁ at F₁, between F₁ and O. Note down and tabulate your observations.

Observation: Nature, position and relative size of the image formed by a convex lens for various positions of the object.

Activity 10.13 (Page 179)

• Take a concave lens. Place it on a lens stand.
• Place a burning candle on one side of the lens.
• Look through the lens from the other side and observe the image. Try to get the image on a screen, if possible. If not. observe the image directly through the lens.
• Note down the nature, relative size and approximate position of the image.
• Move the candle away from the lens. Note the change in the size of the image. What happens to the size of the image when the candle is placed too far away from the lens.

Observation; Nature, position and relative size of the image formed by a concave lens for various positions of the object.

Class 10 Science Chapter 10 Light Reflection and Refraction Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
What is the unit of power of a lens ?
Answer:
Dioptre.

Question 2.
What is the relationship between radius of curvature (R) and focal length) (f).
Answer:
R = 2f.

Question 3.
An incident ray makes an angle of 60° with the surface of a plane mirror. What is the angle of reflection ?
Answer:
Angle of reflection = 90° – 60° = 30°

Question 4.
Given an example of a converging lens.
Answer:
Convex lens.

Question 5.
Give an example of a diverging lens.
Answer:
Concave lens.

Short Answer Type Questions

Question 1.
Why does diamond sparkle ?
Answer:
Diamond sparkle due to total internal reflation of light. Diamond has very high value of refractive index i.e., 2.47 hence the critical angle for diamond air interface is only 24.4°. When a light ray falls at any face of diamond crystal at an angle more than critical angle, it comes back in the same medium and strike an other face. So this way light trapped inside and sparkling takes place.

Question 2.
Why the sun appears red while rising or setting ?
Answer:
While rising or setting, the sun is on the horizon and the sun light has to travel much larger distance through the air than at mid day, so, larger amount of red light is scattered at the rising or setting of sun and the light received by us is purely in red colour, due to which it appears red.

Question 3.
What are optically denser and optically rarer medium ?
Answer:
A medium which has larger refractive index is optically denser and which has lower value of refractive index is optically rarer medium.

For example diamond is denser because is has refractive index 2.42 while air is rarer because it has refractive index 1.0003.

Long Answer Type Question

Question 1.
What are the new cartesian sign conventions for the reflection by spherical mirrors ?
Answer:

• The object is on the left side of the miror. i. e., light is incident on the mirror from the left hand side of the mirror.
• All the distances parallel to the principal axis are measured from the pole of the spherical mirror.
• The distances measured in the direction of incident lights are taken as positive.
• The distances measured in the direction opposite to the direction of incident light are taken as negative.
• The heights measured upwards and perpendicular to the principal axis of the mirror are taken as positive.
• The heights measured downwards and perpendicular to the principal axis of the mirror are taken as negative.

Multiple Choice Questions

Question 1.
One Dioptre is the power of lens of focal length.
(a) 5 cm
(b) 1000 cm
(c) 1 m
(d) 12 m
Answer:
(c) 1 m

Question 2.
When ray of light enters a glass slab from air:
(a) its frequency decreases
(b) its frequency increases
(c) its wavelength decreaces
(d) its wavelength increaces
Answer:
(c) its wavelength decreaces

Question 3.
The refractive index of glas is 1.5, the velocity of light in glass is:
(a) 2.5 × 10¹⁰ cm/sec
(b) 2 × 10¹⁰ cm/sec
(c) 3.5 × 10¹⁰ cm/sec
(d) 1.5 × 10¹⁰ cm/sec
Answer:
(b) 2 × 10¹⁰ cm/sec

Question 4.
The focal length of lens whose power is -1.5 D is:
(a) +2.5 m
(b) -66.6 cm
(c) -2.5 m
(d) +66.6 cm
Answer:
(b) -66.6 cm

Question 5.
The magnification of a concave mirror.
(a) Does not depend on its focal length.
(b) Is equal to the ratio of the object distance to the image distance.
(c) Is equal to the ratio of image distance of the object distance.
(d) Does depend on its focal length
Answer:
(c) Is equal to the ratio of image distance of the object distance.