NCERT Solutions for Class 10 Science Chapter 9 Light Reflection and Refraction

Topic 10.2 Spherical Mirrors, Page No- 142

Q.1 Define the principal focus of a concave mirror.

Answer:

It is the point on the principal axis where a beam of light parallel to the principal axis after reflection actually meets.

F represents the focal length

Q.2 The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Answer:

$\text{ focal length }=\frac{\text{ Radius }}{2}$

So,

$\text{ focal length }=\frac{20}{2}=10\mathrm{cm}$

Hence the Focal length of the spherical mirror is 10 cm.

Q.3 Name a mirror that can give an erect and enlarged image of an object.

Answer:

Convex mirrors usually give a virtual and erect image.

The concave mirror gives a virtual and enlarged image only when the object is between the pole and the focus.

Q.4 Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Answer :

Convex mirrors are preferred as a rearview mirror in vehicles as:

i) It forms an erect (Rigidly upright or straight.) image of an object hence object becomes easily identified.

ii) It forms a diminished image of the object thus increasing the field of vision.

iii) An object that is far away from us is seen closer which helps us to take early decisions while driving.

Topic 10.2.4 Mirror formula and magnification, Page No- 145

Q.1 Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Answer:

As we know for the convex mirror,

Radius $(R)=$ Focus $(f)\times 2$
So,

$f=\frac{R}{2}$

putting the given values,

$f=\frac{30}{2}=15\mathrm{cm}$

Hence Focus of the convex mirror is 15 cm.

Q. 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:

Here Given Magnification:

m= -3 (real image)

Given magnification $\mathrm{m}=3$
and object distance $u=10\mathrm{cm}$ and we know,
Magnification $\mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}$

$\begin{array}{rl}& -\mathrm{v}=3\times -10\\ & \mathrm{v}=30\mathrm{cm}\end{array}$

Thus the image is formed at a distance of 30 cm.

The image is 30 cm in front of the mirror.

Topic 10.3 Refraction of light, Page No- 150

Q.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:

As we know when the light goes from a rare medium to a denser medium the light bends towards the normal. So when the light goes from air to water it will bend toward the normal.

Q. 2 Light enters from air to glass having a refractive index of 1.50. What is the speed of light in the glass? The speed of light in a vacuum is $3\times {10}^8$ .

Answer:

As we know, from the definition of the refractive index that

Refractive Index :

$\mu =\frac{c}{v}$

Where $=$ Velocity of light in the medium and
$c=$ Speed of light in vacuum
So putting those given values we,

$\begin{array}{rl}& 1.5=\frac{3\times {10}^8}{v}\\ & v=\frac{3\times {10}^8}{1.5}\\ & v=2\times {10}^8\mathrm{m}/\mathrm{s}\end{array}$

Hence the speed of the light in the glass is $2\times {10}^8\mathrm{m}/\mathrm{s}$.

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

Answer :

As we know optical density is the tendency to hold(absorb) the light. So,

more refractive index = more absorbing power = more optical density.

It can be observed from Table 10.3 that diamond and air respectively have the highest and lowest refractive index. Therefore, diamond has the highest optical density and air has the lowest optical density.

Q.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:

As we can see from the table:

Refractive index of kerosene = 1.44

Refractive index of terbutaline = 1.47

Refractive index of water = 1.33

We know from the definition of refractive index, that the speed of light is higher in a medium with the lower refractive index.

So, the light travels fastest in water relative to kerosene and turpentine.

speed of light---> water > kerosene > turpentine

Q.5 The refractive index of diamond is 2.42. What is the meaning of this statement?

Answer:

Refractive Index shows the comparison of light speeds in two mediums. Light may travel from rarer to denser medium or from denser to rarer medium. When we say the refractive index of a diamond is 2.42, it means light is travelling from a rarer to a denser medium, and the speed of light in the air (vacuum) is 2.42 times the speed of light in a diamond. On the other hand, if the light travels from denser to rarer medium, that is, from diamond to air the refractive index will be the reciprocal of 2.42.

Topic 10.3.8 Power of Lenses, Page No- 158

Q.1 Define 1 dioptre of power of a lens.

Answer:

A lens whose focal length is 1 metre is said to have the power of 1 dioptre.

Q.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:

Since the image formed by the lens is real and inverted it is formed on the side opposite to the one where the object is placed.

Image position v = 50 cm

Let the object position be u.

Since the image formed is inverted and the size image is equal to the size of the object, magnification (m) = -1.

magnification $(m)=-1$.

$\begin{array}{rl}& m=\frac{v}{u}\\ & -1=\frac{v}{u}\\ & u=-v\\ & u=-50\mathrm{cm}\end{array}$

The needle is placed 50 cm in front of the lens
Using the lens formula we have

$\begin{array}{rl}& \frac{1}{f}=\frac{1}{v}-\frac{1}{u}\\ & \frac{1}{f}=\frac{1}{50}-\frac{1}{-50}\\ & \frac{1}{f}=\frac{2}{50}\\ & f=25\mathrm{cm}\end{array}$

Power of the lens P is given by

$\begin{array}{rl}P& =\frac{1}{f}\\ P& =\frac{1}{25\times {10}^{-2}}\\ P& =4D\end{array}$

The power of the lens P is 4 Dioptre.

Q.3 Find the power of a concave lens of focal length 2 m.

Answer:

The focal length of the lens is f = -2 m. (The focal length of a concave lens is negative)

The power of the lens P is given by

$\begin{array}{rl}& P=\frac{1}{f}\\ & P=\frac{1}{-2}\\ & P=-0.5\mathrm{D}\end{array}$

The Power of the lens is - 0.5 Dioptre.

NCERT Class 10 Science Chapter 9 Light Reflection and Refraction: Exercise Solutions

Solving the Exercise Problem is a good habit, as it makes you more confident and also decreases the stress for students. So, in this, we will cover All NCERT questions with their proper solution.

Page No- 159

Q1. Which one of the following materials cannot be used to make a lens?

(a) Water (b) Glass (c) Plastic (d) Clay

Answer:

Clay cannot be used to make a lens.

(d) is the correct answer.

Q 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:

The position of the object should be between the pole of the mirror and its principal focus.

(d) is the correct answer.

Q 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:

An object should be placed at twice the focal length in front of a convex lens to get a real image of the size of the object

(b) is the correct answer.

Q 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.

Answer:

The mirror and the lens are both concave.

(a) is the correct answer.

Q 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.

Answer:

The mirror is likely to be either plane or convex.

(d) is the correct answer.

Q 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:

I would prefer to use a convex lens of focal length 5 cm while reading small letters found in a dictionary because a convex lens gives a magnified image if the object is in between focus and radius of curvature and the magnification will be high for shorter focal length

(c) is the correct answer.

Q 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:

$\begin{array}{rl}\frac{1}{f}& =\frac{1}{v}+\frac{1}{u}\\ v& =\frac{fu}{u-f}\end{array}$

For the image to be erect it has to be virtual and therefore as per convention $v$ has to be positive

$\begin{array}{rl}& \mathrm{v}>0\\ & \frac{fu}{u-f}>0\\ & \mathrm{u}<0\\ & \mathrm{f}<0\\ & \mathrm{fu}>0\\ & \mathrm{u}-\mathrm{f}>0\\ & \mathrm{u}>\mathrm{f}\end{array}$

Therefore, the object must be placed between the pole and the focus of the mirror. i.e $-15\mathrm{cm}<\mathrm{u}<0$

The image formed would be virtual and larger than the object.

Q 8. (a) Name the type of mirror used in the following situations.

• Headlights of a car.

Support your answer with reason.

Answer:

A concave mirror is used for the headlights of a car as they can produce parallel beams of high intensity which can travel through large distances if the source of light is placed at the focus of the mirror.

Q 8.b) Name the type of mirror used in the following situations.

• Side/rear-view mirror of a vehicle

Support your answer with reason.

Answer:

The convex mirror would be used as a side/rear-view mirror of a vehicle as the image produced will be erect, virtual and diminished. A convex mirror also has a larger field of view as compared to a plane mirror of the same size.

Q 8.c) Name the type of mirror used in the following situations.

• Solar furnace.

Support your answer with reason.

Answer:

The concave mirror would be used in a solar furnace to concentrate the incident rays from the sun on a solar panel.

Q 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:

The lens would produce a complete image. Only the brightness of the image would be diminished.

Verification by experiment:

Apparatus required: a convex lens, a screen, and a candlelight

procedure:

1. Place the screen behind a lens placed vertically

2. Move the candle to obtain a clear full-length image of a candle on the screen

3. Cover half of the lens with the black paper without disturbing the position of the lens

4. Note the observations

Observation: The size of the image is the same but the brightness reduces

Q 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 nature of the image formed.

Answer:

Object distance u = -25 cm.

Focal length = 10 cm

Let image distance be v

$\begin{array}{rl}& \frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ & \frac{1}{10}=\frac{1}{v}+\frac{1}{-25}\\ & \frac{1}{v}=\frac{1}{10}+\frac{1}{25}\\ & v=16.66\mathrm{cm}\end{array}$

The position of the image is 16.66 cm on the other side of the lens.
Object size $\mathrm{O}=5\mathrm{cm}$.
Let the image size be I
Magnification is m

$\begin{array}{rl}m& =\frac{v}{u}\\ \frac{I}{O}& =\frac{16.66}{-25}\\ I& =5\times -\frac{16.66}{25}\\ I& =-3.33\mathrm{cm}\end{array}$

The nature of the image is real and its size is -3.33 cm.

The formation of the image is shown in the following ray diagram

Q 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:

Focal length, f = -15 cm

Image distance, v = -10 cm ( As in the case of the concave lens image is formed on the same side as the object is placed)

Let the object distance be u.

As per the lens formula

$\begin{array}{rl}& \frac{1}{f}=\frac{1}{v}-\frac{1}{u}\\ & \frac{1}{u}=\frac{1}{v}-\frac{1}{f}\\ & \frac{1}{u}=\frac{1}{-10}-\frac{1}{-25}\\ & v=-30\mathrm{cm}\end{array}$

The object is placed at a distance of 30 cm from the lens.

Q 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:

Object distance, u = -10 cm

Focal length, f = 15 cm

Let the image distance be v

As per the mirror formula

$\begin{array}{rl}& \frac{1}{f_1}=\frac{1}{v}+\frac{1}{u}\\ & \frac{1}{15}=\frac{1}{v}+\frac{1}{-10}\\ & v=6\mathrm{cm}\\ & \text{ Magnification }=-\frac{v}{u}\\ & =-\frac{6}{-30}\\ & =0.2\end{array}$

The image formed is virtual, erect, diminished and is formed 6 cm behind of the mirror.

Q 13. The magnification produced by a plane mirror is +1. What does this mean?

Answer:

It means the image formed is virtual, erect and of the same size as the object.

Q 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:

The radius of curvature, R = 30 cm

Focal length, f = R/2 = 15 cm

Object distance, u = -20 cm

Let the image distance be $v$

$\begin{array}{rl}& \frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ & \frac{1}{15}=\frac{1}{v}+\frac{1}{-20}\\ & v=8.57\mathrm{cm}\end{array}$

Since v is positive image is formed behind the mirror.

$\begin{array}{rl}& \text{ Magnification }=-\frac{v}{u}\\ & =-\frac{8.57}{-20}\\ & =0.428\end{array}$

Object size $=5\mathrm{cm}$

$\begin{array}{rl}& \text{ Image size }=\text{ Object size }\times \text{ Magnification }\\ & =5\times 0.428\\ & =2.14\mathrm{cm}\end{array}$

A virtual, erect and diminished image of size 2.14 cm would be formed 8.57 cm behind the mirror.

Q 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 sharply focussed image can be obtained? Find the size and nature of the image.

Answer:

Object distance, u = -27 cm

Focal length, f = -18 cm

Let the image distance be v

$\begin{array}{rl}& \frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ & \frac{1}{-18}=\frac{1}{v}+\frac{1}{-27}\\ & v=-54\mathrm{cm}\end{array}$

The negative sign shows the image is formed in front of the mirror and is real.
Object size $=7\mathrm{cm}$

$\begin{array}{rl}& \text{ Image size }=\text{ magnification }\times \text{ Object size }\\ & =-\frac{v}{u}\times O\\ & =-\frac{-54}{-27}\times 7\\ & =-14\mathrm{cm}\end{array}$

A real, inverted and magnified image of size 14 cm is formed in front of the mirror.

Q 16. Find the focal length of a lens of power – 2.0 D. What type of lens is this?

Answer:

$\begin{array}{rl}& f=\frac{1}{P}\\ & f=\frac{1}{-2}\\ & f=-0.5\mathrm{m}\\ & f=-50\mathrm{cm}\end{array}$

The lens is concave because it's focal length is negative and is equal to -50 cm.

Q 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:

$\begin{array}{rl}& f=\frac{1}{P}\\ & f=\frac{1}{1.5}\\ & f=0.6667\mathrm{m}\\ & f=66.67\mathrm{cm}\end{array}$

The less is convex because its focal length is positive and therefore is converging.

Class 10 Science NCERT Chapter 9: Higher Order Thinking Skills (HOTS) Questions

Q1. There is a point object and a plane mirror. If a mirror is moved by 10cm away from the object then the image will move by?

Answer:

As we learn

Image formation from plane mirror -

1) Distance of object from mirror = Distance of image from the mirror.

2) Line joining a point object and its image is normal to the reflecting surface

3) The size of the image is the same as that of the object.

4) For a real object the image is virtual and for a virtual object, the image is real.

-

After moving the mirror by 10cm the object distance = x +10 cm

Image distance from m = x+ 10cm

Movement of the image from m

= (x+20) -x = 20 cm

Hence, the answer is 20cm

Q2. A light ray is an incident on a glass sphere of refractive index $\mu =\sqrt{3}$ at an angle of incidence 60⁰ as shown the total deviation after two refraction is:

Answer:

At point $A$ :

$\begin{array}{rl}& \text{ 1. }\sin {60}^{\circ }=\sqrt{3}\sin r\\ & \Rightarrow r={30}^{\circ }\end{array}$

from symmetry

$r^{\prime }=r={30}^0$

Apply Snell's law at B.

$\begin{array}{rl}& \text{ 1. }\sin e=\sqrt{3}\sin r^{\prime }=\frac{\sqrt{3}}{2}\\ & \Rightarrow e={60}^{\circ }\\ & {\delta }_1={60}^{\circ }-{30}^0={30}^0\\ & {\delta }_2=e-r^1={60}^{\circ }-{30}^{\circ }={30}^{\circ }\\ & \therefore \text{ total deviation }={60}^{\circ }\end{array}$

Q3. An object is at a distance of 10 cm from a mirror, and an image of the object is at a distance of 30 cm from the mirror from the same side as the object. Then the nature of the mirror and its power is -

Answer:

Optical power of a mirror -

$P=-\frac{1}{f}$

- wherein
$f=$ focal length with a sign and is in meters unit of power = diopter

$\begin{array}{rl}& \frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ & u=-10\mathrm{cm},\mathrm{v}=-30\mathrm{cm}\\ & \frac{1}{f}=\frac{-1}{10}-\frac{1}{30}=-\frac{4}{30}\text{ or }\mathrm{f}=-7.5\mathrm{cm}\end{array}$

$\therefore$ mirror is concave

$p=\frac{-100}{f}D=\frac{100}{7.5}D=\frac{40}{3}D$

Q4. An object is at a distance of 10 cm from a mirror and the image of the object is at a distance of 30 cm from the mirror on the same side as the object. The nature of the mirror and its focal length are:

Answer:

Sign Convention
1) All distances are measured from the pole.
2) Distance measured in the direction of incident rays is taken as positive.
3) Distance measured in the direction opposite to that of incident rays is taken as negative.
4) Distance above the principal axis as positive and below the principal axis as negative.
$u=-10\mathrm{cm}$
$v=-30\mathrm{cm}$ from sign convention
from mirror formula

$\begin{array}{rl}& \frac{1}{f}=\frac{1}{v}+\frac{1}{u}=-\frac{1}{30}-\frac{1}{10}=\frac{-4}{30}\\ & \mathrm{f}=-7.5\mathrm{cm}\end{array}$

since focal length is negative hence its mirror concave.

Q5. The number of images formed by two plane mirrors inclined at an angle 72⁰ of an object placed asymmetrically between mirrors is-

Answer:

as we learn
Combination of two plane Mirror -
No. of the image formed
a) If $\frac{{360}^{\circ }}{\theta }$ even the number

Number of images $=\frac{{360}^{\circ }}{\theta }-1$
b) If an odd number

Number of images $=\frac{{360}^{\circ }}{\theta }-1$
If the object is placed on the angle bisector.
b) If $\frac{{360}^{\circ }}{\theta }$ an odd number

Number of images $=\frac{{360}^{\circ }}{\theta }$
If the object is not placed on the angle bisector.

$n=\frac{{360}^0}{\theta }=\frac{{360}^0}{72}=5$

since n is odd and it is not placed on an angle bisector hence number of images is $\mathrm{n}=5$
Hence, the answer is 5.