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

Topic 9.2.2 Representation of Images Formed by Spherical Mirrors Using Ray Diagrams

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: 10 cm.

Explanation:

$
\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 the 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 make early decisions while driving.

Topic 9.2.4 Mirror formula and Magnification

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

Answer: 15 cm

Explanation:

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 a 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{aligned}
& -\mathrm{v}=3 \times-10 \\
& \mathrm{v}=30 \mathrm{~cm}
\end{aligned}
$

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

The image is 30 cm in front of the mirror.

Topic 9.3 Refraction of light

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 rarer 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: 2 \times 10^8

Explanation:

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{aligned}
& 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{aligned}
$

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

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 a 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 media. Light may travel from a rarer to a denser medium or from a denser to a 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 9.3.8 Power of a Lens

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: 4 diopters.

Explanation:

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

magnification $(m)=-1$.

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

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

$
\begin{aligned}
& \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{aligned}
$

Power of the lens P is given by

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

The power of the lens P is 4 diopters.

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

Answer: - 0.5 Dioptre.

Explanation:

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

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

The Power of the lens is - 0.5 Dioptre.

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 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 between the focus and radius of curvature, and the magnification will be high for a 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 distances 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{aligned}
\frac{1}{f} & =\frac{1}{v}+\frac{1}{u} \\
v & =\frac{f u}{u-f}
\end{aligned}
$

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

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

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 a 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 a 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 than 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 a 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 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{aligned}
& \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{aligned}
$

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

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{aligned}
& \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{aligned}
$

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{aligned}
& \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{aligned}
$

The image formed is virtual, erect, and diminished and is formed 6 cm behind 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{aligned}
& \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\
& \frac{1}{15}=\frac{1}{v}+\frac{1}{-20} \\
& v=8.57 \mathrm{~cm}
\end{aligned}
$

Since v is positive image is formed behind the mirror.

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

Object size $=5 \mathrm{~cm}$

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

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 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 focused 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{aligned}
& \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\
& \frac{1}{-18}=\frac{1}{v}+\frac{1}{-27} \\
& v=-54 \mathrm{~cm}
\end{aligned}
$

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

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

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:

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

The lens is concave because its 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:

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

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