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

(a) Headlights of a Car — Concave Mirror

A concave mirror is used in the headlights of a car. When the bulb is placed at the focus of the concave mirror, the reflected light travels as a strong, parallel beam over a long distance. This provides a bright and focused beam of light on the road ahead.

(b) Side/Rear-View Mirror of a Vehicle — Convex Mirror

A convex mirror is used as a side or rear-view mirror in vehicles. There are two main reasons for this:

• A convex mirror always forms an erect, virtual, and diminished image of any object placed in front of it, making it easier for the driver to see vehicles behind.
• It has a much wider field of view compared to a plane mirror of the same size, allowing the driver to see a larger area of the road.

(c) Solar Furnace — Concave Mirror

A large concave mirror is used in a solar furnace. It converges the parallel rays of sunlight at its principal focus. This concentration of solar energy at a single point generates a very high temperature, which is used for heating purposes in the furnace.

 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: A convex lens forms a complete image of an object, even if its one half is covered with black paper. Only the brightness of the image will be reduced. It can be explained by considering the following two cases.
Case I : When the upper half of the lens is covered
In this case, a ray of light coming from the object will be refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the following figure.

Case II: When the lower half of the lens Is covered
In this case, a ray of light coming from the object is refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the given figure.

However, since less light passes through the lens, the image formed will be less bright, but its size and position remain unchanged.

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:
Object height, \( h = 5 \, \text{cm} \)
Object distance, \( u = -25 \, \text{cm} \)
Focal length of convex lens, \( f = +10 \, \text{cm} \)

We know that, 

\( \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \)
\( ⇒ \frac{1}{10} = \frac{1}{v} – \left(\frac{1}{-25}\right) \)
\( ⇒ \frac{1}{10} = \frac{1}{v} + \frac{1}{25} \)
\( ⇒ \frac{1}{v} = \frac{1}{10} – \frac{1}{25} \)

\( ⇒ \frac{1}{v} = \frac{5 – 2}{50} = \frac{3}{50} \)
\( v= \frac{50}{3} \, \text{cm} \)
\( v = 16.6 \, \text{cm} \)

The image is formed at 16.6 cm on the other side of the lens.

Also, for a converging lens, 

\( m = \frac{v}{u} \)
\( m = \frac{16.6}{-25} = -0.66\, \text{cm} \) 

Again, we know, 

\( m = \frac{h′}{h} \)

So, \( h’ = m \times h = -0.66 \times 5 \approx -3.3 \, \text{cm} \)

Nature of image: Image is real, inverted and diminished.

• Negative magnification → inverted
• Image formed on the other side → real
• Magnitude less than 1 → diminished

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:
Focal length, \( f = -15 \, \text{cm} \)
Image distance, \( v = -10 \, \text{cm} \)

We know that,

\( \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \)
\( ⇒ \frac{1}{-15} = \frac{1}{-10} – \frac{1}{u} \)
\( ⇒-\frac{1}{15} = -\frac{1}{10} – \frac{1}{u} \)
\( ⇒ -\frac{1}{15} + \frac{1}{10} = -\frac{1}{u} \)​

\( ⇒ \frac{-2 + 3}{30} = -\frac{1}{u} \)
\( ⇒ \frac{1}{30} = -\frac{1}{u} \)
\( ∴ u = -30 \, \text{cm} \)

The negative value of u indicates that the object is placed 30 cm in front of the lens. This is shown in the following ray diagram.

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:
Object distance, \( u = -10 \, \text{cm} \)
Focal length, \( f = +15 \, \text{cm} \)

We know that,

\( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)
\( ⇒ \frac{1}{15} = \frac{1}{v} + \frac{1}{-10} \)
\( ⇒ \frac{1}{15} = \frac{1}{v} – \frac{1}{10} \)
\( ⇒ \frac{1}{v} = \frac{1}{15} + \frac{1}{10} \)

\( ⇒ \frac{1}{v} = \frac{2 + 3}{30} \)$$\text{( ⇒ frac{1}{v} = frac{5}{30} ) ( ⇒ frac{1}{v} = frac{1}{6} ) ( ∴ v = +6 , text{cm} )}$$

Because ν is +ve, so a virtual image is formed at a distance of 6 cm behind the mirror.

Magnification, \( m = \frac{-v}{u} \)
\( m = \frac{-6}{-10} = 0.6 \) 

The positive value of m shows that the image is erect, and its value is less than 1, which shows that the image formed is diminished.

So, the nature of the is Virtual, erect, and diminished.

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

Answer: The magnification of a plane mirror is +1, which means:

• The positive sign (+) indicates that the image formed is virtual and erect.
• The value 1 indicates that the size of the image is equal to the size of the object.

Thus, a plane mirror forms an image that is virtual, erect, and of the same size as the object.

14. An object 5 cm is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position, nature and size of the image.

Answer: Given:
Object height, \( h = 5 \, \text{cm} \)
Object distance, \( u = -20 \, \text{cm} \)
Radius of curvature, \( R = 30 \, \text{cm} \)

We know, \( R = 2f \)
\( f = \frac{R}{2} = \frac{30}{2} = +15 \, \text{cm} \)

We know that,

\(  \frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
\( ⇒  \frac{1}{v} + \frac{1}{-20} =\frac{1}{15} \)
\(⇒ \frac{1}{v} – \frac{1}{20} = \frac{1}{15}\)
\( ⇒ \frac{1}{v} = \frac{1}{15} + \frac{1}{20} \)​
\( ⇒ \frac{1}{v} = \frac{4 + 3}{60} \)$$\text{( ⇒ frac{1}{v} = frac{7}{60} ) ( ⇒v= frac{60}{7} ) ( ∴v= 8.57 cm )}$$

So, the image is formed 8.57 cm behind the mirror.

Magnification:

\( m = \frac{-v}{u} \)
\( ⇒ m= \frac{-8.57}{-20} \)
\( ⇒ m = 0.43\, \text{cm} \)

Again, we know that,

\( m = \frac{h′}{h} \)

$$\text{( ⇒ h’ = 0.43 times 5 , text{cm} ) ( ∴ h’ = 2.14 , text{cm} )}$$

• Position: The image forms 8.57 cm behind the mirror.
• Size: The image is 2.14 cm tall. The image is diminished because the value of magnification is less than 1 (m < 1).
• Nature: The image is virtual, erect, and diminished. (Image is virtual since v is positive. Image is erect due to the magnification m is positive. )

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 focused image can be obtained? Find the size and nature of the image.

Answer: Given:
Object height, \( h = 7 \, \text{cm} \)
Object distance, \( u = -27 \, \text{cm} \)
Focal length, \( f = -18 \, \text{cm} \)

We know that, 

\( \frac{1}{v} + \frac{1}{u} = \frac{1}{f}  \)
\( ⇒  \frac{1}{v} + \frac{1}{-27} = \frac{1}{-18} \)
\( ⇒ \frac{1}{v} = -\frac{1}{18} + \frac{1}{27} \)​

\( ⇒ \frac{1}{v} = \frac{-3 + 2}{54} \)$$\text{( ⇒ frac{1}{v} = -frac{1}{54} ) ( ∴v = -54 , text{cm} )}$$

The negative sign confirms the image forms in front of the mirror (real image). So the screen should be placed 54 cm in front of the mirror.

Magnification:

$$\backslash (\Rightarrow \; m\; =\; \backslash frac\{-(-54)\}\{-27\}\; \backslash )\backslash (\; \Rightarrow \; m\; =\; \backslash frac\{54\}\{-27\}\; =\; -2\; \backslash )$$

Again, we know that,

\( m = \frac{h′}{h} \)

$$\text{( ∴h’ = -2 times 7 = -14 , text{cm} )}$$

• Position: The screen should be placed 54 cm in front of the mirror.
• Size: The image is 14 cm tall. The image is enlarged because the value of magnification is less than 1 (m › 1).
• Nature: The image is real, inverted, and enlarged. (Image is real since v is negative. Image is inverted due to the magnification m is negative. )

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

Answer: Given:
Power of lens, \( P = -2.0 \, \text{D} \)

We know:

\( P = \frac{1}{f} \)
So, \( f = \frac{1}{P} \)
\( ⇒ f = \frac{1}{-2.0} \)$$\text{( ⇒ f = -0.5 , text{m} )}$$
\( ∴ f = -50 \, \text{cm} \)

Since the focal length is negative, the lens is a concave (diverging) lens.

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: Given:
Power of lens, \( P = +1.5 \, \text{D} \)

We know:

\( P = \frac{1}{f} \)
So, \( f = \frac{1}{P} \)
\( ⇒ f = \frac{1}{1.5} \)$$\text{( ⇒ f = .66 , text{m} )}$$
\( ∴ f = .66 \, \text{cm} \)

Since the focal length is positive, the lens is a converging (convex) lens.