CBSE Class X Science Chapter 9 Light – Reflection and Refraction Questions and Answers

Chapter 9: Light - Reflection and Refraction

Section 9.2: In-Text Questions (Page 142)

Q1: Define the principal focus of a concave mirror.

Light rays that are parallel to the principal axis and fall on a concave mirror meet or intersect at a single point on its principal axis after reflection. This specific point on the principal axis is called the principal focus of the concave mirror.

Figure 9.2 (a): Rays parallel to the principal axis meeting at the principal focus (F) of a concave mirror.

Q2: The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

The radius of curvature ($R$) of a spherical mirror is related to its focal length ($f$) by the formula:

$$R = 2f$$

Given: $R = 20\text{ cm}$

$$f = \frac{R}{2} = \frac{20\text{ cm}}{2} = 10\text{ cm}$$

Therefore, the focal length of the spherical mirror is 10 cm.

Q3: Name a mirror that can give an erect and enlarged image of an object.

A concave mirror can give an erect and enlarged (magnified) image of an object when the object is placed between its pole ($P$) and principal focus ($F$).

Q4: Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Convex mirrors are preferred as rear-view mirrors in vehicles due to the following reasons:

• They always form an erect, though diminished, image of objects behind the vehicle.
• They provide a wider field of view because they are curved outwards, enabling the driver to see a much larger area of traffic than a plane mirror would allow.

Section 9.2.4: In-Text Questions (Page 145)

Q1: Find the focal length of a convex mirror whose radius of curvature is 32 cm.

The relationship between focal length ($f$) and radius of curvature ($R$) is:

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

Given: $R = +32\text{ cm}$

$$f = \frac{+32\text{ cm}}{2} = +16\text{ cm}$$

The focal length of the convex mirror is +16 cm.

Q2: 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?

Given:

• Object distance ($u$) = $-10\text{ cm}$ (By Sign Convention)
• Magnification ($m$) = $-3$ (Negative for real and inverted image)

Formula:

$$m = -\frac{v}{u}$$

Substituting the given values:

$$-3 = -\frac{v}{-10}$$

$$-3 = \frac{v}{10}$$

$$v = -30\text{ cm}$$

The image is located at a distance of 30 cm in front of the mirror (on the same side as the object).

Section 9.3.2: In-Text Questions (Page 150)

Q1: 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?

The light ray bends towards the normal.

Reason: Water is optically denser than air. When a ray of light travels obliquely from an optically rarer medium (air) to an optically denser medium (water), its speed decreases, causing it to slow down and bend towards the normal.

Q2: 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 $3 \times 10^8\text{ m s}^{-1}$.

Given:

• Refractive index of glass ($n_g$) = $1.50$
• Speed of light in vacuum ($c$) = $3 \times 10^8\text{ m s}^{-1}$

Formula:

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

where $v$ is the speed of light in glass.

$$v = \frac{c}{n_g} = \frac{3 \times 10^8\text{ m s}^{-1}}{1.50} = 2 \times 10^8\text{ m s}^{-1}$$

The speed of light in glass is $2 \times 10^8\text{ m s}^{-1}$.

Q3: Find out, from Table 9.3, the medium having highest optical density. Also find the medium with lowest optical density.

According to Table 9.3:

• Medium with the highest optical density: Diamond, as it has the highest refractive index of $2.42$.
• Medium with the lowest optical density: Air, as it has the lowest refractive index of $1.0003$.

Q4: You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 9.3.

From Table 9.3, the refractive indices ($n$) of the given media are:

• Water: $n = 1.33$
• Kerosene: $n = 1.44$
• Turpentine oil: $n = 1.47$

Light travels fastest in a medium with the lowest refractive index because speed of light is inversely proportional to the refractive index ($v = c/n$). Therefore, light travels fastest in water.

Q5: The refractive index of diamond is 2.42. What is the meaning of this statement?

This statement means that the ratio of the speed of light in air (or vacuum) to the speed of light in diamond is equal to $2.42$. It also indicates that light travels $2.42$ times slower in diamond than in air/vacuum.

Section 9.3.8: In-Text Questions (Page 158)

Q1: Define 1 dioptre of power of a lens.

1 dioptre is the power of a lens whose focal length is $1\text{ metre}$ ($1\text{ D} = 1\text{ m}^{-1}$).

Q2: 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.

A convex lens forms a real, inverted image equal to the size of the object when the object is placed at twice the focal length ($2F_1$), and the image is formed at $2F_2$.

1. Position of the needle:

Image distance ($v$) = $+50\text{ cm}$

Since size of image = size of object, $u = -v = -50\text{ cm}$.

Thus, the needle is placed at a distance of 50 cm in front of the lens.

2. Power of the lens:

Distance $2f = 50\text{ cm} \implies f = 25\text{ cm} = +0.25\text{ m}$

$$P = \frac{1}{f\text{ (in metres)}} = \frac{1}{+0.25\text{ m}} = +4.0\text{ D}$$

The power of the lens is +4 D.

Q3: Find the power of a concave lens of focal length 2 m.

For a concave lens, focal length is negative.

Given: $f = -2\text{ m}$

$$P = \frac{1}{f} = \frac{1}{-2\text{ m}} = -0.5\text{ D}$$

The power of the concave lens is -0.5 D.

Chapter End Exercises (Pages 159-160)

Q1: Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay

Correct Answer: (d) Clay

Explanation: A lens must be made of a transparent material to transmit light rays. Clay is opaque and does not allow light to pass through it.

Q2: 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.

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

Q3: 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.

Correct Answer: (b) At twice the focal length

Q4: 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.

Correct Answer: (a) both concave.

Explanation: By the New Cartesian Sign Convention, focal length is negative for both a concave mirror and a concave lens.

Q5: 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.

Correct Answer: (d) either plane or convex.

Explanation: Both plane mirrors and convex mirrors always produce virtual and erect images regardless of the distance of the object.

Q6: 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.

Correct Answer: (c) A convex lens of focal length 5 cm.

Explanation: A convex lens is used as a magnifying glass. A shorter focal length provides higher power and greater magnification.

Q7: 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.

• Range of object distance: Between $0\text{ cm}$ and $15\text{ cm}$ (i.e., between pole $P$ and focus $F$, or less than $15\text{ cm}$).
• Nature of image: Virtual and erect.
• Relative size: Larger than the object (magnified).

Figure 9.7 (f): Image formation by a concave mirror when the object is placed between P and F.

Q8: 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.

(a) Headlights of a car: Concave mirror.
Reason: When a light source is placed at the focus of a concave mirror, it produces a powerful parallel beam of light to illuminate the road ahead clearly.

(b) Side/rear-view mirror of a vehicle: Convex mirror.
Reason: It always produces an erect, diminished image and offers a wider field of view to see background traffic safely.

(c) Solar furnace: Concave mirror.
Reason: Large concave mirrors concentrate parallel rays of sunlight to a sharp focal point, producing high heat energy.

Q9: 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: Yes, the lens will still produce a complete image of the object.

Experimental Verification: Take a convex lens and cover its lower half with a black paper. Mount it on a stand and place a lighted candle in front of it. Adjust a paper screen on the other side to capture the image. A full image of the candle flame appears on the screen.

Explanation: Light rays from every point of the object pass through the uncovered upper portion of the lens and refract to intersect and form the complete image. However, since the total intensity of light passing through the lens is halved, the brightness/intensity of the image will be reduced.

Q10: 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.

Given:

• Height of object ($h$) = $+5\text{ cm}$
• Object distance ($u$) = $-25\text{ cm}$
• Focal length ($f$) = $+10\text{ cm}$ (for converging/convex lens)

Calculation:

Using the Lens Formula:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} - \frac{1}{-25} = \frac{1}{10}$$

$$\frac{1}{v} = \frac{1}{10} - \frac{1}{25} = \frac{5 - 2}{50} = \frac{3}{50}$$

$$v = \frac{50}{3} \approx +16.67\text{ cm}$$

Using Magnification Formula:

$$m = \frac{h'}{h} = \frac{v}{u}$$

$$h' = h \times \left(\frac{v}{u}\right) = 5 \times \left(\frac{50/3}{-25}\right) = 5 \times \left(-\frac{2}{3}\right) = -\frac{10}{3} \approx -3.33\text{ cm}$$

Results:

• Position: Image is formed at a distance of $16.67\text{ cm}$ on the other side of the lens.
• Nature: Real and inverted (indicated by the negative sign of $h'$).
• Size: Diminished, height is $3.33\text{ cm}$.

Figure 9.16 (b): Ray diagram showing image formation by a convex lens when object is placed beyond $2F_1$.

Q11: 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.

Given:

• Focal length ($f$) = $-15\text{ cm}$ (Concave lens)
• Image distance ($v$) = $-10\text{ cm}$ (Virtual image formed on same side as object)

Calculation:

Using Lens Formula:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{u} = \frac{1}{v} - \frac{1}{f}$$

$$\frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} = -\frac{1}{10} + \frac{1}{15} = \frac{-3 + 2}{30} = -\frac{1}{30}$$

$$u = -30\text{ cm}$$

The object is placed at a distance of 30 cm in front of the concave lens.

Figure 9.17 (b): Ray diagram showing image formation by a concave lens.

Q12: 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.

Given:

• Object distance ($u$) = $-10\text{ cm}$
• Focal length ($f$) = $+15\text{ cm}$ (Convex mirror)

Calculation:

Using Mirror Formula:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - \frac{1}{-10} = \frac{1}{15} + \frac{1}{10} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$$

$$v = +6\text{ cm}$$

Results:

• Position: Image is formed at a distance of $6\text{ cm}$ behind the mirror.
• Nature: Virtual and erect.

Q13: The magnification produced by a plane mirror is +1. What does this mean?

The magnification value of +1 means:

• Positive sign (+): Indicates that the image is virtual and erect.
• Numerical value (1): Indicates that the size of the image is exactly equal to the size of the object.

Q14: 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.

Given:

• Object height ($h$) = $+5.0\text{ cm}$
• Object distance ($u$) = $-20\text{ cm}$
• Radius of curvature ($R$) = $+30\text{ cm}$
• Focal length ($f$) = $R/2 = +15\text{ cm}$

Calculation:

Using Mirror Formula:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} = \frac{1}{15} - \frac{1}{-20} = \frac{1}{15} + \frac{1}{20} = \frac{4 + 3}{60} = \frac{7}{60}$$

$$v = \frac{60}{7} \approx +8.57\text{ cm}$$

Using Magnification Formula:

$$m = -\frac{v}{u} = \frac{h'}{h}$$

$$h' = -h \times \left(\frac{v}{u}\right) = -5.0 \times \left(\frac{60/7}{-20}\right) = +5.0 \times \frac{3}{7} = +\frac{15}{7} \approx +2.14\text{ cm}$$

Results:

• Position: Image is formed $8.57\text{ cm}$ behind the mirror.
• Nature: Virtual and erect.
• Size: Diminished, height is $2.14\text{ cm}$.

Q15: 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.

Given:

• Object size ($h$) = $+7.0\text{ cm}$
• Object distance ($u$) = $-27\text{ cm}$
• Focal length ($f$) = $-18\text{ cm}$ (Concave mirror)

Calculation:

Using Mirror Formula:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} = \frac{1}{-18} - \frac{1}{-27} = -\frac{1}{18} + \frac{1}{27} = \frac{-3 + 2}{54} = -\frac{1}{54}$$

$$v = -54\text{ cm}$$

Using Magnification Formula:

$$h' = -h \times \left(\frac{v}{u}\right) = -7.0 \times \left(\frac{-54}{-27}\right) = -7.0 \times 2 = -14.0\text{ cm}$$

Results:

• Screen Distance: The screen should be placed at $54\text{ cm}$ in front of the mirror.
• Nature: Real and inverted.
• Size: Enlarged, height is $14.0\text{ cm}$.

Q16: Find the focal length of a lens of power -2.0 D. What type of lens is this?

Given: Power ($P$) = $-2.0\text{ D}$

$$f = \frac{1}{P} = \frac{1}{-2.0\text{ D}} = -0.5\text{ m} = -50\text{ cm}$$

The focal length is -0.5 m (or -50 cm). Since the focal length and power are negative, it is a concave lens (diverging lens).

Q17: 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?

Given: Power ($P$) = $+1.5\text{ D}$

$$f = \frac{1}{P} = \frac{1}{+1.5\text{ D}} = +\frac{10}{15}\text{ m} = +\frac{2}{3}\text{ m} \approx +0.67\text{ m} = +66.7\text{ cm}$$

The focal length of the prescribed lens is +0.67 m. Since the power and focal length are positive, it is a converging lens (convex lens).