Class 10 Science Ch 9 – Light Solutions
NCERT CLASS 10 SCIENCE • CHAPTER 9 • LIGHT • FULL EXERCISES Q1-Q17

Light

Full Solutions Q1 – Q17

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
Reasoning: A lens works by refraction, which requires a transparent material for light to pass through. Clay is opaque, so light cannot pass through it.
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 principal focus and 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.
Reasoning: A concave mirror forms a virtual, erect, and magnified image only when the object is placed very close to the mirror, specifically within its focal length ($u < f$).
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
Reasoning: For a convex lens, placing the object at $2F_1$ (the center of curvature) produces a real, inverted image at $2F_2$ that is exactly the same size as the object.
[Image of convex lens ray diagram object at 2F]
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) both convex.
(c) the mirror is concave and the lens is convex.
(d) the mirror is convex, but the lens is concave.
Answer: (a) both concave.
Reasoning: By sign convention, a negative focal length indicates a concave mirror and a concave lens. Positive focal length indicates convex.
Question 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: (d) either plane or convex.
Reasoning: Plane mirrors always form erect images. Convex mirrors also always form erect (and diminished) images regardless of distance. Concave mirrors form inverted images at most distances.
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: (c) A convex lens of focal length 5 cm.
Reasoning:
1. To read small letters, we need a magnifying glass, which is a convex lens.
2. Magnification is inversely proportional to focal length ($P = 1/f$). A smaller focal length gives higher power and magnification.
Therefore, a convex lens with a smaller focal length (5 cm) is preferred.
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.
Range: Between 0 cm and 15 cm (Between Pole and Focus).
Nature: Virtual and Erect.
Size: Larger than the object (Magnified).
Ray Diagram: The object is placed inside the focal point ($F$). Rays diverge after reflection and appear to meet behind the mirror.
[Image of ray diagram concave mirror object between P and F]
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.
(a) Headlights of a car: Concave Mirror.
Reason: The light bulb is placed at the focus. Rays from the focus reflect to form a strong parallel beam of light.
(b) Side/rear-view mirror: Convex Mirror.
Reason: It always forms erect images and has a wider field of view, allowing drivers to see more traffic behind them.
(c) Solar furnace: Concave Mirror.
Reason: Sun rays (parallel rays) converge at the focus, generating intense heat at that point.
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: Yes, the lens will produce a complete image of the object.
Explanation:
• Every part of a lens allows light to pass through and converge to form an image.
• Covering half the lens blocks half the light rays, but the remaining rays still converge at the same point.
Result: The full image is formed, but its intensity (brightness) is reduced by half.
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.
Given:
Height ($h$) = +5 cm
Object distance ($u$) = -25 cm
Focal length ($f$) = +10 cm (Converging/Convex)
Lens Formula: $$\frac{1}{v} – \frac{1}{u} = \frac{1}{f}$$ $$\frac{1}{v} – \frac{1}{-25} = \frac{1}{10}$$ $$\frac{1}{v} = \frac{1}{10} – \frac{1}{25}$$ $$\frac{1}{v} = \frac{5 – 2}{50} = \frac{3}{50}$$ $$v = \frac{50}{3} = +16.67 \text{ cm}$$
Magnification ($m$): $$m = \frac{h’}{h} = \frac{v}{u}$$ $$h’ = h \times (\frac{16.67}{-25}) = 5 \times (-0.66) = -3.33 \text{ cm}$$
Result: The image is Real, Inverted, 3.33 cm tall, and formed 16.67 cm on the other side.
[Image of convex lens object beyond 2F]
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.
Given:
Focal length ($f$) = -15 cm (Concave)
Image distance ($v$) = -10 cm (Concave lens always forms virtual image on the same side)
Lens Formula: $$\frac{1}{v} – \frac{1}{u} = \frac{1}{f}$$ $$\frac{1}{-10} – \frac{1}{u} = \frac{1}{-15}$$ $$-\frac{1}{u} = -\frac{1}{15} + \frac{1}{10}$$ $$-\frac{1}{u} = \frac{-2 + 3}{30} = \frac{1}{30}$$ $$u = -30 \text{ cm}$$
Result: The object is placed 30 cm from the lens.
[Image of concave lens ray diagram]
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.
Given:
Object distance ($u$) = -10 cm
Focal length ($f$) = +15 cm (Convex)
Mirror Formula: $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$ $$\frac{1}{v} + \frac{1}{-10} = \frac{1}{15}$$ $$\frac{1}{v} = \frac{1}{15} + \frac{1}{10}$$ $$\frac{1}{v} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$$ $$v = +6 \text{ cm}$$
Nature: Since $v$ is positive, the image is formed behind the mirror. It is Virtual and Erect.
Question 13
The magnification produced by a plane mirror is +1. What does this mean?
Answer:
• The sign (+) indicates that the image is Virtual and Erect.
• The value (1) indicates that the size of the image is equal to the size of the object.
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, nature and size of the image.
Given:
$h = 5.0$ cm, $u = -20$ cm, $R = +30$ cm
Focal length ($f$) = $R/2 = +15$ cm
Mirror Formula: $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$ $$\frac{1}{v} – \frac{1}{20} = \frac{1}{15}$$ $$\frac{1}{v} = \frac{1}{15} + \frac{1}{20} = \frac{4 + 3}{60} = \frac{7}{60}$$ $$v = \frac{60}{7} \approx +8.57 \text{ cm}$$
Magnification: $$m = -\frac{v}{u} = -\frac{8.57}{-20} \approx +0.428$$ $$h’ = m \times h = 0.428 \times 5 = 2.14 \text{ cm}$$
Result: Image is Virtual, Erect, diminished (2.14 cm), formed 8.57 cm behind the mirror.
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.
Given:
$h = +7$ cm, $u = -27$ cm, $f = -18$ cm
Mirror Formula: $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$ $$\frac{1}{v} – \frac{1}{27} = \frac{1}{-18}$$ $$\frac{1}{v} = -\frac{1}{18} + \frac{1}{27} = \frac{-3 + 2}{54} = -\frac{1}{54}$$ $$v = -54 \text{ cm}$$
Screen Position: 54 cm in front of the mirror.
Magnification: $$m = -\frac{v}{u} = -\frac{-54}{-27} = -2$$ $$h’ = m \times h = -2 \times 7 = -14 \text{ cm}$$
Result: Image is Real, Inverted, and Enlarged (14 cm tall).
Question 16
Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Given: Power ($P$) = -2.0 D
Formula: $P = \frac{1}{f \text{ (in meters)}}$
Calculation: $$f = \frac{1}{P} = \frac{1}{-2.0} = -0.5 \text{ meters} = -50 \text{ cm}$$
Type: Since focal length is negative, it 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?
Given: Power ($P$) = +1.5 D
Calculation: $$f = \frac{1}{P} = \frac{1}{+1.5} = \frac{10}{15} = \frac{2}{3} \text{ meters}$$ $$f = +0.67 \text{ meters} = +66.7 \text{ cm}$$
Type: Since power/focal length is positive, it is a Converging (Convex) Lens.
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