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

Subject: Science
Class: 10
Board: NCERT / CBSE

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

Answer: Clay.

Reason: A lens must be transparent so light can pass through and refract; clay is opaque and does not transmit light.

Q2. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should the object be?

Answer: Between the pole (P) of the mirror and its principal focus (F).

Reason: For a concave mirror, when the object is between P and F, reflected rays diverge and the extensions behind the mirror meet at a virtual, erect and magnified image.

Q3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?

Answer: At twice the focal length (at 2f) on the object side.

Reason: For a convex lens, when object is at 2f, the image forms at 2f on the other side, real, inverted and same size.

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:

Answer: Both are concave.

Reason: Negative focal length indicates concave mirror or diverging lens in the usual sign convention.

Q5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be:

Answer: Either a plane mirror or a convex mirror.

Reason: Plane and convex mirrors always produce erect images (plane: virtual, same size; convex: virtual, diminished).

Q6. Which lens would you prefer to use while reading small letters in a dictionary?

Answer: A convex lens of short focal length (e.g., 5 cm).

Reason: A short focal length convex lens (magnifying glass) gives a larger magnified virtual image useful for reading small letters.

Q7. We wish to obtain an erect image of an object using a concave mirror of focal length 15 cm. What should be the object distance range? Nature and size of the image?

Answer: Place the object between the pole (P) and the principal focus (F) — i.e., object distance < 15 cm from the mirror.
Image will be virtual, erect and enlarged.
Sketch hint: draw concave mirror, object between P and F; draw one ray parallel to axis (reflects through F) and one ray toward centre (reflects back), extend reflected rays backwards to locate image behind mirror.

Q8. Name the type of mirror used in: (a) headlights of a car, (b) side/rear-view mirror of a vehicle, (c) solar furnace. Give reason.

(a) Headlights: Concave mirror — it produces parallel beams of light by reflecting light from the bulb placed at or near focus.
(b) Side/rear-view mirror: Convex mirror — it gives a wider field of view and erect, diminished images.
(c) Solar furnace: Concave mirror — focuses parallel sunlight to a point to produce high temperature.

Q9. One-half of a convex lens is covered with black paper. Will the lens produce a complete image of the object?

Answer: Yes — a complete image will be formed, but it will be dimmer.

Reason: Every part of the lens contributes to the full image; covering half reduces the amount of light reaching the image (less brightness) but does not produce a half image.

Q10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Find position, size and nature of the image. (Show ray diagram)

Solution:

Use the lens formula: 1/f = 1/v + 1/u

Given: f = 10 cm, u = 25 cm (object distance). Compute 1/v = 1/f − 1/u = 1/10 − 1/25 = 0.1 − 0.04 = 0.06

So v = 1/0.06 = 50/3 cm ≈ 16.67 cm.

Magnification m = v/u = (50/3)/25 = 2/3 ⇒ Image size = object × m = 5 × 2/3 = 10/3 cm ≈ 3.33 cm.

Answer: Image is real, inverted, formed at ≈ 16.7 cm on the other side of lens, size ≈ 3.33 cm.

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?

Solution:

Sign convention (cartesian): For concave lens, f = −15 cm; virtual image at v = −10 cm (on same side as object). Lens formula: 1/f = 1/v + 1/u.

Substitute: −1/15 = −1/10 + 1/u ⇒ 1/u = −1/15 + 1/10 = (−2 + 3)/30 = 1/30 ⇒ u = 30 cm.

Answer: Object is 30 cm in front of the lens. (Image is virtual, erect and smaller.)

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 image.

Solution (concise):

For convex mirror, image is virtual, erect and diminished behind the mirror. Use formula (magnitude): v = u f / (u + f).

Compute: v = (10 × 15)/(10 + 15) = 150/25 = 6 cm (behind the mirror).

Answer: Virtual, erect, diminished image located 6 cm behind the mirror.

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

Answer: Magnification +1 means the image size equals the object size and the image is erect (the + sign indicates upright orientation). For a plane mirror, image distance = object distance and image is virtual and same size.

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.

Solution:

Radius R = 30 cm ⇒ f = R/2 = 15 cm. Using v = u f / (u + f): v = (20 × 15)/(20 + 15) = 300/35 = 60/7 ≈ 8.571 cm (behind mirror).

Magnification m = v/u = (60/7)/20 = 3/7 ≈ 0.4286 ⇒ Image size = 5 × 3/7 = 15/7 ≈ 2.1429 cm.

Answer: Image at ≈ 8.57 cm behind mirror, virtual, erect, diminished, size ≈ 2.14 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 should a screen be placed to get a sharp focused image? Find size and nature of the image.

Solution:

Use v = u f / (u – f) for concave mirror (object beyond f). Here u = 27, f = 18 ⇒ u − f = 9.

v = (27 × 18)/9 = 486/9 = 54 cm. So screen at 54 cm in front of mirror.

Magnification m = v/u = 54/27 = 2 ⇒ Image size = 7 × 2 = 14 cm.

Answer: Screen at 54 cm; image is real, inverted, magnified, size = 14 cm.

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

Solution: Power P = 1/f (in metres) ⇒ f = 1/P = 1/(−2.0) = −0.5 m = −50 cm.

Answer: f = −0.5 m (−50 cm). Negative focal length ⇒ diverging (concave) lens.

Q17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length. Is the lens diverging or converging?

Solution: f = 1/P = 1/1.5 = 2/3 m ≈ 0.6667 m = 66.67 cm.

Answer: f ≈ 0.667 m (66.7 cm). Positive power ⇒ converging (convex) lens.

Latest Posts

Frequently Asked Questions (FAQs)

Q1. What is the difference between reflection and refraction of light?

Reflection is the bouncing back of light when it strikes a smooth surface like a mirror. Refraction is the bending of light when it passes from one medium to another, such as from air to glass.

Q2. What are the laws of reflection of light?

(1) The incident ray, reflected ray, and the normal at the point of incidence all lie in the same plane.
(2) The angle of incidence (i) is equal to the angle of reflection (r).

Q3. What is the mirror formula for spherical mirrors?

The mirror formula is 1/f = 1/v + 1/u, where f = focal length, v = image distance, and u = object distance.

Q4. How does the refractive index affect the bending of light?

When light enters a medium with a higher refractive index, it slows down and bends towards the normal. When it enters a medium with a lower refractive index, it bends away from the normal.

Q5. What are the applications of reflection and refraction in daily life?

Reflection: Used in mirrors, car headlights, solar cookers, and reflecting telescopes.
Refraction: Used in lenses, eyeglasses, microscopes, cameras, and optical fibers.

Latest Posts

Frequently Asked Questions (FAQs)

Q1. What is the difference between reflection and refraction of light?

Q2. What are the laws of reflection of light?

Q3. What is the mirror formula for spherical mirrors?

Q4. How does the refractive index affect the bending of light?

Q5. What are the applications of reflection and refraction in daily life?

Leave a Comment Cancel reply