Complete Question–Answer Solutions (All Exercises)

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Exercise 10.1 (Page 168)

Q1. Define the principal focus of a concave mirror.

Answer: The principal focus of a concave mirror is the point on its principal axis where a beam of light rays travelling parallel to the axis actually meets after being reflected by the mirror. Because a concave mirror curves inward, it converges parallel rays to this single point.

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

Answer: For any spherical mirror, the radius of curvature is twice the focal length: R = 2f.

So f = R/2 = 20/2 = 10 cm. The focal length of the mirror is 10 cm.

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

Answer: A concave mirror can give an erect, enlarged (magnified) image — but only when the object is placed close to the mirror, between the pole and the principal focus.

(A convex mirror, by contrast, always gives an erect but diminished image, never enlarged.)

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

Answer: A convex mirror always forms an erect, virtual, and diminished image of objects behind the vehicle. Because the image is smaller than the actual object, a convex mirror can fit a much wider field of view into the same mirror size compared to a plane or concave mirror.

This lets a driver see a larger stretch of the road behind them at a glance, which is why convex mirrors are preferred for rear-view use.

Exercise 10.2 (Page 171)

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

Answer: Using R = 2f: f = R/2 = 32/2 = 16 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?

Answer: Object distance, u = −10 cm. Since the image is real, magnification is negative: m = −3.

Using m = −v/u: −3 = −v/(−10), which gives v = −30 cm.

The negative sign shows the image is real and inverted, formed on the same (reflecting) side of the mirror, 30 cm in front of it.

Refraction of Light (Page 176)

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?

Answer: It bends towards the normal. Water is optically denser than air, so light slows down as it enters water. Whenever light passes from a rarer medium into a denser one at an angle, this drop in speed makes the ray bend closer to the normal at the point of incidence.

Q2. Light enters from air to glass, having a refractive index 1.50. What is the speed of light in the glass? (Speed of light in vacuum = 3 × 10⁸ m/s)

Answer: Refractive index n = speed of light in vacuum ÷ speed of light in the medium, so speed in medium v = c/n.

v = (3 × 10⁸) / 1.50 = 2 × 10⁸ m/s. Light travels at 2 × 10⁸ m/s inside the glass.

Q3. Find out, from the standard refractive-index table, the medium with the highest optical density and the one with the lowest.

Answer: Air has the lowest optical density (refractive index ≈ 1.0003), and diamond has the highest optical density (refractive index = 2.42).

Optical density rises with refractive index — a denser medium slows light down more and bends it more sharply, which is exactly the behaviour diamond is famous for.

Q4. Among kerosene, turpentine and water, in which does light travel fastest?

Answer: Speed of light in a medium is inversely proportional to its refractive index — a lower refractive index means light travels faster.

Refractive indices: water ≈ 1.33, kerosene ≈ 1.44, turpentine ≈ 1.47. Water has the lowest value, so light travels fastest in water among these three.

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

Answer: It means light travels 2.42 times slower in diamond than it does in vacuum — its speed in diamond is (1/2.42) of its speed in vacuum.

This large refractive index is also why diamond bends light so strongly, giving it its characteristic sparkle.

Lenses (Page 184)

Q1. Define 1 dioptre of power of a lens.

Answer: One dioptre (1 D) is the power of a lens whose focal length is exactly 1 metre. Power and focal length are related by P = 1/f (f in metres), so a shorter focal length always means a higher-power lens.

Q2. A convex lens forms a real and inverted image of a needle at 50 cm from it, equal in size to the object. Where is the needle placed, and what is the power of the lens?

Answer: A real, inverted image that is the same size as the object forms only when the object sits at 2F (twice the focal length) — and in that case the image also forms at 2F on the other side.

So the object (needle) is placed 50 cm in front of the lens: u = −50 cm, v = +50 cm.

Using the lens formula 1/v − 1/u = 1/f: 1/50 − (−1/50) = 1/f → 2/50 = 1/f → f = 25 cm = 0.25 m.

Power P = 1/f = 1/0.25 = 4 D.

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

Answer: A concave lens has a negative focal length by convention: f = −2 m.

P = 1/f = 1/(−2) = −0.5 D.

Exercise (Pages 185–186)

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

Answer: (d) Clay. A lens must be transparent so it can transmit and bend light; clay is opaque and blocks light entirely, so it cannot function as a lens.

Q2. The image formed by a concave mirror is virtual, erect and larger than the object. Where is the object placed? (a) Between focus and centre of curvature (b) At centre of curvature (c) Beyond centre of curvature (d) Between pole and focus

Answer: (d) Between the pole of the mirror and its principal focus. Only in this range does a concave mirror produce a virtual, erect, magnified image.

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

(a) At focus (b) At twice the focal length (c) At infinity (d) Between optical centre and focus

Answer: (b) At twice the focal length (2F). At this position, a convex lens forms a real, inverted image that is also located at 2F on the other side and is exactly the same size as the object.

Q4. A spherical mirror and a thin spherical lens both have a focal length of −15 cm. What are they likely to be?

(a) Both concave (b) Both convex (c) Mirror concave, lens convex (d) Mirror convex, lens concave

Answer: (a) Both are likely concave. By the standard sign convention, a concave mirror has a negative focal length, and so does a concave (diverging) lens — a negative f for both instruments points to both being concave.

Q5. No matter how far you stand from a mirror, your image appears erect. What kind of mirror is it?

(a) Plane (b) Concave (c) Convex (d) Either plane or convex

Answer: (d) Either plane or convex. Both types always produce an erect image of a real object, regardless of the object’s distance — unlike a concave mirror, whose image can flip from erect to inverted depending on where the object is placed.

Q6. Which lens would you prefer for reading small letters in a dictionary?

(a) Convex, f = 50 cm (b) Concave, f = 50 cm (c) Convex, f = 5 cm (d) Concave, f = 5 cm

Answer: (c) A convex lens of focal length 5 cm. A short-focal-length convex lens gives high magnifying power (a simple magnifying glass), which is exactly what’s needed to read small print clearly.

Q7. We want an erect image using a concave mirror of focal length 15 cm. What object-distance range is needed? What is the nature and size of the image?

Answer: The object must be placed between the pole and the focus — that is, anywhere from 0 cm to 15 cm from the mirror.

In this range, a concave mirror always produces a virtual, erect image that is larger than the object.

Q8. Name the type of mirror used for:

(a) car headlights (b) vehicle side/rear-view mirrors (c) solar furnaces, with reasons.

Answer: (a) Concave mirror — placing the bulb at the mirror’s principal focus makes it reflect the diverging light as a strong, parallel beam, which is ideal for headlights.

(b) Convex mirror — it always gives an erect, diminished image with a wide field of view, letting drivers see more of the traffic behind them.

(c) Concave mirror — it converges parallel rays of sunlight to its principal focus, concentrating enough energy at one point to generate very high temperatures.

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

Answer: Yes — the lens still forms the complete image of the object, because every uncovered point on the remaining half still receives light from every point of the object and helps form its share of the full image.

The only change is that the image becomes dimmer (lower intensity/brightness), since less light is now reaching the screen — this can be confirmed by projecting the image of a distant object onto a screen with and without the covering.

Q10. An object 5 cm tall is held 25 cm from a converging (convex) lens of focal length 10 cm. Find the position, size and nature of the image.

Answer: Given: h₀ = 5 cm, u = −25 cm, f = +10 cm (convex lens).

Lens formula: 1/v − 1/u = 1/f → 1/v = 1/10 + (1/−25 flipped) → 1/v = 1/10 − 1/25 = (5 − 2)/50 = 3/50 → v = 50/3 ≈ 16.7 cm.

Magnification m = v/u = 16.7/(−25) ≈ −0.67. Image height = m × h₀ ≈ −3.3 cm.

The image forms about 16.7 cm behind the lens, is inverted, real, and diminished, with a height of about 3.3 cm.

Q11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed?

Answer: Given: f = −15 cm (concave), v = −10 cm (concave lenses always form a virtual image on the same side as the object).

Lens formula: 1/v − 1/u = 1/f → −1/10 − 1/u = −1/15 → 1/u = −1/10 + 1/15 = (−3 + 2)/30 = −1/30 → u = −30 cm.

The object is placed 30 cm in front of the (concave) lens.

Q12. An object is placed 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Answer: Given: f = +15 cm (convex), u = −10 cm.

Mirror formula: 1/v + 1/u = 1/f → 1/v = 1/15 − (−1/10) = 1/15 + 1/10 = 5/30 = 1/6 → v = 6 cm.

The image forms 6 cm behind the mirror. Since v is positive, the image is virtual; magnification m = −v/u = −6/(−10) = 0.6 (positive and less than 1), so the image is erect and diminished.

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

Answer: The positive sign shows the image is virtual and erect (upright, same orientation as the object). A magnification of exactly 1 means the image is exactly the same size as the object — which is always true for a plane mirror.

Q14. An object 5 cm tall is placed 20 cm from a convex mirror of radius of curvature 30 cm. Find the position, nature and size of the image.

Answer: Given: u = −20 cm, R = 30 cm → f = R/2 = +15 cm (convex), h₀ = 5 cm.

Mirror formula: 1/v = 1/f − 1/u = 1/15 − (−1/20) = 1/15 + 1/20 = 7/60 → v ≈ 8.6 cm.

m = −v/u = −8.6/(−20) ≈ 0.43. Image height = m × h₀ ≈ 2.1 cm.

The image is virtual, erect, and diminished, formed about 8.6 cm behind the mirror, roughly 2.1 cm tall.

Q15. An object 7.0 cm tall is placed 27 cm in front of a concave mirror of focal length 18 cm. Where should a screen be placed for a sharp image? Find its size and nature.

Answer: Given: u = −27 cm, f = −18 cm (concave), h₀ = 7 cm.

Mirror formula: 1/v = 1/f − 1/u = −1/18 − (−1/27) = −1/18 + 1/27 = (−3 + 2)/54 = −1/54 → v = −54 cm.

The screen should be placed 54 cm in front of the mirror to catch a sharp, real image.

Magnification m = −v/u = −(−54)/(−27) = −2. Image height = m × h₀ = −14 cm, so the image is real, inverted, and enlarged, about 14 cm tall.

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

Answer: f = 1/P = 1/(−2) = −0.5 m.

A negative focal length always indicates a concave (diverging) lens.

Q17. A doctor prescribes a corrective lens of power +1.5 D. Find its focal length. Is it diverging or converging?

Answer: f = 1/P = 1/1.5 ≈ 0.67 m.

A positive focal length (and positive power) means the lens is convex — it is a converging lens.

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Quick Chapter Summary

• Reflection follows two laws: the angle of incidence equals the angle of reflection, and the incident ray, reflected ray, and normal all lie in the same plane.

• Concave mirrors can form either real or virtual images depending on object distance; convex mirrors always form virtual, erect, diminished images.

• Refraction is the bending of light as it passes between media of different optical density, caused by a change in the speed of light.

• Refractive index measures how strongly a medium bends light; a higher value means slower light and stronger bending.

• Convex lenses converge light and can form real or virtual images; concave lenses diverge light and always form virtual, erect, diminished images.

• Power of a lens (in dioptres) is the reciprocal of its focal length in metres — shorter focal length means higher power.

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