Samacheer Kalvi 10th Science Chapter 2 Numerical Problems — Optics with Solutions

Numerical Problems — Chapter 2: Optics

Important Formulas:

Sign Convention:

• Object distance u = always negative (object on left)
• Real image: v = positive; Virtual image: v = negative
• Convex lens: f = positive; Concave lens: f = negative

Problem 1. The refractive index of glass is 1.5. Find the speed of light in glass.

• Refractive index of glass, n = 1.5
• Speed of light in vacuum, c = 3 × 10⁸ m/s
• Speed in glass, v = ?

n = c/v → v = c/n

v = (3 × 10⁸) / 1.5 = 2 × 10⁸ m/s

Problem 2. A ray of light passes from air into water. The angle of incidence is 45°. If the refractive index of water is 1.33, find the angle of refraction.

• Angle of incidence, i = 45°
• Refractive index of water, n = 1.33
• Angle of refraction, r = ?
• (n_air = 1)

By Snell's Law: n_air × sin i = n_water × sin r

1 × sin 45° = 1.33 × sin r

sin r = sin 45° / 1.33 = 0.7071 / 1.33 = 0.5317

r = sin⁻¹(0.5317) = 32.1°

Problem 3. An object is placed 30 cm in front of a convex lens of focal length 10 cm. Find the image distance and magnification.

• Object distance, u = −30 cm (negative by sign convention)
• Focal length, f = +10 cm (convex lens)
• Image distance, v = ?

Using lens formula: 1/f = 1/v − 1/u

1/10 = 1/v − 1/(−30)

1/10 = 1/v + 1/30

1/v = 1/10 − 1/30 = 3/30 − 1/30 = 2/30

v = 15 cm (positive → real image, on other side)

Magnification: m = v/u = 15/(−30) = −0.5

Negative sign means image is inverted. Magnitude 0.5 means image is diminished (smaller).

Problem 4. An object is placed 6 cm in front of a concave lens of focal length 12 cm. Find the image position.

• Object distance, u = −6 cm
• Focal length, f = −12 cm (concave lens → negative)
• Image distance, v = ?

1/f = 1/v − 1/u

1/(−12) = 1/v − 1/(−6)

−1/12 = 1/v + 1/6

1/v = −1/12 − 1/6 = −1/12 − 2/12 = −3/12

v = −4 cm (negative → virtual image, on same side as object)

Magnification: m = v/u = (−4)/(−6) = +0.67

Positive means erect image; 0.67 < 1 means diminished.

Problem 5. The focal length of a convex lens is 25 cm. Find its power.

• Focal length, f = +25 cm = +0.25 m (convert to metres)
• Power, P = ?

P = 1/f = 1/0.25 = +4 D

Problem 6. A person uses a lens of power −2.5 D. What type of lens is it and what is its focal length?

• Power, P = −2.5 D
• Focal length, f = ?
• Type of lens?

f = 1/P = 1/(−2.5) = −0.4 m = −40 cm

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

Problem 7. Two thin lenses of focal lengths 20 cm and 40 cm are placed in contact. Find the combined focal length and power.

• f₁ = 20 cm = 0.20 m → P₁ = 1/0.20 = +5 D
• f₂ = 40 cm = 0.40 m → P₂ = 1/0.40 = +2.5 D

Method 1: Using power

P_total = P₁ + P₂ = 5 + 2.5 = +7.5 D

f_total = 1/P_total = 1/7.5 = 0.133 m ≈ 13.3 cm

Method 2: Using focal length formula

1/f = 1/f₁ + 1/f₂ = 1/20 + 1/40 = 2/40 + 1/40 = 3/40

f = 40/3 = 13.3 cm

Problem 8. Find the critical angle for glass-air interface given that the refractive index of glass is 1.5.

• Refractive index of glass, n = 1.5
• Critical angle, θ_c = ?

sin θ_c = 1/n (for glass-air interface, n_air = 1)

sin θ_c = 1/1.5 = 0.667

θ_c = sin⁻¹(0.667) = 41.8° ≈ 42°

Problem 9. An object 5 cm tall is placed 20 cm in front of a convex lens of focal length 15 cm. Find: (a) Image distance (b) Image height (c) Nature of the image

• Object height, h = +5 cm
• Object distance, u = −20 cm
• Focal length, f = +15 cm

Step 1: Find v (image distance)

1/f = 1/v − 1/u

1/15 = 1/v − 1/(−20) = 1/v + 1/20

1/v = 1/15 − 1/20 = 4/60 − 3/60 = 1/60

v = +60 cm (real image, other side)

Step 2: Find magnification

m = v/u = 60/(−20) = −3

Step 3: Find image height

h' = m × h = −3 × 5 = −15 cm

Negative sign means image is inverted.

• Image distance = 60 cm (on other side)
• Image height = 15 cm (inverted, so 3× larger)
• Nature: Real, inverted, magnified

Quick Summary Table for Image Formation by Convex Lens

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