Reflection and Refraction of Light — Practice Quiz
NIOS · Class 12 · Physics
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Quick Quiz: Reflection and Refraction of Light
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An object is placed 20 cm in front of a concave mirror of focal length 15 cm. What is the image distance, and what is the nature of the image?
A ray of light passes from water (μ = 4/3) to glass (μ = 3/2). If the angle of incidence in water is 30°, what is the angle of refraction in glass?
The refractive index of diamond is 2.42. A ray of light inside diamond strikes a diamond-air interface. What is the critical angle, and what happens if light strikes at 30°?
A biconvex lens has radii of curvature R₁ = +20 cm and R₂ = –30 cm, and is made of glass with μ = 1.5. What is the focal length of this lens?
Sample Questions
A convex lens of focal length 40 cm is kept in contact with a concave lens of focal length 25 cm. What is the power of the combination?
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–1.5 D
Step 1: Power of convex lens: P₁ = 1/f₁ = 1/0.40 m = +2.5 D. Step 2: Power of concave lens: P₂ = 1/f₂ = 1/(–0.25 m) = –4 D. Step 3: Total power P = P₁ + P₂ = 2.5 + (–4) = –1.5 D. Step 4: Negative power confirms the combination acts as a diverging (concave) lens system. Step 5: Note that focal lengths must be converted to metres before calculating power. Option A has wrong sign. Option C uses 1/F where F was wrongly calculated.
A point object is placed in air at a distance of 20 cm from a convex spherical glass surface (μ = 1.5, R = 10 cm). Using the refraction formula, where is the image formed?
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v = +60 cm inside glass
Step 1: Formula for refraction at spherical surface: μ₂/v – μ₁/u = (μ₂ – μ₁)/R. Step 2: Here μ₁ = 1 (air), μ₂ = 1.5 (glass), u = –20 cm (object on left), R = +10 cm (convex surface). Step 3: 1.5/v – 1/(–20) = (1.5 – 1)/10 → 1.5/v + 1/20 = 0.5/10 = 0.05. Step 4: 1.5/v = 0.05 – 0.05 = 0.05 – 0.05 = 0 ... let us redo: 1.5/v = 0.05 – 0.05 = 0.05 – 1/20 = 0.05 – 0.05 = 0. Re-checking: 1.5/v = 1/20 → wait: 1.5/v = (0.5/10) – (1/20) = 0.05 – 0.05 = wait—1/20=0.05. So 1.5/v = 0.05 – 0.05 = 0? That gives ∞. Recalculate: 1.5/v = (μ₂–μ₁)/R – μ₁/u correction: μ₂/v = (μ₂–μ₁)/R + μ₁/u = 0.5/10 + 1/(–20) = 0
A convex lens (μ_glass = 1.5) has focal length 20 cm in air. It is immersed in a liquid of refractive index 1.65. What happens to the lens?
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It behaves as a concave lens with focal length –110 cm
Step 1: When a lens is immersed in a medium, the effective refractive index is μ_eff = μ_glass/μ_liquid = 1.5/1.65 = 10/11 ≈ 0.909. Step 2: Since μ_eff < 1, the term (μ_eff – 1) becomes negative (= –1/11). Step 3: Using lens maker's formula, 1/f_liquid = (μ_eff – 1)(1/R₁ – 1/R₂). The sign of 1/f flips, so the lens now diverges light. Step 4: In air, 1/20 = (0.5)(1/R₁ – 1/R₂) → (1/R₁ – 1/R₂) = 1/10. Step 5: In liquid, 1/f_liq = (–1/11)(1/10) = –1/110 → f_liq = –110 cm. Negative focal length confirms concave (diverging) behavior. Option D is a misconception — focal length depends on the surround
Using Newton's formula for a lens, if the object is placed 5 cm beyond the first focal point and the image is formed 45 cm beyond the second focal point, what is the focal length of the lens?
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f = 15 cm
Step 1: Newton's formula states: x₁ × x₂ = f², where x₁ = distance of object from first focal point, x₂ = distance of image from second focal point. Step 2: Given x₁ = 5 cm, x₂ = 45 cm. Step 3: f² = x₁ × x₂ = 5 × 45 = 225 cm². Step 4: f = √225 = 15 cm. Step 5: This formula is very useful experimentally as it directly gives f from easily measurable distances. Note that both option A and C are essentially the same answer — f = 15 cm. Option B and D result from incorrect arithmetic (e.g., using f = (x₁+x₂)/2 = 25 cm, which is wrong).
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