Refraction of Light at spherical surfaces: Lenses
ICSE · Class 12 · Physics
Most important questions from Refraction of Light at spherical surfaces: Lenses for ICSE Class 12 Physics board exam 2026. MCQs, short answer, and long answer questions with marks.
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Sample Questions
A glass sphere (n = 1.5) of radius 10 cm has a small air bubble at its centre. An observer looks at the bubble from outside along a diameter. Where does the bubble appear to be?
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At the centre of the sphere
Step 1: The bubble is at the centre, 10 cm from each surface. For refraction at the nearer surface (glass to air), object is in glass. Use formula: n1/v - n2/u = (n1 - n2)/R. Here light goes from glass (n2 = 1.5) to air (n1 = 1). Step 2: Object distance u = –10 cm (bubble at centre, 10 cm from surface, measured into glass). R = –10 cm (concave surface as seen from inside going outward; the surface curves away). Step 3: 1/v – 1.5/(–10) = (1 – 1.5)/(–10) → 1/v + 0.15 = 0.05 → 1/v = –0.10 → v = –10 cm. So image appears 10 cm inside glass from the surface — same as object position! Step 4: Using c
A convex lens of focal length 20 cm produces a real image that is 4 times the size of the object. If the object is 5 cm tall, what is the distance between the object and its image?
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125 cm
Step 1: For a real image 4 times the size, magnification m = –4 (negative for real, inverted). So v/u = –4 → v = –4u. Step 2: Using lens formula: 1/v – 1/u = 1/f → 1/(–4u) – 1/u = 1/20 → –1/(4u) – 1/u = 1/20 → (–1 – 4)/(4u) = 1/20 → –5/(4u) = 1/20 → u = –5×20/4 = –25 cm. Step 3: v = –4u = –4×(–25) = +100 cm. So image is 100 cm on the other side. Step 4: Distance between object and image = |v| + |u| = 100 + 25 = 125 cm. Wait — but option 100 cm is marked correct. If the question asks for image distance alone (not object-to-image), then v = 100 cm. The distance between object and image = u + v =
A plano-convex lens (n = 1.5, R = 10 cm) is placed on a horizontal plane mirror. A pin placed 40 cm above the lens has its image coinciding with itself. What is the focal length of the lens system (lens + mirror)?
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40 cm
Step 1: When a pin's image coincides with itself, the light rays after passing through the lens reflect from the plane mirror and retrace their path. This means the rays from the lens fall normally on the mirror — i.e., they emerge as parallel rays from the lens (after refraction). Step 2: For rays to emerge parallel from the lens, the object must be at the focal point of the lens. So f_lens = 40 cm (the pin-to-lens distance). Step 3: Verify using lens maker's formula for plano-convex: 1/f = (n–1)(1/R1 – 1/R2). With curved surface up: R1 = +10 cm, R2 = ∞ (plane). 1/f = (1.5–1)(1/10 – 0) = 0.5/
In the displacement method of finding focal length of a convex lens, the distance between the object and screen is D = 100 cm, and the two positions of the lens give image sizes I1 = 9 cm and I2 = 1 cm. What is the size of the object?
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3 cm
Step 1: In the displacement method, the object size is given by: Object size = √(I1 × I2), where I1 and I2 are the sizes of the two images formed in the two positions of the lens. Step 2: Object size = √(9 × 1) = √9 = 3 cm. Step 3: This formula comes from the property that m1 × m2 = 1 (the product of the two magnifications equals 1 when the object and screen are fixed). Step 4: So m1 = I1/O and m2 = I2/O, with m1 × m2 = 1 → (I1/O)(I2/O) = 1 → O² = I1 × I2 → O = √(I1 × I2) = √9 = 3 cm. Option 5 cm is wrong (simple average), option 4 cm results from (I1+I2)/2 approach — both are common errors. T
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