A lens is placed between a source of light and a wall. It forms images of area A₁ and A₂ on the wall, for its two different positions. The area of the source of light is (a) (A₁ - A₂)/(2) (b) 1/A₁ + 1/A₂ (c) √(A₁ A₂) (d) (A₁ + A₂)/(2)

Answer: (c). In the displacement method the two magnifications are reciprocal: m₁ m₂ = 1. Linear size: O = √(I₁ I₂) Areas scale as size squared: A₀ = √(A₁ A₂) So the answer is (c) √(A₁ A₂).

A lens is placed between a source of light and a wall. It forms… | NEET 1995 (AIPMT)

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Spherical Surfaces & Thin Lenses

A lens is placed between a source of light and a wall. It forms images of area $A_1$ and $A_2$ on the wall, for its two different positions. The area of the source of light is

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Answer: (c)

In the displacement method the two magnifications are reciprocal: $m_1 m_2 = 1$ .
Linear size: $O = \sqrt{I_1 I_2}$
Areas scale as size squared: $A_0 = \sqrt{A_1 A_2}$
So the answer is (c) $\sqrt{A_1 A_2}$ .

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