A thin symmetric biconvex lens is made of glass of refractive index 1.5. The magnitudes of the radii of curvature of both surfaces are 20 cm. The left surface is silvered externally, so the light passes through the lens, reflects and passes through the lens again. At what distance from the lens should an object be placed so that the final image is formed at the position of the object?
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Question

A thin symmetric biconvex lens is made of glass of refractive index 1.5. The magnitudes of the radii of curvature of both surfaces are 20 cm. The left surface is silvered externally, so the light passes through the lens, reflects and passes through the lens again. At what distance from the lens should an object be placed so that the final image is formed at the position of the object?

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G1 Physics · Accepted Answer

For the symmetric biconvex lens, 1 ÷ f(L) = (μ-1)[1 ÷ R₁-1 ÷ R₂]. With μ = 1.5, R₁ = +20 cm, R₂ = -20 cm, we get 1 ÷ f(L) = 0.5[1 ÷ 20+1 ÷ 20] = 1 ÷ 20. Thus, f(L) = 20 cm. The silvered surface acts as a concave mirror of radius 20 cm, so f(M) = -R ÷ 2 = -10 cm. Because light passes through the lens twice, the lens power contributes twice. The equivalent power is therefore 1 ÷ f(eq) = 1 ÷ f(M) - 2 ÷ f(L). Hence, 1 ÷ f(eq) = -1 ÷ 10 - 2 ÷ 20 = -1 ÷ 10 - 1 ÷ 10 = -1 ÷ 5. Therefore, f(eq) = -5 cm.…