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Q.

A thin lens made of glass of refractive index 1.5 has a front surface + 11 D power and back surface – 6 D. If this lens is submerged in a liquid of refractive index 1.6, the resulting power of the lens is

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a

– 0.625 D

b

+ 0.625 D

c

– 0.5 D

d

+ 0.5 D

answer is C.

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Detailed Solution

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$Total power P = P_{1} + P_{2} = 11 - 6 = 5 D Also \frac{f_{l}}{f_{a}} = \frac{(_{a} μ_{g} - 1)}{(μ_{g} - 1)} \Rightarrow \frac{P_{a}}{P_{l}} = \frac{(_{a} μ_{g} - 1)}{(μ_{g} - 1)} \Rightarrow \frac{5}{P_{l}} = \frac{(1 .5 - 1)}{(1 .5 / 1 .6 - 1)} \Rightarrow P_{l} = - 0 .625 D$

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