The power of a lens having refractive index $1.25$ is $+ 3$ diapters. When placed in a liquid, its power is $- 2$ diopters. The refractive index of the liquid is

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$1.2$

$1.4$

$1.5$

$1.6$

answer is C.

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Detailed Solution

$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = 3$

$3 = (1.25 - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) ............ (1)$

$a n d - 2 = (\frac{1.25}{μ} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$ ……………..(2)

$- \frac{3}{2} = \frac{0.25 μ}{1.25 - μ} \Rightarrow - 0.5 μ = 3.75 - 3 μ$ $\Rightarrow μ = 3.75 / 2.5 = 1.5$

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