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Q.

Power of a convex lens is + 5D ( $μ_{g} = 1.5$ ). When this lens is immersed in a liquid of refractive index $μ$ , it acts like a divergent lens of focal length 100 cm. Then refractive index of the liquid will be

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a

7/4

b

5/3

c

6/5

d

4/3

answer is A.

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Detailed Solution

$\frac{P_{a}}{P_{l}} = \frac{μ_{g} - 1}{\frac{μ_{g} - μ_{l}}{μ_{l}}} = \frac{5}{- 1} = - 5 \frac{P_{a}}{P_{l}} = \frac{\frac{3}{2} - 1}{\frac{(3 / 2)}{μ_{l}} - 1} = - 5 \frac{1}{2} = - 5 (\frac{3}{2 μ_{l}} - 1) \frac{1}{2} = - \frac{15}{2 μ_{l}} + 5; \frac{15}{2 μ_{l}} = \frac{9}{2}; μ_{l} = \frac{15}{9} = \frac{5}{3}$

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