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Q.

A convex lens of refractive index 1.5 and focal length 20 cm in air is immersed in water.The change in focal length of lens will be ____cm (Refractive index of water $= \frac{4}{3}$ )

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answer is 60.

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Detailed Solution

$\frac{\frac{1}{F a i r} = (μ_{g} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})}{\frac{1}{F w} = (\frac{μ_{g}}{μ_{l s}} - 1) (\frac{1}{R_{1}} - 1 / R_{2})} \frac{F w}{20} = \frac{1.5 - 1}{(\frac{3 / 2}{4 / 3} - 1)} \Rightarrow F_{w} = 20 \times \frac{1 / 2}{1 / 8} = 80 c m$

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