Bottom of a glass beaker is made of a thin equi-convex lens having bottom side silver polished as shown in the figure. Water is filled in the beaker up to a height of h=4 m. The image of point object, floating at middle point of beaker at the surface of water coincides with it . Find out the radius of curvature of the lens in meter. $({}_{a}μ_{g} = 3 / 2, {}_{a}μ_{w} = 4 / 3)$

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answer is 5.

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Detailed Solution

Power of equivalent mirror ${P^{'}}_{m} = 2 P_{l e n s} + P_{M}$

$\frac{1}{f_{l e n s}} = (\frac{3 / 2}{4 / 3} - 1) (\frac{1}{R} + \frac{1}{R}) = \frac{1}{8} \times \frac{2}{R} = \frac{1}{4 R}$

And $P_{l e n s} = \frac{1}{4 R}$

$P_{m}' = \frac{1}{- f_{m}^{'}} = 2 \times \frac{1}{4 R} + \frac{2}{R} = \frac{5}{2 R} \Rightarrow f_{m}^{'} = - \frac{2}{5} R$

As image coincides with object hence $h = 2 \times \frac{2}{5} R = 4 m \Rightarrow R = 5 m$

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