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The bottom of glass beaker is made of a thin equiconvex lens having bottom side silvered. Now the water is filled in the beaker upto the height 4m. The image of the point object floating at the middle point of beaker at the surface of water coincides with it. Then the radius of curvature of lens (in m) (refractive index of glass $μ = 1.5$ , refractive index of water $μ = 1.33$ )

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answer is 5.

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Detailed Solution

Power of equivalent mirror is $p_{m}^{1} = 2 p_{l} + p_{m}$

$p_{l} = \frac{1}{f_{l}} = (\frac{3 / 2}{4 / 3} - 1) (\frac{1}{R} + \frac{1}{R}) = \frac{1}{4 R}$

$\frac{- 1}{f_{m}^{1}} = 2 (\frac{1}{4 R}) + \frac{2}{R} = \frac{5}{2 R}$

$f_{m}^{1} = - \frac{2 R}{5}$

$h = 2 |f m^{1}|$

$4 = 2 (\frac{2}{5} R)$

$R = 5 m$

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