A double convex lens of focal length 30 cm is made of glass. When it is immersed in a liquid of refractive index 1.4, the focal length is found to be 126 cm. The critical angle between glass and the liquid is

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$\sin^{- 1} (\frac{4}{5})$

$\sin^{- 1} (\frac{7}{13})$

$\sin^{- 1} (\frac{7}{8})$

$\sin^{- 1} (\frac{3}{4})$

answer is D.

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Detailed Solution

$f = 30 cm, μ_{l} = 1.4, f^{l} = 12$

$\frac{f^{l}}{f} = \frac{(μ_{g} - 1)}{(\frac{μ_{g}}{μ_{l}} - 1)}$

$\Rightarrow μ_{g} = \frac{8}{5}$

$_{l} μ_{g} = \frac{8}{5} \times \frac{5}{7} = \frac{8}{7}$

$c = \sin^{- 1} (\frac{7}{8})$

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