Question

A glass slab is immersed in water find critical angle at glass water interface

Answer 1

$n_{glass} = 1.50 \\ n_{water} = 1.33 \\ then \\ \theta_{c} = {sin}^{ - 1} ( \frac{n_{water}}{n_{glass}}) \\ \implies\theta_{c} = {sin}^{ - 1} ( \frac{1.33}{1.50}) = {sin}^{ - 1}(0.886666) \\ \implies \boxed{ \theta_{c} =1.09008 \: rad}$
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Answer 2

Explanation:

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