Critical angle at air glass interface is θ1 and that at water air interface is θ2. If μg=32 and μw=43, then critical angle at water-glass interface is

θ1=sin-11μg=sin-123θ2=sin⁡−11μw=sin⁡−134∴ θ=sin−1μwμg=sin⁡−14323=sin⁡−189>θ2

θ1=sin-11μg=sin-123θ2=sin⁡−11μw=sin⁡−134∴ θ=sin−1μwμg=sin⁡−14323=sin⁡−189>θ2

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Critical angle at air glass interface is $θ_{1}$ and that at water air interface is $θ_{2}$ . If $μ g = \frac{3}{2} and μ_{w} = \frac{4}{3}$ , then critical angle at water-glass interface is

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less than $θ_{1}$

less than $θ_{2}$

between $θ_{1}$ and $θ_{2}$

greater than $θ_{2}$

answer is B.

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$θ_{1} = \sin^{- 1} \frac{1}{μ_{g}} = \sin^{- 1} \frac{2}{3}$

$θ_{2} = \sin^{- 1} \frac{1}{μ_{w}} = \sin^{- 1} \frac{3}{4}$

$∴ θ = \sin^{- 1} \frac{μ_{w}}{μ_{g}} = \sin^{- 1} \frac{4}{3} \frac{2}{3} = \sin^{- 1} \frac{8}{9} > θ_{2}$

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