State the approximate value of the critical angle for

(a) glass-air surface

(b) water-air surface.

(a) As we know,

$$\begin{gathered} {}_a{\mu }_g={\displaystyle \frac{1}{\text{sin C}}} \\ =\text{cosec C} \end{gathered}$$aμg = \dfrac{1}{\text {sin C}} \\[0.5em] = \text {cosec C}

Refractive index is

${}_a{\mu }_g={\displaystyle \frac{3}{2}}$aμg = \dfrac{3}{2}

Substituting the values in the formula we get,

$\dfrac{3}{2} = \dfrac{1}{\text {sin C}} \\[1em] \text {sin C} = \dfrac{1}{\dfrac{3}{2}} \\[1em] \text {sin C} = \dfrac{2}{3}$

As,

$\text {sin 42°} = \dfrac {2}{3}$

Hence, critical angle for glass air surface is 42°.

(b) As we know,

${}_a{\mu }_w={\displaystyle \frac{1}{\text{sin C}}}=\text{cosec C}$aμw = \dfrac{1}{\text {sin C}} = \text {cosec C} \\[0.5em]

Refractive index is

${}_a{\mu }_w={\displaystyle \frac{4}{3}}$aμw = \dfrac {4}{3} \\[0.5em]

Substituting the values in the formula we get,

$\dfrac{4}{3} = \dfrac{1}{\text {sin C}} \\[1em] \text {sin C} = \dfrac{1}{\dfrac{4}{3}} \\[1em] \text {sin C} = \dfrac{3}{4} \\[1em] \text {sin 49°} = \dfrac{3}{4} \\[0.5em]$

Hence, critical angle for water air surface is 49°.