91Ó°ÊÓ

$n_1=1.52$

Using Snell’s law, we can determine the critical angle. Using this, we can find the largest value for $\mathrm{ϕ}$if the prism is immersed in air and in water.

Formula:

$n_1\sin {\mathrm{θ}}_1=n_2\sin {\mathrm{θ}}_2$

As the ray is travelling perpendicular to the surface, it does not refract on the initial face of bc.

Due to this,thetotal internal reflection occurs at the critical angle.

Snell’s law gives that

$n_1\sin {\mathrm{θ}}_1=n_2\sin {\mathrm{θ}}_2$

Sincetheprism is immersed in air,

$n_2=1$And ${\mathrm{θ}}_2=90°$

$\begin{array}{c}1.52\sin {\mathrm{θ}}_1=1\sin 90°\\ \sin {\mathrm{θ}}_1=\frac{1}{1.52}\\ {\mathrm{θ}}_1={\sin }^{−1}0.65\\ {\mathrm{θ}}_1=41.1°\end{array}$

Since it is a prism,

$\begin{array}{c}{\mathrm{θ}}_1+\mathrm{ϕ}=90°\\ \mathrm{ϕ}=90−{\mathrm{θ}}_1\\ \mathrm{ϕ}=90−41.1°\\ \mathrm{ϕ}=48.9°\end{array}$

Similarly,

Ifthe prism is immersed in water,

$n_2=1.33$and ${\mathrm{θ}}_2=90°$

$\begin{array}{c}1.52\sin {\mathrm{θ}}_1=1\sin 90°\\ \sin {\mathrm{θ}}_1=\frac{1.33}{1.52}\\ {\mathrm{θ}}_1={\sin }^{−1}0.875\\ {\mathrm{θ}}_1=61.0°\end{array}$

Since it is a prism,

$\begin{array}{c}{\mathrm{θ}}_1+\mathrm{ϕ}=90°\\ \mathrm{ϕ}=90−{\mathrm{θ}}_1\\ \mathrm{ϕ}=90−61\\ \mathrm{ϕ}=29°\end{array}$