Question

In the figure, a ray of light is perpendicular to the face $\mathit{a}\mathit{b}$of a glass prism$\mathbf{(}\mathbf{n}\mathbf{=}\mathbf{1}\mathbf{.52}\mathbf{)}$

.(a) Find the largest value for the angle $\mathit{\varphi }$so that the ray is totally reflected at face ac if the prism is immersed in air, and

(b) Find the largest value for the angle $\mathit{\varphi }$so that the ray is totally reflected at face ac if the prism is immersed in water.

Short answer

1. Largest value for $\varphi$so that the ray is totally reflected at face ac if the prism is immersed in air will be $48.9°$.
2. Largest value for $\varphi$ so that the ray is totally reflected at face ac if the prism is immersed in water will be$29°$ .

Step-by-step solution

Given

$n_1=1.52$

Understanding the concept

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

Formula:

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

(a) Calculate the largest value for   so that the ray is totally reflected at face ac if the prism is immersed in air

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

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Sincetheprism is immersed in air,

$n_2=1$And ${\theta }_2=90°$

$\begin{array}{c}1.52\sin {\theta }_1=1\sin 90°\\ \sin {\theta }_1=\frac{1}{1.52}\\ {\theta }_1={\sin }^{-1}0.65\\ {\theta }_1=41.1°\end{array}$

Since it is a prism,

$\begin{array}{c}{\theta }_1+\varphi =90°\\ \varphi =90-{\theta }_1\\ \varphi =90-41.1°\\ \varphi =48.9°\end{array}$

(b) Calculate the Largest value for  ϕ so that the ray is totally reflected at face ac if the prism is immersed in water

Similarly,

Ifthe prism is immersed in water,

$n_2=1.33$and ${\theta }_2=90°$

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

Since it is a prism,

$\begin{array}{c}{\theta }_1+\varphi =90°\\ \varphi =90-{\theta }_1\\ \varphi =90-61\\ \varphi =29°\end{array}$