Question

There’s one angle of incidence $\beta$onto a prism for which the light inside an isosceles prism travels parallel to the base and emerges at angle $\beta$.

a. Find an expression for $\beta$in terms of the prism’s apex angle $\alpha$and index of refraction $n$.

b. A laboratory measurement finds that $\beta$= 52.2° for a prism shaped like an equilateral triangle. What is the prism’s index of refraction?

Short answer

a) $\beta ={\sin }^{-1}\left[n\sin \frac{\alpha }{2}\right]$

b) Refractive index is equal to 1.58

Step-by-step solution

Step 1. Given information is :a)Angle of incidence of light ray is βPrism's apex angle is αRefractive index of air n1 = 1.0Refractive index of Prism n2 = nb) β=52.2° for a prism that is shaped as an equilateral triangle.

We need to find out :

a) Expression of $\beta$in terms of $\alpha andn$

b) Refractive index of prism with$\beta =52.2°$

Step 2. Part a) Using Snell's Law for Air-Prism Interface

$$\begin{gathered} n_1\sin \beta =n_2\sin \left(90°-\theta \right) \\ Puttingvalues, \\ \left(1\right)\sin \beta =n\sin \left(90°-\theta \right) \\ \beta ={\sin }^{-1}\left[n\sin \left(90°-\theta \right)\right]1. \end{gathered}$$

Using Figure, we can write as :

$$\begin{gathered} \left(90°-\theta \right)+\left(90°-\theta \right)+\alpha =180° \\ 180°-2\theta +\alpha =180° \\ \theta =\frac{\alpha }{2} \end{gathered}$$

Substituting the value in equation 1.

role="math" localid="1648363429911" $$\begin{gathered} \beta ={\sin }^{-1}\left[n\sin \frac{\alpha }{2}\right] \\ \sin \beta =n\sin \frac{\alpha }{2} \\ n=\frac{\sin \beta }{\sin {\displaystyle \frac{\alpha }{2}}} \\ Puttingvalues, \\ n=\frac{\sin 52.2°}{\sin {\displaystyle \frac{60°}{2}}} \\ n=1.58 \end{gathered}$$