Question

A light ray is incident on an equilateral prism of refractive index 5/3. What should be the angle of incidence such that the ray gazes on the second surface?

• A. $si{n}^{-1}\left(\frac{1}{3}\right)$
• B. $si{n}^{-1}\left(\frac{2}{3}\right)$
• C. $si{n}^{-1}\left(\frac{2}{5}\right)$
• D. $si{n}^{-1}\left(\frac{4}{5}\right)$

Answer 1

The correct option is B $si{n}^{-1}\left(\frac{2}{3}\right)$


For the first surface,
$\frac{sin\theta }{sin{r}_1}=\frac{5}{3}..\left(1\right)$
For the second surface,
$\frac{sin{r}_2}{sin90}=\frac{3}{5}$
$\Rightarrow sin{r}_2=\frac{3}{5}$
$\Rightarrow r_2={37}^{\circ }$
$\Rightarrow r_1={60}^{\circ }-{37}^{\circ }={23}^{\circ }$
$\Rightarrow \frac{sin\theta }{sin{23}^{\circ }}=\frac{5}{3}$
$sin\theta =\frac{5}{3}\times sin{23}^{\circ }$
$=\frac{5}{3}\times .4$
$=\frac{2}{3}$
$\Rightarrow \theta =si{n}^{-1}\left(\frac{2}{3}\right)$