Question

A ray of light is incident normally on one face of a ${30}^o-{60}^o-{90}^o$ prism of refractive index $\frac{5}{3}$ immersed in water of refractive index $\frac{4}{3}$ as shown.

• A. the exit angle ${\theta }_2$ of the ray is $si{n}^{-1}\left(\frac{5}{8}\right)$
• B. the exit angle ${\theta }_2$ of the the ray is $si{n}^{-1}\left(\frac{5}{4\sqrt{3}}\right)$
• C. total internal reflection at point P ceases if the refractive index of water is increased to $\frac{5}{2\sqrt{3}}$ by dissolving some substance
• D. Total internal reflection at point P ceases if the refractive index of water is increased to $\frac{5}{3}$ by dissolving some substance.

Answer 1

Answer: (A), (C)

The correct options are
A the exit angle ${\theta }_2$ of the ray is $si{n}^{-1}\left(\frac{5}{8}\right)$
C total internal reflection at point P ceases if the refractive index of water is increased to $\frac{5}{2\sqrt{3}}$ by dissolving some substance

From geometry, we find that ${\theta }_1={60}^o$ and $\theta ={30}^o$

Now snell's law at point Q, $\frac{4}{3}sin{\theta }_2=\frac{5}{3}sin\theta ⟹sin{\theta }_2=\frac{5}{8}$

Also, for total internal reflection to be ceased at point P, then ${\theta }^{\prime }$ must be equal to ${90}^o$.

Now let new refractive index of water be $\frac{5}{2\sqrt{3}}$, snell's law at point P $\frac{5}{3}sin{\theta }_1=\frac{5}{2\sqrt{3}}sin{\theta }^{\prime }$

As ${\theta }_1={60}^o⟹sin{\theta }^{\prime }=1⟹{\theta }^{\prime }={90}^o$