Q. A rectangular glass slab $ABCD$, of refractive index $n_{1}$, is immersed in water of refractive index $n_{2}(n_{1} > n_{2})$. A ray of light in incident at the surface $AB$ of the slab as shown. The maximum value of the angle of incidence $\alpha_{\max}$ such that the ray comes out only from the other surface $CD$ is given by
Explanation
(a) Ray comes out from $CD,$ means rays after refraction from $AB$ get, total intensity reflected at $AD$
$\frac{n_{1}}{n_{2}} = \frac{{\sin\alpha}_{\max}}{\sin r_{1}} \Rightarrow \alpha_{\max} = \sin^{- 1}\left\lbrack \frac{n_{1}}{n_{2}}\sin r_{1} \right\rbrack\ \ \ \ldots(i)$
Also $r_{1} + r_{2} = 90{^\circ} \Rightarrow r_{1} = 90 - r_{2} = 90 - C$
$\Rightarrow r_{1} = 90 - \sin^{- 1}\left( \frac{1}{\ _{2}\mu_{1}} \right) \Rightarrow r_{1} = 90 - \sin^{- 1}\left( \frac{n_{2}}{n_{1}} \right)\ \ \ldots(ii)$
Hence from equation (i) and (ii)
$\alpha_{\max} = \sin^{- 1}\left\lbrack \frac{n_{1}}{n_{2}}\sin\left( 90 - \sin^{- 1}\frac{n_{2}}{n_{1}} \right) \right\rbrack$
$= \sin^{- 1}\left\lbrack \frac{n_{1}}{n_{2}}\cos{(\sin^{- 1}\left( \frac{n_{2}}{n_{1}} \right)} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.