question_answer

Monochromatic light falls on a right-angled prism at an angle of incidence \[45{}^\circ .\] The emergent light is found to slide along the face AC. Find the refractive index of material of prism.

Answer:

(i) \[\mu =\sqrt{3/2}\] Since the emergent light slides along the face AC, angle of emergence is\[90{}^\circ \], as shown. It implies that angle of incidence of the ray that falls on face AC is equal to the critical angle \[{{\theta }_{C.}}\] \[\therefore \]                  \[{{r}_{2}}={{\theta }_{C}}\]                           ?(i) From the prism theory, we know             \[{{r}_{1}}+{{r}_{2}}=A=90{}^\circ \] \[\therefore \]                  \[{{r}_{2}}=90{}^\circ -{{r}_{1}}\]                 ?(ii) From Eqs. (i) and (ii),             \[90{}^\circ -{{r}_{1}}={{\theta }_{C}}\] \[\therefore \]      \[\sin (90{}^\circ -{{r}_{1}})=sin{{\theta }_{C}}\]             \[\cos {{r}_{1}}=\sin {{\theta }_{C}}\] But       \[\sin {{\theta }_{C}}=\frac{1}{\mu }\] \[\therefore \]      \[\cos {{r}_{1}}=\frac{1}{\mu }\] and \[\sin {{r}_{1}}=\sqrt{1-{{\cos }^{2}}{{r}_{1}}}=\sqrt{1-\frac{1}{{{\mu }_{2}}}}\]             Applying Snell?s law at the boundary AB, \[1\sin 45{}^\circ =\mu \sin {{r}_{1}}=\mu \sqrt{1-\frac{1}{{{\mu }^{2}}}}\] \[\Rightarrow \]   \[\frac{1}{\sqrt{2}}=\mu \sqrt{\frac{{{\mu }^{2}}-1}{{{\mu }^{2}}}}=\sqrt{{{\mu }^{2}}-1}\] \[\Rightarrow \]   \[\frac{1}{2}={{\mu }^{2}}-1\Rightarrow {{\mu }^{2}}=\frac{3}{2}\] \[\therefore \]                  \[\mu =\sqrt{\frac{3}{2}}\]