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, \[1\sin 45{}^\circ =\mu \sin {{r}_{1}}=\mu \sqrt{1-\frac{1}{{{\mu }^{2}}}}\] \[\Rightarrow \] \[\frac{1}{\sqrt{2}}=\mu \sqrt{\frac{{{\mu }^{2}}-1}{{{\mu }^{2}}}}=\sqrt{{{\mu }^{2}}-1}\] \[\Rightarrow \] \[\frac{1}{2}={{\mu }^{2}}-1\Rightarrow {{\mu }^{2}}=\frac{3}{2}\] \[\therefore \] \[\mu =\sqrt{\frac{3}{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec