question_answer

A light ray falls on square glass slab as shown in the diagram. The index of refraction of the glass, if total internal reflection is to occur at the vertical face, is equal to:     [JEE Main Online Paper ( Held On 23  April 2013 )

A) \[\left( \sqrt{2}+1 \right)/2\]                                    

B) \[\sqrt{\frac{5}{2}}\]

C) \[\frac{3}{2}\]                                  

D) \[\sqrt{\frac{3}{2}}\]

Correct Answer: D

Solution :

                  At point A by Snell? slaw \[\mu =\frac{\sin {{45}^{o}}}{\sin r}\Rightarrow \sin r=\frac{1}{\mu \sqrt{2}}\]                   ?(I     At point B, for total internal reflection, \[\sin {{i}_{1}}=\frac{1}{\mu }\] From figure,  \[{{i}_{1}}={{90}^{o}}-r\] \[\therefore \]\[(\sin {{90}^{o}}-r)=\frac{1}{\mu }\]\[\Rightarrow \cos r=\frac{1}{\mu }\]                             ?(ii ) \[=\sqrt{\frac{2{{\mu }^{2}}-1}{2{{\mu }^{2}}}}\]                                                             ...(ii) \[\frac{1}{\mu }=\sqrt{\frac{2{{\mu }^{2}}-1}{2{{\mu }^{2}}}}\]   Squaring both sides and then solving, we get \[\mu =\sqrt{\frac{3}{2}}\]                  \[\sqrt{\frac{3}{2}}\]