A) \[n>\sqrt{2}\]
B) \[n=1\]
C) \[n=1.1\]
D) \[n=1.3\]
Correct Answer: A
Solution :
[a] Let a ray of light enter at A and the refracted beam is AB. This is incident at an angle \[\theta .\] For no refraction at the lateral face, \[\theta >C\] or, \[\sin \,\,\theta >\sin \,\,C\]But \[\theta +r=90{}^\circ \] \[\Rightarrow \] \[\theta =(90{}^\circ -r)\] \[\therefore \] \[\sin \,\,(90{}^\circ -r)>\sin \,\,C\] or \[\cos \,\,r>\sin \,\,C\] ?(1) From Snell s law, \[n=\frac{\sin \,i}{\sin \,r}\]\[\Rightarrow \]\[\sin r=\frac{\sin \,i}{n}\] \[\therefore \]\[\cos \,r=\sqrt{1-{{\sin }^{2}}r}=\sqrt{\left( 1-\frac{{{\sin }^{2}}i}{{{n}^{2}}} \right)}\] \[\therefore \] equation (1) gives \[\sqrt{1-\frac{{{\sin }^{2}}i}{{{n}^{2}}}}>\sin \,\,C\]\[\Rightarrow \]\[1-\frac{{{\sin }^{2}}i}{{{n}^{2}}}>{{\sin }^{2}}C\] Also, \[\sin \,\,C=\frac{1}{n}\] \[\therefore \]\[1-\frac{{{\sin }^{2}}i}{{{n}^{2}}}>\frac{1}{{{n}^{2}}}\]or \[1>\frac{{{\sin }^{2}}i}{{{n}^{2}}}+\frac{1}{{{n}^{2}}}\] or \[\frac{1}{{{n}^{2}}}({{\sin }^{2}}i+1)<1\] or \[{{n}^{2}}>({{\sin }^{2}}i+1)\] Maximum value of \[\sin i=1\] \[\therefore \]\[{{n}^{2}}>2\]\[\Rightarrow \]\[n>\sqrt{2}\]You need to login to perform this action.
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