A) \[n>\sqrt{2}\]
B) n = 1
C) n = 1.1
D) n = 1.3
Correct Answer: A
Solution :
Key Idea: The first idea is that for no refraction at its lateral face, angle of incidence should be greater than critical angle. Let a light ray enters at A and refracted beam is AB. At the lateral face, the angle of incidence is \[\theta \]. For no refraction at this face, \[\theta >C\] \[\sin \theta >\sin \,C\] but \[\theta +r={{90}^{o}}\] \[\Rightarrow \theta =({{90}^{o}}-r)\] \[\therefore \sin ({{90}^{o}}-r)>\sin \,C\] \[or\cos r>\sin C\] ...(i) Key Idea: The second idea is that in Eq. (i), the substitution for cos r can be found from Snell's law. Now from Snell's law. Now from Snell?s law, \[n=\frac{\sin i}{\sin r}\Rightarrow \sin r=\frac{\sin i}{n}\] \[\therefore \cos r=\sqrt{1-{{\sin }^{2}}r}=\sqrt{\left( 1-\frac{{{\sin }^{2}}i}{{{n}^{2}}} \right)}\] \[\therefore Eq.\,(i)\,gives,\] \[\sqrt{1-\frac{{{\sin }^{2}}i}{{{n}^{2}}}}>\sin \,C\] \[\Rightarrow 1-\frac{{{\sin }^{2}}i}{{{n}^{2}}}>{{\sin }^{2}}C\] Also \[\sin C=\frac{1}{n}\] \[\therefore 1-\frac{{{\sin }^{2}}i}{{{n}^{2}}}>\frac{1}{{{n}^{2}}}\] or \[1>\frac{1}{{{n}^{2}}}+\frac{{{\sin }^{2}}i}{{{n}^{2}}}\] or \[\frac{1}{{{n}^{2}}}({{\sin }^{2}}i+1)<1\] or \[{{n}^{2}}>{{\sin }^{2}}i+1\] The maximum value of sin i is 1. So, \[\therefore {{n}^{2}}>2\] or \[n>\sqrt{2}\]You need to login to perform this action.
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