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 \[\cos \text{ }r\] 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 \[n>\sqrt{2}\] |
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