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, \[\theta >C.\] i.e., \[\sin \theta >\sin C\] 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, \[n=\frac{\sin i}{\sin r}\] \[\Rightarrow \] \[\sin \,r=\frac{\sin i}{n}\] \[\therefore \] \[\cos \,r=\sqrt{1-{{\sin }^{2}}r}=\sqrt{\left( 1-\frac{{{\sin }^{2}}i}{{{n}^{2}}} \right)}\] \[\therefore \] \[\sqrt{1-\frac{{{\sin }^{2}}i}{{{n}^{2}}}}>\sin \,C\] Also \[\sin C=\frac{1}{n}\] \[\therefore \] \[1-\frac{{{\sin }^{2}}i}{{{n}^{2}}}>\frac{1}{{{n}^{2}}}\]or \[{{n}^{2}}>{{\sin }^{2}}i+1\] The maximum value of \[\sin i\]is 1. So,You need to login to perform this action.
You will be redirected in
3 sec