A)
B)
C)
D)
Correct Answer: A
Solution :
The incident ray is incident normally on face\[AC\], so angle of refraction at face \[AC\] is also zero. Angle of incident i, which the ray makes with normal to face \[AB\] is\[{{45}^{o}}\]. Since this angle is greater than critical\[(={{\sin }^{-1}}\frac{1}{1\cdot 5}=41\cdot {{8}^{o}})\], the ray suffers total internal reflection. Now\[,\]\[i=r\]. The reflected ray from the face \[AB\] is now incident on face\[BC\], at angle of incidence again \[{{45}^{o}}\] and for the same reason it is turned back as shown in the above figure.You need to login to perform this action.
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