A) \[45{}^\circ \]
B) \[60{}^\circ \]
C) \[90{}^\circ \]
D) \[180{}^\circ \]
Correct Answer: A
Solution :
In the position of minimum deviation angle of incidence is equal to angle of emergence. Let a ray of monochromatic light \[PQ\] be incident on face\[AB\]. \[PQRS\] is path of light ray, where \[i\] is angle of incidence \[r\] angle of refraction, \[r\]angle of incidence and\[i\]angle of emergence. In position of minimum deviation \[i=i,\,\,r=\delta =\delta m\] \[2r=A\]or\[r=\frac{A}{2}\] Given \[A={{60}^{o}}\], \[r=\frac{60}{2}={{30}^{o}}\] Also from Snells law \[\mu =\frac{\sin i}{\sin r}=\frac{\sin i}{\sin {{30}^{o}}}\] \[\sqrt{2}=\frac{\sin i}{\sin {{30}^{o}}}\] \[\Rightarrow \] \[\sin i=\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}\] \[\Rightarrow \] \[i={{45}^{o}}\]You need to login to perform this action.
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