MCQ
When light rays are incident on a prism at an angle of $45^o ,$ the minimum deviation is obtained. If refractive index of the material of prism is $\sqrt 2 $, then the angle of prism will be......$^o$
- A$30$
- B$40$
- C$50$
- ✓$60$
✓
Answer
Correct option: D.
$60$
d
(d) $\frac{{\sin \frac{{A + \delta \,m}}{2}}}{{\sin \frac{A}{2}}} = \mu ,$
(d) $\frac{{\sin \frac{{A + \delta \,m}}{2}}}{{\sin \frac{A}{2}}} = \mu ,$
But $\frac{{A + {\delta _m}}}{2} = i = {45^o}$
So $\frac{{\sin {{45}^o}}}{{\sin (A/2)}} = \sqrt 2 \Rightarrow \frac{1}{2} = \sin \frac{A}{2} \Rightarrow A = {60^o}$
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