Optics Ques 105
- A ray of light undergoes deviation of $30^{\circ}$ when incident on an equilateral prism of refractive index $\sqrt{2}$. The angle made by the ray inside the prism with the base of the prism is …..
(1992, 1M)
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Answer:
Correct Answer: 105.$(30^\circ)$
Solution:
Formula:
- Let $\delta _m$ be the angle of minimum deviation. Then
$ \begin{aligned} \mu & =\frac{\sin (\frac{A+\delta _m}{2})}{\sin (A / 2)}\left(A=60^{\circ} \text { for an equilateral prism }\right) \\ \therefore \sqrt{2} & =\frac{\sin (\frac{60^{\circ}+\delta _m}{2})}{\sin (\frac{60^{\circ}}{2})} \end{aligned} $
Solving this we get $\delta _m=30^{\circ}$
The given deviation is also $30^{\circ}$ (i.e. $\delta _m$ )
Under minimum deviation, the ray inside the prism is parallel to base for an equilateral prism.
BETA
