Question:
A ray of light passing through a prism $(\mu=\sqrt{3})$ suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. Then, the angle of prism is (in degrees)
Solution:
At minimum deviation $r_{1}=r_{2}=\frac{A}{2}$
Also given $\mathrm{i}=2 \mathrm{r}_{1}=\mathrm{A}$
Now $1 \cdot \sin \mathrm{i}=\sqrt{3} \sin \mathrm{r}_{1}$
$1 \sin A=\sqrt{3} \sin \frac{A}{2}$
$\Rightarrow \quad 2 \sin \frac{A}{2} \cos \frac{A}{2}=\sqrt{3} \sin \frac{A}{2}$
$\Rightarrow \quad \cos \frac{A}{2}=\frac{\sqrt{3}}{2} \Rightarrow \frac{A}{2}=30^{\circ}$
$\Rightarrow \quad A=60^{\circ}$
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