The angle of minimum deviation for a prism is $30^\circ$ and the angle of prism is $60^\circ$. The refractive index of the material of the prism is:

(a) 2
(b) $\sqrt{2}$
(c) 1.5
(d) $\dfrac{1}{\sqrt{2}}$

Question

The angle of minimum deviation for a prism is $30^\circ$ and the angle of prism is $60^\circ$. The refractive index of the material of the prism is:

(a) 2
(b) $\sqrt{2}$
(c) 1.5
(d) $\dfrac{1}{\sqrt{2}}$

Accepted Answer

>

> [!TLDR]
> $n = \dfrac{\sin 45^\circ}{\sin 30^\circ} = \dfrac{1/\sqrt2}{1/2} = \sqrt2$: option (B).
>
> The refractive index of a prism at minimum deviation is
> $$n = \frac{\sin\left(\dfrac{A + D_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)},$$
> where $A$ is the prism angle and $D_m$ the angle of minimum deviation.
>
> With $A = 60^\circ$ and $D_m = 30^\circ$: …

Q.The angle of minimum deviation for a prism is $30^\circ$ and the angle of prism is $60^\circ$ . The refractive index of the material of the prism is:

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Q.The angle of minimum deviation for a prism is 30∘30^\circ30∘ and the angle of prism is 60∘60^\circ60∘. The refractive index of the material of the prism is:

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Q.The angle of minimum deviation for a prism is $30^\circ$ and the angle of prism is $60^\circ$ . The refractive index of the material of the prism is:

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