Question

The refractive index of prism is µ = √3 and the ratio of the angle of minimum deviation to the angle of prism is one. The value of angle of prism is ______o.

Answer 1

For $\delta_\text{min}$:

$i = e \quad \text{and} \quad r_1 = r_2 = \frac{A}{2}$

$\delta_\text{min} = 2i - A$

Given, $\frac{\delta_\text{min}}{A} = 1$:

$\frac{2i - A}{A} = 1$

Simplifying:

$2i - A = A \implies 2i = 2A \implies i = A$

Using Snell's law:

$1 \cdot \sin i = \mu \cdot \sin r \implies \sin i = \mu \cdot \sin \left(\frac{A}{2}\right)$

Substituting $i = A$:

$\sin A = \mu \cdot \sin \left(\frac{A}{2}\right)$

Expanding $\sin A$:

$2 \sin \frac{A}{2} \cos \frac{A}{2} = \sqrt{3} \cdot \sin \frac{A}{2}$

Dividing by $\sin \frac{A}{2}$:

$2 \cos \frac{A}{2} = \sqrt{3} \implies \cos \frac{A}{2} = \frac{\sqrt{3}}{2}$

$\frac{A}{2} = 30^\circ \implies A = 60^\circ$

Answer 2

For $\delta_\text{min}$:

$i = e \quad \text{and} \quad r_1 = r_2 = \frac{A}{2}$

$\delta_\text{min} = 2i - A$

Given, $\frac{\delta_\text{min}}{A} = 1$:

$\frac{2i - A}{A} = 1$

Simplifying:

$2i - A = A \implies 2i = 2A \implies i = A$

Using Snell's law:

$1 \cdot \sin i = \mu \cdot \sin r \implies \sin i = \mu \cdot \sin \left(\frac{A}{2}\right)$

Substituting $i = A$:

$\sin A = \mu \cdot \sin \left(\frac{A}{2}\right)$

Expanding $\sin A$:

$2 \sin \frac{A}{2} \cos \frac{A}{2} = \sqrt{3} \cdot \sin \frac{A}{2}$

Dividing by $\sin \frac{A}{2}$:

$2 \cos \frac{A}{2} = \sqrt{3} \implies \cos \frac{A}{2} = \frac{\sqrt{3}}{2}$

$\frac{A}{2} = 30^\circ \implies A = 60^\circ$