Question 13: Ray Optics

Question

Angle of minimum deviation is equal to the half of the angle of prism in an equilateral prism. The refractive index of the prism is $\_\_\_\_$

Accepted Answer

For a prism with angle A and refractive index $\mu$, the relationship between the angle of minimum deviation ( $\delta_{\mathrm{m}}$ ) and refractive index is given by:
$$ \mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)} $$
For equilateral prism the angle of the prism is $\mathrm{A}=60^{\circ}$.
The angle of minimum deviation is half the angle of prism $\left(\delta_m=\frac{A}{2}\right)$.
Therefore, $\delta_m=\frac{60^{\circ}}{2}=30^{\circ}$
Substituting $\mathrm{A}=60^{\circ}$ and $\delta_{\mathrm{m}}=30^{\circ}$ into the formula:
$$ \mu=\frac{\sin \left(\frac{60^{\circ}+30^{\circ}}{2}\right)}{\sin \left(\frac{60^{\circ}}{2}\right)} $$
$$\Rightarrow $$ $\mu=\frac{\sin \left(45^{\circ}\right)}{\sin \left(30^{\circ}\right)}$
$$\Rightarrow $$ $\mu=\frac{1 / \sqrt{2}}{1 / 2}=\frac{2}{\sqrt{2}}$
$$\Rightarrow $$ $$ \mu=\sqrt{2} $$
Therefore, the refractive index of the prism is $\sqrt{2}$.
Thus the correct option is (C).

Answer

1.5

Answer

$\sqrt{3}$

Answer

$\sqrt{2}$

Answer

1.65

Ray Optics

Angle of minimum deviation is equal to the half of the angle of prism in an equilateral prism. The refractive index of the prism is \_\_\_\_

Choose the correct answer: