Question

When light of wavelength $\lambda$ is incident on an equilateral prism kept in its minimum deviation position, it is found that the angle of deviation equals the angle of the prism itself. The refractive index of the material of the prism for the wavelength $\lambda$ is, then

• A. $\sqrt{3}$
• B. $\frac{\sqrt{3}}{2}$
• C. $2$
• D. $\sqrt{2}$

Answer 1

The correct option is A $\sqrt{3}$

In an equilateral prism, the angle of prism is,
$\Rightarrow A=\frac{{180}^{\circ }}{3}={60}^{\circ }$
The refractive index ${\mu }_p$ of the prism is given by
${\mu }_p=\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}...\left(1\right)$
The minimum angle of deviation,
${\delta }_m=A$ (given)
$\Rightarrow {\mu }_p=\frac{\sin \left(\frac{{60}^{\circ }+{60}^{\circ }}{2}\right)}{\sin \left(\frac{{60}^{\circ }}{2}\right)}$
or, ${\mu }_p=\frac{\sin {60}^{\circ }}{\sin {30}^{\circ }}$
$\therefore {\mu }_p=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$

Why this question? Caution: Here angle of prism $(A={60}^{\circ })$ is not very small, hence we can not apply ${\delta }_m=(\mu -1)A$.