Question

For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index
(1) Lies between 2^1/2 and 1
(2) Lies between 2 and 2^1/2
(3) Is less than 1
(4) Is greater than 2

Answer 1

Answer on Question #38971, Physics, Optics

For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index

(1) Lies between $2^{\wedge}1/2$ and 1

(2) Lies between 2 and $2^{\wedge}1/2$

(3) Is less than 1

(4) Is greater than 2

Solution:

The refractive index $n$, angle of minimum deviation $\delta$ of light through a prism with refracting angle $A$ are related by the following formula:

$$n = \frac {\sin \left(\frac {A + \delta}{2}\right)}{\sin \left(\frac {A}{2}\right)}$$

We have $\delta = A$

$$n = \frac {\sin \left(\frac {A + A}{2}\right)}{\sin \left(\frac {A}{2}\right)} = \frac {\sin \left(2 \frac {A}{2}\right)}{\sin \left(\frac {A}{2}\right)} = \frac {2 \sin \left(\frac {A}{2}\right) \cos \left(\frac {A}{2}\right)}{\sin \left(\frac {A}{2}\right)} = 2 \cos \left(\frac {A}{2}\right)$$

When $\delta = A$, these angles can change in range from 0 to 90 degrees.

$\cos (0{}^{\circ}) = 1$, then $n = 2$

$$\cos \left(\frac {90{}^{\circ}}{2}\right) = \cos (45{}^{\circ}) = \frac {\sqrt {2}}{2}, \text{ then } n = \sqrt {2}.$$

Thus $n$ lies between 2 and $\sqrt{2}$.

Answer. (2) Lies between 2 and $2^{\wedge}1/2$.