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

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

Accepted Answer

1.  **Prism Formula:** The refractive index $\mu$ of a prism is given by the formula:
    $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}ight)}{\sin\left(\frac{A}{2}ight)}$
2.  **Given Condition:** The angle of minimum deviation ($\delta_m$) is equal to the refracting angle ($A$) of the prism. So, $\delta_m = A$.
3.  **Substitution:**
    $\mu = \frac{\sin\left(\frac{A + A}{2}ight)}{\sin\left(\frac{A}{2}ight)} = \frac{\sin(A)}{\sin\left(\frac{A}{2}ight)}$
4.  **Trigonometric Identity:** Using $\sin(A) = 2\sin\left(\frac{A}{2}ight)\cos\left(\frac{A}{2}ight)$, we get:
    $\mu = \frac{2\sin\left(\frac{A}{2}ight)\cos\left(\frac{A}{2}ight)}{\sin\left(\frac{A}{2}ight)} = 2\cos\left(\frac{A}{2}ight)$
5.  **Limits for Angle of Prism ($A$):**
    *   For minimum deviation to occur, the angle of incidence $i$ must be $\le 90^\circ$.
    *   We know $i = \frac{A + \delta_m}{2} = \frac{A + A}{2} = A$. Therefore, $A \le 90^\circ$.
    *   Also, for a prism to exist, $A > 0^\circ$.
    *   So, $0^\circ < A \le 90^\circ$, which means $0^\circ < \frac{A}{2} \le 45^\circ$.
6.  **Limits for Refractive Index ($\mu$):**
    *   When $\frac{A}{2} 	o 0^\circ$, $\cos\left(\frac{A}{2}ight) 	o 1$, so $\mu 	o 2(1) = 2$.
    *   When $\frac{A}{2} = 45^\circ$, $\cos(45^\circ) = \frac{1}{\sqrt{2}}$, so $\mu = 2\left(\frac{1}{\sqrt{2}}ight) = \sqrt{2}$.
7.  **Conclusion:** The refractive index $\mu$ must lie between $\sqrt{2}$ and $2$.

Step-by-Step Solution

Practice All PHYSICS Questions