Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

lies between $\sqrt{2}$ and $1$

lies between $2$ and $\sqrt{2}$

is less than $1$

is greater than $2$

Prism Formula: The refractive index $\mu$ of a prism is given by the formula: $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$
Given Condition: The angle of minimum deviation ( $\delta_m$ ) is equal to the refracting angle ( $A$ ) of the prism. So, $\delta_m = A$ .
Substitution: $\mu = \frac{\sin\left(\frac{A + A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin(A)}{\sin\left(\frac{A}{2}\right)}$
Trigonometric Identity: Using $\sin(A) = 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)$ , we get: $\mu = \frac{2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = 2\cos\left(\frac{A}{2}\right)$
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 $0^\circ < \frac{A}{2} \le 45^\circ$ .
Limits for Refractive Index ( $\mu$ ): When $\frac{A}{2} \to 0^\circ$ , $\cos\left(\frac{A}{2}\right) \to 1$ , so $\mu \to 2(1) = 2$ . When $\frac{A}{2} = 45^\circ$ , $\cos(45^\circ) = \frac{1}{\sqrt{2}}$ , so $\mu = 2\left(\frac{1}{\sqrt{2}}\right) = \sqrt{2}$ .
Conclusion: The refractive index $\mu$ must lie between $\sqrt{2}$ and $2$ .

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