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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
A
is greater than 2
B
lies between $\sqrt2$ and 1
C
lies between 2 and $\sqrt2$
D
is less than 1
Detailed Solution
$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$
With $\delta_m = A$: $\mu = \frac{\sin A}{\sin\frac{A}{2}} = \frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{\sin\frac{A}{2}} = 2\cos\frac{A}{2}$
At minimum deviation $i = \frac{A + \delta_m}{2} = A$, and i must be less than $90^\circ$, so A lies between $0^\circ$ and $90^\circ$.
For A → $0^\circ$, $\mu$ → 2; for A = $90^\circ$, $\mu = 2\cos45^\circ = \sqrt2$.
So $\mu$ lies between 2 and $\sqrt2$.
With $\delta_m = A$: $\mu = \frac{\sin A}{\sin\frac{A}{2}} = \frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{\sin\frac{A}{2}} = 2\cos\frac{A}{2}$
At minimum deviation $i = \frac{A + \delta_m}{2} = A$, and i must be less than $90^\circ$, so A lies between $0^\circ$ and $90^\circ$.
For A → $0^\circ$, $\mu$ → 2; for A = $90^\circ$, $\mu = 2\cos45^\circ = \sqrt2$.
So $\mu$ lies between 2 and $\sqrt2$.