Geometrical OpticsHard

Question

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index $n$ of the prism satisfies.

Options

A.$\sqrt{2} < n < 2\sqrt{2}$
B.$1 <$ n $< 2$
C.$n \geq 2$
D.$\sqrt{2} < n < 2$

Solution

$\delta_{\text{min~}} = 2i - A \Rightarrow i = \delta_{\text{min~}} = A$

Also, $\mu = \frac{sin\left( \frac{\delta_{\text{min~}} + A}{2} \right)}{sin\left( \frac{A}{2} \right)}$

$$\Rightarrow \mu = \frac{sinA}{sin\frac{A}{2}} = 2cos\left( \frac{\text{ }A}{2} \right)$$

$$\begin{array}{r} 1 < \mu < 2\#(1) \end{array}$$

$${\delta_{\text{min~}} = 2i - A }{A = 2i - A \Rightarrow i = A }$$$i < 90^{\circ}$ (grazing incidence)

$${A < 90^{\circ} }{\mu = 2cos(\text{ }A/2) }{\&\text{ }A < 90^{\circ}}$$

$$\begin{array}{r} \mu > \sqrt{2}\#(2) \end{array}$$

from (i) & (2)

$$\sqrt{2} < \mu < 2$$

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