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A ray of light falls upon a $60^{\circ}$ prism $(\mu=\sqrt{2})$ and it suffers minimum deviation. The angle of incidence for this ray must be
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Verified Answer
The correct answer is:
$45^{\circ}$
Given, angle of prism, $A=60^{\circ}$
$$
\mu=\sqrt{2}
$$
We know that,
$\mu=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin A / 2}$
$\Rightarrow \quad \sqrt{2}=\frac{\sin \left(\frac{60^{\circ}+\delta_{m}}{2}\right)}{\sin \frac{60}{2}}$
$\Rightarrow \quad \sin \left(\frac{60^{\circ}+\delta_{m}}{2}\right)=\sqrt{2} \sin 30^{\circ}$
$\Rightarrow \quad \sin \left(\frac{60^{\circ}+\delta_{m}}{2}\right)=\frac{1}{\sqrt{2}}$
$\Rightarrow \quad \sin \left(\frac{60^{\circ}+\delta_{m}}{2}\right)=\sin 45^{\circ}$
$\Rightarrow \quad \frac{60^{\circ}+\delta_{m}}{2}=45^{\circ}$
$\delta_{m}=30^{\circ}$
In the case of minimum deviation, angle of incidence,
$i=\frac{A+\delta_{m}}{2}=\frac{60^{\circ}+30^{\circ}}{2}=45^{\circ}$
$$
\mu=\sqrt{2}
$$
We know that,
$\mu=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin A / 2}$
$\Rightarrow \quad \sqrt{2}=\frac{\sin \left(\frac{60^{\circ}+\delta_{m}}{2}\right)}{\sin \frac{60}{2}}$
$\Rightarrow \quad \sin \left(\frac{60^{\circ}+\delta_{m}}{2}\right)=\sqrt{2} \sin 30^{\circ}$
$\Rightarrow \quad \sin \left(\frac{60^{\circ}+\delta_{m}}{2}\right)=\frac{1}{\sqrt{2}}$
$\Rightarrow \quad \sin \left(\frac{60^{\circ}+\delta_{m}}{2}\right)=\sin 45^{\circ}$
$\Rightarrow \quad \frac{60^{\circ}+\delta_{m}}{2}=45^{\circ}$
$\delta_{m}=30^{\circ}$
In the case of minimum deviation, angle of incidence,
$i=\frac{A+\delta_{m}}{2}=\frac{60^{\circ}+30^{\circ}}{2}=45^{\circ}$
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