CBSE 2024, Set 1

Question

The refractive index of the material of a prism is $\sqrt{2}$. If the refracting angle of the prism is $60^{\circ}$, find the (1) Angle of minimum deviation, and (2) Angle of incidence.

Edvaya · Accepted Answer

The correct answer is Subjective. **SOLUTION** Given parameters: * Refractive index, $\mu = \sqrt{2}$ * Prism angle, $A = 60^{\circ}$ **(1) Angle of minimum deviation ($\delta_m$):** Using the standard prism formula: $$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}ight)}{\sin\left(\frac{A}{2}ight)}$$ $$\sqrt{2} = \frac{\sin\left(\frac{60^{\circ} + \delta_m}{2}ight)}{\sin\left(\frac{60^{\circ}}{2}ight)}$$ $$\sqrt{2} = \frac{\sin\left(\frac{60^{\circ} + \delta_m}{2}ight)}{\sin(30^{\circ})}$$ Since $\sin(30^{\circ}) = \frac{1}{2}$: $$\sqrt{2} 	imes \frac{1}{2} = \sin\left(\frac{60^{\circ} + \delta_m}{2}ight)$$ $$\frac{1}{\sqrt{2}} = \sin\left(\frac{60^{\circ} + \delta_m}{2}ight)$$ We know that $\sin(45^{\circ}) = \frac{1}{\sqrt{2}}$. Therefore: $$\frac{60^{\circ} + \delta_m}{2} = 45^{\circ}$$ $$60^{\circ} + \delta_m = 90^{\circ}$$ $$\delta_m = 30^{\circ}$$ **(2) Angle of incidence ($i$):** At the position of minimum deviation, the angle of incidence is calculated as: $$i = \frac{A + \delta_m}{2}$$ $$i = \frac{60^{\circ} + 30^{\circ}}{2}$$ $$i = \frac{90^{\circ}}{2} = 45^{\circ}$$

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