A ray of light is incident at an angle of 75^ | vedclass

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(B) Let the angle of incidence be $i = 75^{\circ}$.The deviation of the reflected ray is $\delta_r = 180^{\circ} - 2i$.The deviation of the refracted

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$\frac{\sqrt{3} + 1}{2}$

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(B) Let the angle of incidence be $i = 75^{\circ}$.The deviation of the reflected ray is $\delta_r = 180^{\circ} - 2i$.The deviation of the refracted ray is $\delta_t = i - r$,where $r$ is the angle of refraction.According to the problem,the deviations are equal in magnitude but opposite in direction,so $\delta_r = -\delta_t$ (or $|\delta_r| = |\delta_t|$).Thus,$180^{\circ} - 2i = -(i - r) \implies 180^{\circ} - 2i = r - i \implies r = 180^{\circ} - i$.Since $i = 75^{\circ}$,$r = 180^{\circ} - 75^{\circ} = 105^{\circ}$. This is physically impossible for a single interface refraction as $r$ must be less than $90^{\circ}$.Re-evaluating the condition: The deviation of the reflected ray is $\delta_r = 180^{\circ} - 2i$ and the refracted ray is $\delta_t = i - r$. If they are equal,$180^{\circ} - 2i = i - r \implies r = 3i - 180^{\circ}$.For $i = 75^{\circ}$,$r = 3(75^{\circ}) - 180^{\circ} = 225^{\circ} - 180^{\circ} = 45^{\circ}$.Using Snell's Law: $\mu = \frac{\sin i}{\sin r} = \frac{\sin 75^{\circ}}{\sin 45^{\circ}}$.$\sin 75^{\circ} = \sin(45^{\circ} + 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$.$\mu = \frac{(\sqrt{3} + 1) / 2\sqrt{2}}{1 / \sqrt{2}} = \frac{\sqrt{3} + 1}{2}$.

$A$ ray of light is incident at an angle of $75^{\circ}$ into a medium having refractive index $\mu$. The reflected and the refracted rays are found to suffer equal deviations in opposite directions. The value of $\mu$ is:

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