The refractive index of the medium with a certain region, x > 0, y > 0, changes with y. A thin light ray travelling in the x-direction in medium having refractive index $μ_{0} = 1$ strikes another medium of refractive index $μ$ at right angles and moves through the medium along a circular arc of radius R as shown in the figure. The material with the greatest known refractive index is diamond, but even the refractive index of this material does not reach the value $μ_{\max}$ = 2.5. It is this limit that sets the maximum angular size of the arc the light ray can cover. Angular size of arc is the angle subtendly by the arc at the centre.

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The variant of refractive index $μ$ with y is given as $μ = \frac{R}{R - y}$

If the maximum angular size of the arc of light is $θ_{\max}$ then $\sin θ_{\max} = 2 / 5$

The unit vector in the direction of refracted light at $y = \frac{R}{2} i s \frac{1}{2} i + \frac{\sqrt{3}}{2} j$

If the maximum angular size of the arc of light is $θ_{\max}$ then $\cos θ_{\max} = 2 / 5$

answer is A, B, D.

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Detailed Solution

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$μ_{0} \sin 90^{o} = μ \sin (90^{o} - θ)$
$μ = \frac{μ_{0}}{\cos θ} = \frac{1}{\cos θ} = \frac{R}{R - y}$
$a t y = \frac{R}{2}$
$μ = \frac{R}{R / 2} = 2$
$\cos θ = \frac{1}{2}$
$θ = 60^{o}$
$\hat{R} = \hat{i} \cos 60^{o} + \hat{j} \sin 60^{o}$
$\hat{R} = \frac{\hat{i}}{2} + \frac{\sqrt{3}}{2} \hat{j}$
$\cos θ = \frac{1}{μ}$
$\cos θ_{\max} = \frac{1}{2.5} = \frac{2}{5}$