An opaque sphere of radius a is just immersed in a transparent liquid as shown in figure. A point source is placed on the vertical diameter of the sphere at a distance a/2 from the top of the sphere. One ray originating from the point source after refraction from the air liquid interface forms tangent to the sphere. The angle of refraction for that particular ray is 30º. The refractive index of the liquid is

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An opaque sphere of radius a is just immersed in a transparent liquid as shown in figure. A point source is placed on the vertical diameter of the sphere at a distance a/2 from the top of the sphere. One ray originating from the point source after refraction from the air liquid interface forms tangent to the sphere. The angle of refraction for that particular ray is 30º. The refractive index of the liquid is

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

$\frac{3}{\sqrt{5}}$

$\frac{4}{\sqrt{5}}$

$\frac{4}{\sqrt{7}}$

answer is D.

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Detailed Solution

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Given    $∠ N E D = 30 °$
   $∴ ∠ B E D = 120 °$
BCDE is cyclic quadrilateral
$∴ ∠ B C D = 60 °$
The line CE will be angle bisector of $∠ B C D$
   $∴ BE = a tan30º = \frac{a}{\sqrt{3}}$
$∴ n o w \tan i = \frac{B E}{A B} = \frac{a \sqrt{3}}{a / 2} = \frac{2}{\sqrt{3}}$
$∴ \sin i = \frac{2}{\sqrt{7}}$ now by snell’s law

$1 \times \sin i = n \sin r \Rightarrow \frac{2}{\sqrt{7}} = n \times \frac{1}{2}$