A body of elliptical cross – section is made up of material of refractive index $n_{2} .$ The body is surrounded by a liquid of refractive index $n_{1} .$ The figure here shows a cross –section of this body. If any ray parallel to major axis of the cross – section passes through focus of the cross – section (as shown), then the eccentricity of the elliptical cross – section is

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$\frac{1}{\sqrt{n_{1} n_{2}}}$

$\frac{n_{1} n_{2}}{n_{2} - n_{1}}$

$\frac{2}{n_{1} + n_{2}}$

$\frac{n_{1}}{n_{2}}$

answer is A.

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Detailed Solution

$n_{1} \sin 90^{\circ} = n_{2} \sin r \Rightarrow e = \frac{n_{1}}{n_{2}}$

$\sin r = \frac{a e}{a}$

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