A person looking through a telescope focuses lens at a point on the edge of the bottom of an empty cylindrical vessel. Next he fills the entire vessel with a liquid of refractive index $μ$ , without disturbing the telescope. Now, he was able to see the mid point of the bottom of the vessel. Determine the radius to depth ratio of the vessel

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$\frac{1}{2} \sqrt{\frac{1 - μ^{2}}{μ^{2} + 1}}$

$\frac{1}{2} \sqrt{\frac{4 - μ^{2}}{μ^{2} - 1}}$

$\frac{1}{2} \sqrt{\frac{4 + μ}{μ + 1}}$

$\frac{1}{2} \sqrt{\frac{4 + μ^{2}}{μ^{2} + 1}}$

answer is B.

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Detailed Solution

$\sin i = \frac{2 R}{\sqrt{4 R^{2} + h^{2}}} Now the refracting rays will be making' r' with the normal to the surface of liquid sinr = \frac{R}{\sqrt{R^{2} + h^{2}}} from snells law, 1 \sin i = μsin r \frac{2 R}{\sqrt{4 R^{2} + h^{2}}} = \frac{μR}{\sqrt{R^{2} + h^{2}}} \frac{R}{h} = \frac{1}{2} \sqrt{\frac{4 - μ^{2}}{μ^{2} - 1}}$