In a vessel, as shown in figure, point P is just visible when no liquid is filled in the vessel through a telescope in the air. When liquid is filled in the vessel completely, point Q is visible without moving the vessel or telescope. Find the refractive index of the liquid.


A. $\dfrac{\sqrt{14}}{3} \\ $
B. $\dfrac{\sqrt{85}}{5} \\ $
C. $\sqrt{2}$
D. $\sqrt{3}$

Hint:In this question, a figure is given in which a point P is visible when no liquid is filled in the vessel. When we fill the liquid, a point Q is visible and we have to find the refractive index of that point. We find the refractive index with the help of Snell's law. By putting the values in the snell’s law, we find out the value of the refractive index.

Formula Used:
We solve this question with the help of snell’s law:-
$\mu \sin i=\sin r$
Here, $\mu$ is the refractive index, $i$ is the angle of incidence and $r$ is the angle of refraction.

Complete step by step solution:
When the vessel is not filled with the liquid, Point P can just be seen.


Then the triangles ABP and BCD are similar.
Thus $\dfrac{PA}{AB}=\dfrac{BC}{CD}$
Hence CD = 4 R
Now after filling in a liquid of refractive index $\mu $, point Q can be seen. Hence the angle of refraction at D is r can be seen in the figure.

Now we apply snell’s law to the refraction at point D,
$\mu \sin i=\sin r$
$\Rightarrow \mu \sin (\angle CDQ)=1\times \sin (\angle CDB)$
Hence $\mu \times \dfrac{1}{\sqrt{5}}=1\times \dfrac{2R}{R\sqrt{20}}$
Solving further, we get
$\mu =\dfrac{2\sqrt{17}}{\sqrt{20}}$
Hence $\mu =\dfrac{\sqrt{85}}{5}$

Hence, option B is the correct answer.

Note: Remember that in the snell’s law, when the ray of light is incident perpendicularly, the speed changes, but the direction remains unaltered. When light passes from a rare medium to a denser medium, it is inclined closer towards the normal. When light rays pass from a dense medium to a rare medium, it is inclined away from the normal.