JEE MAIN - Physics (2025 - 28th January Morning Shift - No. 18)
A hemispherical vessel is completely filled with a liquid of refractive index $\mu$. A small coin is kept at the lowest point $(\mathrm{O})$ of the vessel as shown in figure. The minimum value of the refractive index of the liquid so that a person can see the coin from point E (at the level of the vessel) is _________.
_28th_January_Morning_Shift_en_18_1.png)
$\frac{3}{2}$
$\frac{\sqrt{3}}{2}$
$\sqrt{3}$
$\sqrt{2}$
Explanation
_28th_January_Morning_Shift_en_18_2.png)
from $$\Delta OCB,\,\tan \theta = {{CB} \over {OC}} = {R \over R} = 1$$
$$ \Rightarrow \theta = 45^\circ $$
We know, $$\sin c = {1 \over \mu }$$
for minimum $\mu$, sinc maximum $\Rightarrow$ C maximum
Here, $${c_{\max }} = \theta = 45^\circ $$
So, $$\mu = {1 \over {\sin 45^\circ }} = {1 \over {{1 \over {\sqrt 2 }}}}$$
$$ \Rightarrow \mu = \sqrt 2 $$
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