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Question

Physics Question on Ray optics and optical instruments

A bulb is kept at a depth h inside water of refractive index ′n′' n' . Area of the bright patch on the water surface is

A

πh2n2−1\frac{ \pi h^2}{n^2 - 1}

B

πh2n2−1\frac{ \pi h^2}{\sqrt{ n^2 - 1}}

C

π(n2−1)h2\pi ( n^2 - 1) h^2

D

πh2n2\frac{ \pi h^2}{n^2}

Answer

πh2n2−1\frac{ \pi h^2}{n^2 - 1}

Explanation

Solution

Area of the bright patch on the water surface is seen due to total internal reflection of light.
Apply Snell's law at the interface of water and air.
nsin⁡θc,=1sin⁡ 90∘n \sin \theta_c , = 1 \sin \, 90^{\circ}
sin⁡θc=1n\sin \theta_{c} = \frac{1}{n}
or , rr2+h2=1n\frac{r}{\sqrt{r^{2} + h^{2} } } = \frac{1}{n} or,r2=h2n2−1 r^{2} = \frac{h^{2}}{n^{2} - 1}
or, r2(n2−1)=h2r^{2} \left(n^{2} - 1\right)= h^{2} or, r2=h2n2−1r^{2} = \frac{h^{2}}{n^{2} -1}
Area of bright patch on the water surface
=πr2=πh2n2−1= \pi r^{2} = \frac{\pi h^{2}}{n^{2} - 1 }