A point source of light is placed at the bottom of a tank filled with water of refractive index $\mu$ to a depth $d$. The area of the surface of water through which light from the source can emerge is:

(B) $\dfrac{\pi d^{2}}{(\mu^{2}-1)}$

(A) $\dfrac{\pi d^{2}}{2(\mu^{2}-1)}$

(C) $\dfrac{\pi d^{2}}{\sqrt{2}\sqrt{\mu^{2}-1}}$

(D) $\dfrac{2\pi d^{2}}{(\mu^{2}-1)}$

CBSE Class 12 · 2022 · Geometrical OpticsMultiple Choice

A point source of light is placed at the bottom of a tank filled with water of refractive index

μ

to a depth

d

. The area of the surface of water through which light from the source can emerge is:

Answer: (B)

Light escapes only within the cone of semi-vertical angle

θ_{c}

. The radius of the illuminated circle is

r = d tan θ_{c} = \frac{d}{μ ^{2} - 1}

, so the area is

A = π r^{2} = \frac{π d ^{2}}{μ ^{2} - 1}

CBSE Class 12 Board · 2022 · Single correct option