Question

A vessel of depth t is half filled with a liquid having refractive index $n_1$ and the other half is filled with water of having refractive index $n_2$. The apparent depth of the vessel as viewed from top is:

• A. $\frac{2t(n_1+n_2)}{n_1{n}_2}$
• B. $\frac{t\left(n_1{n}_2\right)}{(n_1+n_2)}$
• C. $\frac{t(n_1+n_2)}{2n_1{n}_2}$
• D. $\frac{n_1{n}_2}{(n_1+n_2)t}$

Answer 1

Answer: (C)

$\frac{t(n_1+n_2)}{2n_1{n}_2}$

Let apparent depth be $d$.

Refer the ray diagram in the attached figure.

$\tan {\theta }_1=\frac{x}{t/2}...........\left(i\right)$

$\tan {\theta }_2=\frac{y}{t/2}..........\left(ii\right)$

$\tan \theta =\frac{x+y}{d}.............\left(iii\right)$

By Snell's Law,

$n_1\sin {\theta }_1=n_2\sin {\theta }_2=\sin \theta$

For small angles, $\sin \theta \approx \tan \theta$

$\sin {\theta }_1\approx \tan {\theta }_1$

$\sin {\theta }_2\approx \tan {\theta }_2$

Hence, $\tan \theta =n_1\tan {\theta }_1...............\left(iv\right)$

$\tan \theta =n_2\tan {\theta }_2.............\left(v\right)$

Substituting (i), (ii) and (iii) in (iv) and (v), we get

$\frac{x+y}{d}=n_1\frac{2x}{t}...........\left(vi\right)$

$\frac{x+y}{d}=n_2\frac{2y}{t}...........\left(vii\right)$

Eliminating $x$ and $y$ from (vi) and (vii), we get

$d=\frac{t(n_1+n_2)}{2n_1{n}_2}$