Question

A vessel of depth d filled with a liquid of refractive index ${\mu }_1$ up to half its depth and the remaining space is filled with a liquid of refractive index ${\mu }_2$ . The apparent depth while seeing normal to the
free surface of the liquid is

• A. $d(\frac{1}{{\mu }_1}+\frac{1}{{\mu }_2})$
• B. $d({\mu }_1+{\mu }_2)$
• C. $\frac{d}{2}$ $(\frac{1}{{\mu }_1}+\frac{1}{{\mu }_2})$
• D. $\frac{d}{2}$(${\mu }_1+{\mu }_2$)

Answer 1

The correct option is C $\frac{d}{2}$ $(\frac{1}{{\mu }_1}+\frac{1}{{\mu }_2})$

the shift in depth due to first
liquid $=\frac{d}{2}(1-\frac{1}{{\mu }_1})$
the shift in depth due to second
liquid $=\frac{d}{2}(1-\frac{1}{{\mu }_2})$
So,
Apparent depth $=d-d+\frac{d}{2}(\frac{1}{{\mu }_1}+\frac{1}{{\mu }_2})$
$=\frac{d}{2}(\frac{1}{{\mu }_1}+\frac{1}{{\mu }_2})$