Question

A stone is placed at the bottom of lake of depth $10m$. Refractive index of water of lake is $\mu =(1+\frac{y}{10})$ where $y$ is depth of water from surface. A person looks at stone normally, then depth at which stone appears to him in meters is ($ln2=0.693$)

• A. $6.93$
• B. $8.23$
• C. $7.5$
• D. $6$

Answer 1

The correct option is A $6.93$


Due to a rectangular slab of thickness $t$ and refractive index $\mu$, the shift in an obejct's location is $t(1-\frac{1}{\mu })$.

Similarly the shift due to the small element given is,
Shift $\Delta S=\int dS=\int \left(dy\right(1-\frac{1}{\mu }\left)\right)$

$\Delta S={\int }_0^{D}dy-{\int }_0^D\frac{dy}{\mu }=D-{\int }_0^D\frac{dy}{1+\frac{y}{10}}$

$\Delta S=D-D\ln \left(2\right)$
Apparent depth =$D-\Delta S=D\ln \left(2\right)$
$=10\times 0.693=6.93m$