Question

A stone is placed at the bottom of lake of depth 10 m. 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 (ln 2 = 0.693). Write upto two digits after the decimal point.

Answer 1

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