# Two rays of light are incident normally on a tub of height H filled with water. They pass through glass slabs of heights h_1 and h_2. If the speed of light in vacuum is c, what is the time difference between the rays of light to reach the bottom?

## Refractive indices of water and glass are $\frac{4}{3}$ and $\frac{3}{2}$ respectively.

Nov 24, 2016

The situation described in the problem is shown in the above figure
As the rays of light are incident normally , the rays will refracted through the media without deviation.

For slab of height ${h}_{1}$

the total path length of water and glass =H

Path length of glass medium$= {h}_{1}$

Then path length of water medium$= H - {h}_{1}$

For slab of height ${h}_{2}$

the total path length of water and glass =H

Path length of glass medium$= {h}_{2}$

Then path length of water medium$= H - {h}_{2}$

We know refractive index of water

mu_w=4/3="velocity of light in vacuum"/("velocity of light in water"(v_w))

$\implies \frac{4}{3} = \frac{c}{v} _ w \implies {v}_{w} = \frac{3 c}{4}$

Again we know refractive index of glass

mu_g=3/2="velocity of light in vacuum"/("velocity of light in glass"(v_g))

$\implies \frac{3}{2} = \frac{c}{v} _ g \implies {v}_{g} = \frac{2 c}{3}$

Calculation of time to reach $\textcolor{red}{r a {y}_{1}}$ at the bottom from surface

So time taken by the light $r a {y}_{1}$ to pass through water of path length $= H - {h}_{1}$ is given by

${t}_{w} = \frac{H - {h}_{1}}{v} _ w = \frac{4 \left(H - {h}_{1}\right)}{3 c}$

And time taken by the light $r a {y}_{1}$ to pass through glass of path length ${h}_{1}$ is given by

${t}_{g} = {h}_{1} / {v}_{g} = \frac{3 {h}_{1}}{2 c}$

Hence total time ${t}_{1}$ to reach $\textcolor{red}{r a {y}_{1}}$ at the bottom from surface

$\textcolor{b l u e}{{t}_{1} = {t}_{w} + {t}_{g}}$

Calculation of time to reach $\textcolor{red}{r a {y}_{2}}$ at the bottom from surface

Now time taken by the light $r a {y}_{2}$ to pass through water of path length $= H - {h}_{2}$ is given by

$t {'}_{w} = \frac{H - {h}_{2}}{v} _ w = \frac{4 \left(H - {h}_{2}\right)}{3 c}$

And time taken by the light $r a {y}_{2}$ to pass through glass of path length ${h}_{2}$ is given by

$t {'}_{g} = {h}_{2} / {v}_{g} = \frac{3 {h}_{2}}{2 c}$

Hence total time ${t}_{1}$ to reach $\textcolor{red}{r a {y}_{2}}$ at the bottom from surface

$\textcolor{b l u e}{{t}_{2} = t {'}_{w} + t {'}_{g}}$

Hence the time difference between the two rays of light to reach the bottom is given by

$\Delta t = \left\mid {t}_{1} - {t}_{2} \right\mid = \left\mid \left({t}_{w} + {t}_{g}\right) - \left(t ' w + t ' g\right) \right\mid$

$= \left\mid \frac{4 \left(H - {h}_{1}\right)}{3 c} + \frac{3 {h}_{1}}{2 c} - \frac{4 \left(H - {h}_{2}\right)}{3 c} - \frac{3 {h}_{2}}{2 c} \right\mid$

$= \left\mid \frac{8 H - 8 {h}_{1} + 9 {h}_{1} - 8 H + 8 {h}_{2} - 9 {h}_{2}}{6 c} \right\mid$

$= \left\mid \frac{{h}_{1} - {h}_{2}}{6 c} \right\mid$