The Frozen Lake

Since the Refractive Index is a linear function of distance,
If $x$ represents a general depth inside the ice,
Then, by functional analysis, we have,

$\mu = 1.5\left(1+\frac{x}{h}\right)$

Now, By the real definition of Refractive Index, we know,

Speed of Light in a Medium ( $v$ ) $= \frac{c}{\mu}$ , where $c$ is the speed of light in vacuum .

Therefore,
$v = \frac{dx}{dt} = \frac{2c}{3} \frac{h}{h+x}$

Rearranging,
$\left(h+x\right) dx = \frac{2ch}{3} dt$

Integrating both sides,
$\int_0^h \left(h+x\right) dx = \int_0^t \frac{2ch}{3} dt$

Note that $t$ here, is only half the time. So, we will have to double it once we are finished.

$\frac{3h^2}{2} = \frac{2ch}{3} t$

$\Rightarrow t = \frac{9h}{4c}$

Doubling $t$ ,
$\Rightarrow T = 2t = \frac{9h}{2c} = 5\times 10^{-6}$ sec (as given)

Simplifying, and using the values given,
$h = 333.1 \approx \boxed{333m}$

Since refractive index is linearly varying so we can find out mid point ((1.5+3)/2)and assume to be constant over height "h". Since we are observing the ray after 2T time where T denotes total time required before 2nd reflection rest is simple distance-speed calculation

$T = 5. 10^{-6} s$
$t_{1} + t_{2} = 5. 10^{-6}$
$h/v_{1} + h/v_{2} = 5. 10^{-6}$
$h(1/v_{1} + 1/v_{2}) = 5. 10^{-6}$
$h(µ_{1}/c + µ_{2}/c) = 5. 10^{-6}$
$h(µ_{1} + µ_{2}) = 5. 10^{-6}.c$
$h(1.5 + 3) = 5. 10^{-6}.c$
$h = 5. 10^{-6}. 2.99792458 . 10^{8} / 4.5$
$= 3.33 . 10^{2} m$
$= 333.33 m$

As R.I of ice is changing linearly , we can take its R.I as mean of 1.5 and 3 .i.e R.I=2.25. Therefore speed of light in ice is 1.33*10^2 m/s. Thus thickness is 332.5 .e.i=333 metres.