Refraction through Plane Surface and Glass Slab Questions in English

(A) Consider an elemental strip of thickness $dz$ at a height $z$ from the bottom of the cube.
The apparent thickness $dh$ of this elemental strip is given by:
$dh = \frac{dz}{\mu(z)} = \frac{dz}{1 + \frac{z}{t}} = \frac{t}{t + z} dz$
The total apparent depth $h'$ of the cube as seen from the top is the integral of $dh$ from $z = 0$ to $z = t$:
$h' = \int_0^t \frac{t}{t + z} dz = t [\ln(t + z)]_0^t$
$h' = t [\ln(2t) - \ln(t)] = t \ln\left(\frac{2t}{t}\right) = t \ln 2$
The shift in the position of the printed letters is the difference between the real thickness $t$ and the apparent thickness $h'$:
$\text{Shift} = t - h' = t - t \ln 2 = (1 - \ln 2) t$

(D) Initially,with the empty tank,the object distance $u = -45 \ cm$ and the image distance $v = +36 \ cm$. Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{36} - \frac{1}{-45} = \frac{1}{36} + \frac{1}{45} = \frac{5+4}{180} = \frac{9}{180} = \frac{1}{20} \Rightarrow f = 20 \ cm$.
When the liquid is poured to a depth of $40 \ cm$,the apparent depth of the mark is $d' = \frac{40}{\mu}$. The remaining distance from the lens to the liquid surface is $45 - 40 = 5 \ cm$. Thus,the new object distance is $u' = -(\frac{40}{\mu} + 5) \ cm$. The new image distance is $v' = +48 \ cm$.
Using the lens formula again: $\frac{1}{f} = \frac{1}{v'} - \frac{1}{u'}$
$\frac{1}{20} = \frac{1}{48} - \frac{1}{-(\frac{40}{\mu} + 5)} = \frac{1}{48} + \frac{1}{\frac{40}{\mu} + 5}$
$\frac{1}{\frac{40}{\mu} + 5} = \frac{1}{20} - \frac{1}{48} = \frac{12 - 5}{240} = \frac{7}{240}$
$\frac{40}{\mu} + 5 = \frac{240}{7} \approx 34.286$
$\frac{40}{\mu} = 34.286 - 5 = 29.286$
$\mu = \frac{40}{29.286} \approx 1.366$.

(C) The thickness of the glass plate is given as $t$.
The refractive index of the glass is $\mu$.
The velocity of light in vacuum is $c$.
The velocity of light in a medium $(v)$ is related to the refractive index $(\mu)$ and the velocity of light in vacuum $(c)$ by the formula: $\mu = \frac{c}{v}$.
Therefore,the speed of light in the glass plate is $v = \frac{c}{\mu}$.
The time $(T)$ taken to travel a distance $(t)$ at a constant speed $(v)$ is given by $T = \frac{\text{distance}}{\text{speed}} = \frac{t}{v}$.
Substituting the value of $v$: $T = \frac{t}{c / \mu} = \frac{\mu t}{c}$.
Thus,the time taken is $\frac{\mu t}{c}$.

(A) Given: Velocity of light in vacuum $(c)$ = $3 \times 10^{8} \text{ m/s}$,thickness of glass plate $(d)$ = $10 \text{ cm} = 0.1 \text{ m}$,and refractive index $(\mu)$ = $1.5$.
The velocity of light in the glass plate $(v)$ is given by $v = \frac{c}{\mu} = \frac{3 \times 10^{8}}{1.5} = 2 \times 10^{8} \text{ m/s}$.
The time taken $(t)$ to travel through the glass plate is $t = \frac{d}{v}$.
Substituting the values: $t = \frac{0.1 \text{ m}}{2 \times 10^{8} \text{ m/s}} = 0.05 \times 10^{-8} \text{ s} = 0.5 \times 10^{-9} \text{ s}$.
Since $1 \text{ nanosecond} = 10^{-9} \text{ s}$,the time taken is $0.5 \text{ ns}$.