91Ó°ÊÓ

Refractive index of the medium through which pulse $a$ passes is$1.6$.

Refractive index of the medium through which pulse$b$passes is$1.5$.

Refractive index of the medium through which pulse $c$ passes is$1.55$ .

$\mathit{v}$The velocity of light in a medium having a refractive index$\mathit{c}$is given by

$\mathit{v}\mathbf{=}\frac{\mathbf{c}}{\mathbf{n}}$ .....(1)

Here, $\mathit{c}$is the velocity of light in a vacuum, is the velocity of light in the substance,and $\mathit{n}$ in the index of refraction.

From equation (1), the velocity of pulse $a$ in the first layer is

$V_a=\frac{c}{1.6}$

Let the thickness of each layer be$d$ . Thus, the time taken by the first pulse to cross the layer is

$\begin{array}{l}{t}_a=\frac{d}{c/1.6}\\ =\frac{1.6d}{c}\end{array}$

Similarly, the time taken by the second pulse to cross the layer is,

$t_b=\frac{1.5d}{c}$

And the time taken by third pulse to cross the layer is,

$t_c=\frac{1.55d}{c}$

Hence, the rank of the pulses according to their travel time through the plastic layers, greatest first is$t_a>t_c>t_b$ .