Question

In Fig. 35-23, three pulses of light— $\mathit{a}$, $\mathit{b}$, and $\mathit{c}$—of the same wavelength are sent through layers of plastic having the given indexes of refraction and along the paths indicated. Rank the pulses according to their travel time through the plastic layers, greatest first.

Short answer

The first wave takes the largest time to travel through the plastic layers followed by the third wave and the second wave.

Step-by-step solution

Given data:

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$1.55$ .

Dependence of velocity of light on the refractive index:

$\mathit{v}$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.

Determining the time taken by the three pulses to cross the layers:

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$t_a>t_c>t_b$ .