Question

Two waves of light in air, of wavelength $\mathit{\lambda }\mathbf{=}\mathbf{600}\mathbf{.}\mathbf{0}\mathbf{\text{nm}}$, are initially in phase. They then both travel through a layer of plastic as shown in Fig. 35-36, with ${\mathit{L}}_{\mathbf{1}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathit{\mu }\mathit{m}$, ${\mathit{L}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{50}\mathbf{}\mathit{\mu }\mathit{m}$, ${\mathit{n}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{40}$, ${\mathit{n}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{60}$and. (a) What multiple of $\mathit{\lambda }$gives their phase difference after they both have emerged from the layers? (b) If the waves later arrive at some common point with the same amplitude, is their interference fully constructive, fully destructive, intermediate but closer to fully constructive,or intermediate but closer to fully destructive?

Short answer

a. The multiple of which will give their phase difference after both have emerged from the layer is 0.833.

b. The interference is intermediate but closer to fully constructive.

Step-by-step solution

Definition of phase difference.

The difference in angles between two or more electromagnetic waves from a reference wave is called as phase difference of that wave. The phase difference is of three types leading, lagging and zero phases.

Calculation of the phase difference.

a.

For the phase difference calculation, we have to choose a horizontal x-axis with its origin on the left side of the plastic.

Therefore, between $x=0$and $x=L_2$the phase difference can be calculated by using the formulae,

$K_{{\varphi }_1}=\frac{L_2}{\lambda }\left[n_2-n_1\right]$…..(i)

Between $x=L_2$and$x=L_1$ the phase difference can be calculated by using the formulae,

$K_{{\varphi }_2}=\frac{L_1-L_2}{\lambda }\left[1-n_1\right]$…..(ii)

Since the top ray in the figure is now traversing through the air, the air has a refractive index of 1, so$n_2$ will be.

Therefore, by adding the equation (i) and (ii),

$$\begin{gathered} K_{\varphi }=K_{{\varphi }_1}+K_{{\varphi }_2} \\ =\frac{L_2}{\lambda }\left[n_2-n_1\right]+\frac{L_1-L2}{\lambda }\left[1-n_1\right] \\ =\frac{0.600\mu m}{3.50\mu m}\left(1.60-1.40\right)+\frac{4.00\mu m-3.50\mu m}{0.600\mu m}\left(1-1.40\right) \\ =0.833 \end{gathered}$$

Determination of the nature of the wave.

b.

Consider the multiple of $\lambda$that gives their phase difference $0.5$, the two waves will be completely out of phase or destructive in nature, and hence there will be a formation of dark spots. For the multiple of $\lambda$ that gives their phase difference 1, then the two waves will be in phase or constructive in nature, and hence there will be a formation of bright spot. Since$K_{\varphi }=0.833$ is near to,1 so the interference is more nearly constructive.