91Ó°ÊÓ

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.

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,

$K_{{\mathrm{Ï•}}_1}=\frac{L_2}{\mathrm{λ}}\left[n_2-n_1\right]$…..(¾±)

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

$K_{{\mathrm{Ï•}}_2}=\frac{L_1-L_2}{\mathrm{λ}}\left[1-n_1\right]$…..(¾±¾±)

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),

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

b.

Consider the multiple of $\mathrm{λ}$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 $\mathrm{λ}$ 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$K_{\mathrm{ϕ}}=0.833$ is near to,1 so the interference is more nearly constructive.