Suppose that the two waves in Fig. 35-4 have a wavelength $\mathit{\lambda }\mathbf{=}\mathbf{500}\mathbf{\text{nm}}$in air. What multiple of $\mathit{\lambda }$gives their phase difference when they emerge if (a) ${\mathit{n}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}$, ${\mathit{n}}_{\mathbf{2}}\mathbf{=}\mathbf{16}$and $\mathit{L}\mathbf{}\mathbf{=}\mathbf{}\mathbf{8}\mathbf{.}\mathbf{50}\mathit{\mu }\mathit{m}$; (b) ${\mathit{n}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{62}$, ${\mathit{n}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{72}$, and $\mathit{L}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{50}\mathit{\mu }\mathit{m}$; and (c) ${\mathit{n}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{59}$, ${\mathit{n}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{79}$, and $\mathit{L}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{25}\mathit{\mu }\mathit{m}$? (d) Suppose that in each of these three situations, the waves arrive at a common point (with the same amplitude) after emerging. Rank the situations according to the brightness the waves produce at the common point.

Wavelength is the difference between two successive peaks of waves in the same phase. Wavelength is dependent on the medium through which the electromagnetic wave is passing through. The waves which have higher frequency have a shorter wavelength.

a.

The formulae for calculating the phase difference are:

$K_{\varphi }=\frac{L}{\lambda }\left[n_2-n_1\right]$

Here, $\lambda$the light's wavelength and $n_2$,$n_1$the refractive index of the two mediums.

Here, it is given that $n_2=1.60$, $n_1=1.60$and $L=8.50\mu m$.

Therefore the phase difference can be calculated as:

$$\begin{gathered} K_{\varphi }=\frac{8.50\times {10}^{-6}\text{m}}{500\times {10}^{-9}\text{m}}\left[1.60-1.50\right] \\ =1.70 \end{gathered}$$.

b.

Here, it is given that $n_2=1.72$, $n_1=1.62$and $L=8.50\mu m$.

Therefore the phase difference can be calculated as:

$$\begin{gathered} K_{\varphi }=\frac{8.50\times {10}^{-6}\text{m}}{500\times {10}^{-9}\text{m}}\left[1.72-1.62\right] \\ =1.70 \end{gathered}$$.

c.

Here, it is given that $n_2=1.79$, $n_1=1.59$and $L=8.50\mu m$.

Therefore the phase difference can be calculated as:

$$\begin{gathered} K_{\varphi }=\frac{3.25\times {10}^{-6}\text{m}}{500\times {10}^{-9}\text{m}}\left[1.79-1.59\right] \\ =1.30 \end{gathered}$$

d.

It is known that when the phase difference is 0.5, the two light waves are destructive, meaning the two waves are completely out of phase. Hence, there will be a dark spot at the common point of the two waves.When the phase difference is 1, then two light waves are constructive, which means the two waves are in phase and produce the brightest spot at the common point; therefore, all the phase differences near 1 will be brighter than the phase differences which are away from 1, therefore as in step (4) the phase difference is 1.3 and for step (2) and step (3), the phase difference is 1.70. Hence, as 1.3 is nearer to 1, the common point illumination will be the brightest compared 1.70. Therefore, situation (c) will produce the brightest illumination at the common point than situations (b) and (a).