In Fig. 35-4, assume that two waves of light in air, of wavelength $\mathbf{400}\mathbf{}\mathbf{\text{nm}}$, are initially in phase. One travels through a glass layer of index of refraction ${\mathit{n}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{60}$and thickness $\mathit{L}$. The other travels through an equally thick plastic layer of index of refraction ${\mathit{n}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}$. (a) What is the smallest value $\mathit{L}$should have if the waves are to end up with a phase difference of 5.65 rad? (b) If the waves 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?

1. The wavelength of two rays of light is,$\lambda =400\text{nm}$ .
2. The index of refraction of glass layer is, $n_1=1.60$.
3. The thickness of glass layer is, $L$.
4. The index of refraction of thick plastic layer is, $n_2=1.50$.
5. The phase difference between two rays is, ${\varphi }_1-{\varphi }_2=5.65\text{rad}$.

The value of the ‘phase difference’ between two different light waves changeswhen the waves travelthrough different mediums having different values of indexes of refraction.

For two mediums having index of refraction $n_1>n_2$, the value of the phase difference between two light waves is given by,

${\varphi }_1-{\varphi }_2=\left(\frac{n_1}{{\lambda }_1}-\frac{n_2}{{\lambda }_2}\right)L$

Here,$\varphi$ is the wave phase, $\lambda$is the wavelength and $L$is the medium length.

We take the phases of both waves to be zero at the front surfaces of the layers.

The phase of the first wave at the back surface of the glass is given by,

The formula for the phase difference $\left({\varphi }_1-{\varphi }_2\right)$between two waves passing through two different medium shaving same medium length is given by,

$$\begin{gathered} {\varphi }_1-{\varphi }_2=\left(\frac{2\pi }{{\lambda }_1}-\frac{2\pi }{{\lambda }_2}\right) \\ {\varphi }_1-{\varphi }_2=2\pi \left(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right) \end{gathered}$$

Putting wavelengths for each wave,${\lambda }_1=\frac{{\lambda }_{air}}{n_1}$ and ${\lambda }_2=\frac{{\lambda }_{air}}{n_2}$,

$\begin{array}{rcl}{\varphi }_1-{\varphi }_2& =& 2\pi \left(\frac{1}{\frac{{\lambda }_{air}}{n_1}}-\frac{1}{\frac{{\lambda }_{air}}{n_2}}\right)L\\ {\varphi }_1-{\varphi }_2& =& 2\pi \left(\frac{n_1}{{\lambda }_{air}}-\frac{n_2}{{\lambda }_{air}}\right)L\\ {\varphi }_1-{\varphi }_2& =& \frac{2\pi }{{\lambda }_{air}}\left(n_1-n_2\right)L\\ L& =& \frac{\left({\varphi }_1-{\varphi }_2\right){\lambda }_{air}}{2\pi \left(n_1-n_2\right)}\end{array}$

Putting values, ${\lambda }_{air}=400\times {10}^{-9}\text{m}$

$$\begin{gathered} L=\frac{5.65\times 400\times {10}^{-9}\text{m}}{2\pi \left(1.60-1.50\right)} \\ L=\text{3596.901}\times {10}^{-9}\text{m} \\ L=3.60\times {10}^{-6}\text{m} \end{gathered}$$

Hence, the smallest value of $L$is $3.60\times {10}^{-6}\text{m}$.

For the completely constructive interference, the phase difference of waves should be in the integer multiple of $2\pi \text{rad}$and for the completely destructive, the phase difference of waves should be equal to $\pi \text{rad}$.

The value of the phase difference between two rays of light is,

$\pi <5.65\text{rad}<2\pi$

So, the interference is closer to the completely constructive than to completely destructive.

Hence, interference is closer to the completely constructive interference.