In Fig. 35-51a, the waves along rays 1 and 2 are initially in phase, with the same wavelength $\mathit{\lambda }$ in air. Ray 2 goes through a material with lengthLand index of refraction n. The rays are then reflected by mirrors to a common point P on a screen. Suppose that we can varyL from 0 to 2400 nm. Suppose also that, from $\mathit{L}\mathbf{=}\mathbf{0}$ to ${\mathit{L}}_{\mathbf{s}}\mathbf{=}\mathbf{900}\mathbf{}\mathbf{nm}$, the intensity I of the light at point P varies withL as given in Fig. 35-52. At what values of L greater than ${\mathit{L}}_{\mathbf{s}}$is intensity I (a) maximum and (b) zero? (c) What multiple of $\mathit{\lambda }$gives the phase difference between ray 1 and ray 2 at common point P when $\mathit{L}\mathbf{=}\mathbf{1200}\mathbf{}\mathit{n}\mathit{m}$?

The intensity I varies with L from 0 nm to 900 nm.

The quantity I is the wave intensity as a function of the phase difference of two (identical) parent waves. If the two waves are in phase, then the intensity of the combined wave is Io when the two waves are in phase.

The number of the unit of L in graph is 6.

From graph, the intensity is maximum at $L=0$.

The intensity is minimum is at 5 units.

Therefore, the value of L at which intensity is maximum,

$\begin{array}{rcl}\Delta L& =& \frac{900}{6}\times 5\\ & =& 750\text{nm}\end{array}$

Therefore, for $\Delta L=750\text{nm}$, the intensity is minimum.

The length of the material should be equal to $2\Delta L$.

$\begin{array}{rcl}2\Delta L& =& 2\times 750\\ & =& 1500\text{nm}\end{array}$

Hence, at $L=1500\text{nm}$, the intensity is maximum.

The intensity is zero at $L=750\text{nm}$and maximum at $L=1500\text{nm}$. Therefore, the next value of L at which intensity is minimum,

$\begin{array}{rcl}L& =& 1500+750\\ & =& 2250\text{nm}\end{array}$

Hence, the value of L at which the intensity is zero is 2250 nm.

Calculate the multiple of $\lambda$gives the phase difference between ray 1 and ray 2 at common point P when$L=1200\text{nm}$:

For $L=1500\text{nm}$, the phase difference is one wavelength.

Calculate the phase difference at $L=1200\text{nm}$.

$\begin{array}{rcl}PD& =& \frac{1200}{1500}\\ & =& \frac{12}{15}\\ & =& 0.8\text{wavelength}\end{array}$

Hence, the phase difference between the two rays $L=1200\text{nm}$ is $0.8$ wavelength.