Question

Figure 35-22 shows two light rays that are initially exactly in phase and that reflect from several glass surfaces. Neglect the slight slant in the path of the light inthe second arrangement.

(a) What is the path length difference of the rays?

In wavelengths$\mathit{\lambda }$,

(b) what should that path length difference equal if the rays are to be exactly out of phase when they emerge, and

(c) what is the smallest value of that will allow that final phase difference?

Short answer

(a) The path difference of the two rays introduced in the mirror setup is $\mathbf{2}\mathit{d}$.

(b) The path difference should be equal to $\frac{\mathbf{\lambda }}{\mathbf{2}}$ for the rays to be exactly out of phase when they emerge from the mirror setup.

(c) The smallest value of$\mathit{d}$ that will allow that final phasedifference is $\frac{\mathbf{\lambda }}{\mathbf{4}}$.

Step-by-step solution

Given data

Two light rays each of wavelength $\lambda$are reflected in multiple mirrors each at a distance$$" width="9" height="19" role="math">

The path difference between two rays is the actual difference in length of the paths of the two rays. For the two rays to be out of phase the path difference should be half of the wavelength of the rays.

Step 3:Determining the path difference and the value of  d

(a) From the figure the left ray travels a path equal to$5d$inside the mirror setup. The right ray travels a path equal to$3d$inside the mirror setup. Hence their path difference is

$5d-3d=2d$

Thus, their path difference is $2d$.

(b) For the two rays to be out of phase after exiting the mirror setup, their path difference should be equal to$\frac{\lambda }{2}$ .

(c) The value of that will allow this phase difference is obtained by

$\begin{array}{l}2d=\frac{\lambda }{2}\\ d=\frac{\lambda }{4}\end{array}$

Thus, the required value of is $\frac{\lambda }{4}$.