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 wavelengthsλ,

(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

Expert verified

(a) The path difference of the two rays introduced in the mirror setup is 2d.

(b) The path difference should be equal to λ2 for the rays to be exactly out of phase when they emerge from the mirror setup.

(c) The smallest value ofd that will allow that final phasedifference is λ4.

Step by step solution

01

Given data

Two light rays each of wavelength λare 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.

02

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

(a) From the figure the left ray travels a path equal to5dinside the mirror setup. The right ray travels a path equal to3dinside 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 toλ2 .

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

2d=λ2d=λ4

Thus, the required value of is λ4.

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Most popular questions from this chapter

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