Question:

In the adjacent diagram, CP represents a wavefront and AO and iBP, the corresponding two rays. Find the condition of $\theta$ for constructive interference at P between the ray BP and reflected ray OP

Updated On: Jun 14, 2022
  • $cos \, \theta=\frac{3 \lambda}{2d}$
  • $cos \, \theta=\frac{\lambda}{4d}$
  • $sec \theta-cos \, \theta=\frac{\lambda}{d}$
  • $sec \theta-cos \, \theta=\frac{4 \lambda}{d}$
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The Correct Option is B

Solution and Explanation

PR=d
$\therefore \hspace10mm PO=d \, sec \theta$
and $\hspace10mm CO=PO cos \, 2 \, \theta=d \, sec \, \theta cos \, 2 \, \theta$
path difference between the two rays is,
$ \, \, \, \, ? x=PO+OC=(d \, sec \theta+d \, sec \, \theta \, cos \, 2 \, \theta)$
phase difference between the two rays is
$? \phi =\pi$(one is reflected, while another is direct)
Therefore, condition for constructive interference should be
$\hspace20mm ? x=\frac{\lambda}{2},\frac{3 \lambda}{2}...$
or $\hspace10mm d \, sec \, \theta (1+ cos \, 2 \, \theta)=\frac{\lambda}{2}$
or $\hspace15mm \big(\frac{d}{cos \theta}\big)(2 cos^2 \theta)=\frac{\lambda}{2}$
or $\hspace30mm cos \theta=\frac{\lambda}{4d}$
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Concepts Used:

Wave interference

What is Interference?

When two or more waves meet each other then interference happens . Interference is a phenomenon in which 2 or more waves superpose to form a resultant wave of greater, lower or the same amplitude.

There are two types of wave interference:

The principle of superposition of waves refers that when two or more waves of the same type are incident on the same point, the resultant amplitude at that point is equal to the vector sum of the amplitudes of the individual waves. If the crest of a wave meets the crest of another wave of the same frequency at the same point,  sum of individual amplitudes is called as constructive interference.The destructive interference occurs when the maxima of the two waves are at 180 degrees out of phase and a positive displacement of one wave is cancelled exactly by a negative displacement of the other wave.