Question

In the Young’s double slit experiment using a monochromatic light of wavelength\[\lambda \], the path difference (in terms of an integer n) corresponding to any point having half peak intensity is
A. \[\left( 2n+1 \right)\dfrac{\lambda }{2}\]
B. \[\left( 2n+1 \right)\dfrac{\lambda }{4}\]
C. \[\left( 2n+1 \right)\dfrac{\lambda }{8}\]
D. \[\left( 2n+1 \right)\dfrac{\lambda }{16}\]

Answer 1

Hint: In this question we are asked to calculate the path difference for a light which has half the intensity of its maximum intensity. We have an equation relating phase difference and path difference. Therefore, we shall calculate the phase difference for the given monochromatic light. To calculate the phase difference, we will use the equation for intensity.
Formula Used: \[I={{I}_{o}}{{\cos }^{2}}\dfrac{\Delta \phi }{2}\]
Where,
I is the intensity
\[{{I}_{o}}\] is the maximum intensity
\[\Delta \phi \] is the phase difference
\[\Delta x=\dfrac{\lambda }{2\pi }\Delta \phi \]
Where,
\[\lambda \]is the wavelength
\[\Delta x\] is the path difference

Complete answer:
From the equation of intensity
We know,
 \[I={{I}_{o}}{{\cos }^{2}}\dfrac{\Delta \phi }{2}\]
Now, it is said that the intensity of the monochromatic light is half the maximum intensity i.e. \[I=\dfrac{{{I}_{o}}}{2}\]
Therefore,
\[\dfrac{{{I}_{o}}}{2}={{I}_{o}}{{\cos }^{2}}\dfrac{\Delta \phi }{2}\]
On solving we get,
\[\dfrac{1}{2}={{\cos }^{2}}\dfrac{\Delta \phi }{2}\]
Therefore, after taking root
\[\cos \dfrac{\Delta \phi }{2}=\pm \dfrac{1}{\sqrt{2}}\]
Now, from trigonometric ratios we know that,
\[\cos \left( 2n+1 \right)\dfrac{\pi }{4}=\pm \dfrac{1}{\sqrt{2}}\]
Therefore,
\[\dfrac{\Delta \phi }{2}=\left( 2n+1 \right)\dfrac{\pi }{4}\]
Therefore,
\[\Delta \phi =\left( 2n+1 \right)\dfrac{\pi }{2}\]
Now, we also know that phase difference and path difference is given by,
\[\Delta x=\dfrac{\lambda }{2\pi }\Delta \phi \]
After substituting the value of \[\Delta \phi \] in above equation
We get,
\[\Delta x=\dfrac{\lambda }{2\pi }\times \left( 2n+1 \right)\dfrac{\pi }{2}\]
Therefore,
\[\Delta x=\left( 2n+1 \right)\dfrac{\lambda }{4}\]

Therefore, the correct answer is option B.

Note:
Phase difference is used to describe the difference in degrees or radians when more than two alternating waves reach their peak or zero value. It is also known as the maximum possible value of two alternating waves having the same frequency. Path difference is the difference in measured distance travelled by two waves from their source to a given point.