Question:
In Young's double slit experiment intensity at a point is (1/4)
of the maximum intensity. Angular position of this point is
In Young's double slit experiment intensity at a point is (1/4)
of the maximum intensity. Angular position of this point is
Updated On: Apr 22, 2024
- $sin^{-1}\big(\frac{\lambda}{d}\big)$
- $sin^{-1}\big(\frac{\lambda}{2d}\big)$
- $sin^{-1}\big(\frac{\lambda}{3d}\big)$
- $sin^{-1}\big(\frac{\lambda}{4d}\big)$
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The Correct Option is C
Solution and Explanation
$\hspace25mm I=I_{max}cos^2 \big(\frac{\phi}{2}\big)$
$\therefore \hspace20mm \frac{I_{max}}{4}=I_{max}cos^2 \frac{\phi}{2}$
$\hspace20mm cos \frac{\phi}{2}=\frac{1}{2}$
or $\hspace20mm \frac{\phi}{2}=\frac{\pi}{3}$
$\therefore \hspace20mm \phi =\frac{2 \pi}{3}=\big(\frac{2 \pi}{\lambda}\big)\Delta x \hspace20mm$ ...(i)
where $\hspace10mm \Delta x=d \, sin \, \theta$
Substituting in E (i), we get
$\hspace25mm sin \, \theta=\frac{\lambda}{3d}$
or $\hspace 25mm \theta=sin^{-1}\big(\frac{\lambda}{3d}\big)$
$\therefore$ Correct answer is (c).
$\therefore \hspace20mm \frac{I_{max}}{4}=I_{max}cos^2 \frac{\phi}{2}$
$\hspace20mm cos \frac{\phi}{2}=\frac{1}{2}$
or $\hspace20mm \frac{\phi}{2}=\frac{\pi}{3}$
$\therefore \hspace20mm \phi =\frac{2 \pi}{3}=\big(\frac{2 \pi}{\lambda}\big)\Delta x \hspace20mm$ ...(i)
where $\hspace10mm \Delta x=d \, sin \, \theta$
Substituting in E (i), we get
$\hspace25mm sin \, \theta=\frac{\lambda}{3d}$
or $\hspace 25mm \theta=sin^{-1}\big(\frac{\lambda}{3d}\big)$
$\therefore$ Correct answer is (c).
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Concepts Used:
Young’s Double Slit Experiment
- Considering two waves interfering at point P, having different distances. Consider a monochromatic light source ‘S’ kept at a relevant distance from two slits namely S1 and S2. S is at equal distance from S1 and S2. SO, we can assume that S1 and S2 are two coherent sources derived from S.
- The light passes through these slits and falls on the screen that is kept at the distance D from both the slits S1 and S2. It is considered that d is the separation between both the slits. The S1 is opened, S2 is closed and the screen opposite to the S1 is closed, but the screen opposite to S2 is illuminating.
- Thus, an interference pattern takes place when both the slits S1 and S2 are open. When the slit separation ‘d ‘and the screen distance D are kept unchanged, to reach point P the light waves from slits S1 and S2 must travel at different distances. It implies that there is a path difference in the Young double-slit experiment between the two slits S1 and S2.
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