Question:
The maximum intensity of fringes in Youngs experiment is $I$. If one of the slit is closed then the intensity at that place becomes $I_{0}$ . Which of the following relation is true?
The maximum intensity of fringes in Youngs experiment is $I$. If one of the slit is closed then the intensity at that place becomes $I_{0}$ . Which of the following relation is true?
Updated On: Aug 1, 2022
- $I=I_{0}$
- $I=2I_{0}$
- $I=4I_{0}$
- There is no relation between $I$ and $I_{0}$
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The Correct Option is C
Solution and Explanation
Suppose slit width's are equal so they produces waves of equal intensity say I'. Resultant intensity at any point $I_{R}=4 I' \cos ^{2} \phi$ where $\phi$ is the phase difference between the waves at the point of observation.
For maximum intensity, $\phi=0^{\circ}$
$\Rightarrow I_{\max }=4 I'=I$ ...(i)
If one of slit is closed, resultant intensity at the same point will be $I'$ only
ie, $I'=I_{0}$...(ii)
Comparing Eqs. (i) and (ii) we get
$I=4 I_{0}$
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Concepts Used:
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- 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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