Question:

In the Young's double slit experiment, the interference pattern is found to have an intensity ratio between the bright and dark fringes as 9, This implies that

Updated On: Jun 14, 2022
  • the intensities at the screen due to the two slits are 5 units and 4 units respectively
  • the intensities at the screen due to the two slits are 4 units and 1 unit respectively
  • the amplitude ratio is 3
  • the amplitude ratio is 2
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The Correct Option is D

Solution and Explanation

$\frac{I_{max}}{I_{min}}=\frac{(\sqrt {I_1}+\sqrt {I_2})^2}{(\sqrt {I_1}-\sqrt {I_2})^2}=\bigg(\frac{\sqrt{I_1 / I_2}+1}{\sqrt{I_1 / I_2}-1}\bigg)^2=9$ (Given)
Solving this, we have $\frac{I_1}{I_2}=4 \, \, but \, \, \, I \propto A^2$
$\therefore \hspace25mm \frac{A_1}{A_2}=2$
$\therefore$ Correct options are (b) and (d).
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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.

Read More: Young’s Double Slit Experiment