Question:
Two coherent monochromatic light beams of intensities I and
4I are superposed. The maximum and minimum possible
intensities in the resulting beam are
Two coherent monochromatic light beams of intensities I and
4I are superposed. The maximum and minimum possible
intensities in the resulting beam are
Updated On: Aug 1, 2022
- 5I and I
- 5I and 3 I
- 9I and I
- 9I and 3 I
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
$I_{max}=(\sqrt{I_1}+\sqrt{I_2})^2=(\sqrt{4I}+\sqrt I)^2=9I$
$I_{min}=(\sqrt{I_1}-\sqrt{I_2})^2=(\sqrt{4I}-\sqrt I)^2=I$
$\therefore$ Correct option 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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