Question:

In Young's double-slit experiment, in an interference pattern, the second minimum is observed exactly in front of one slit. The distance between the slits is $\text{d}$ and the distance between source and screen is $\text{D}$ . The wavelength of the light source used is

Updated On: Apr 24, 2024
  • $\frac{\text{d}^{2}}{\text{D}}$
  • $\frac{\text{d}^{2}}{2 \text{D}}$
  • $\frac{\text{d}^{2}}{3 \text{D}}$
  • $\frac{\text{d}^{2}}{4 \text{D}}$
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The Correct Option is C

Solution and Explanation

Distance of second minima from central point is $y=\frac{d}{2}$ . We know that minima is given by, $dsin\theta =\left(n - \frac{1}{2}\right)\lambda \frac{dy}{D}=\left(2 - \frac{1}{2}\right)\lambda $ $\frac{d^{ \, }}{2}=\frac{D}{d}\left(\frac{2 \times 2 - 1}{2}\right)\lambda $ $\therefore \frac{d}{2}=\frac{D}{d}\times \frac{3}{2}\lambda $ $\therefore \lambda =\frac{\text{d}^{2}}{3 \text{D}}$
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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