Question:

The slits in Young?? double slit experiment are illuminated by light of wavelength $6000 ?$. If the path difference at the central bright fringe is zero, what is the path difference for light from the slits at the fourth bright fringe?

Updated On: Jul 7, 2022
  • $2.4\times 10^{-6}m$
  • $1.2\times 10^{-6}m$
  • $10^{-6}m$
  • $ 0.5 \times 10^{-6}m$
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The Correct Option is A

Solution and Explanation

The path difference at the centre is zero. The path difference at the $4^{th}$ bright fringe $= 4\lambda$ Here, $\lambda = 6000 \times 10^{-10}\, m$ $\therefore$ Path difference $= 4 \times 6000 \times 10^{-10} $ $= 24 \times 10^{-7}$ $= 2.4 \times 10^{-6}\, m$
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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