Question:

Two slits are made one millimeter apart and the screen is placed one metre away. The fringe separation when blue green light of wavelength $500\, nm$ is used is

Updated On: Jul 7, 2022
  • $5 \times 10^{-4}\,m$
  • $2.5 \times 10^{-3}\,m$
  • $2 \times 10^{-4}\,m$
  • $10 \times 10^{-4}\,m$
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The Correct Option is A

Solution and Explanation

Fringe width, $\beta = \frac{D\lambda}{d} $ Here, $D = 1m, d = 1\times10^{-3} m$, $ \lambda = 500 \,nm = 5\times10^{-7} m$ $\therefore\beta = \frac{1\times5\times10^{-7}}{1\times10^{-3}} $ $= 5\times10^{-4} m $
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Notes on Youngs double slit experiment

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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