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Young's Double Slit Experiment

  1. Nov 9, 2007 #1
    Q.Suppose that the the light rays in young's Double Slit experiment fall at an angle [tex] \theta=\sin^{-1}(\frac{\lambda}{2d})[/tex] .Then prove that the the Point P which is the symmetric point on screen between two slits is the centre of central minimum

    Necessary Formulaes
    for minima the waves should interfere destructively so the path difference [tex]\Delta x=\frac{n\lambda}{2}[/tex]

    i am not so sure of this but we can have a plane [tex]\pi,\pi^{'}[/tex] with inclination [tex]\theta[/tex] to the plane of slits and through the slits.the [tex]\perp[/tex] distance of the Point P from the plane determines it's phase so the phase difference is difference between the length of perpendiculars from P to the planes [tex]=d \sin \theta=\frac{\lambda}{2}[/tex] which produces destructive interference.

    My doubt lies on the fact that how do we apply difference in optical lengths concept here
  2. jcsd
  3. Nov 9, 2007 #2


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    When a plane wave front falls on the slits with an inclination theta,such that it satisfies the condition given in the problem, light coming from the two slits are not in phase. There phase difference is pi, i.e. they are out of phase. When the light from these two slits reach point P, even though they are equidistant from the point P, they produce distructive interference. So there is no need to apply difference in optical length concept.
    Last edited: Nov 9, 2007
  4. Nov 9, 2007 #3
    why are they out of phase even though the distance is the same
  5. Nov 9, 2007 #4
    What a confusingly worded question!

    θ could be the angle between the incident rays and either a) the plane of the slits or b) the normal to the plane of the slits. It has to be b) for P to be on the central mimimum.

    This leads to the path length from the source to the furthest slit being 0.5λ longer than to the nearest slit.

    Using this information you can calculate the directions of the maxima and the minima applying the difference in optical lengths concept.
  6. Nov 10, 2007 #5


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    In this question a parallel beam of light is illuminating the slits and the plane wave front, which is perpendicular to the parallel beam, makes an angle θ with the plane of the slits.This leads to the path length from the source to the furthest slit being 0.5λ longer than to the nearest slit. So the waves coming from the two slits are not in phase but out of phase. When these waves superimpose at a point P equidistance from the two slits produce distructive interference.
  7. Nov 10, 2007 #6
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