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In the arrangement shown in the followin...

In the arrangement shown in the following figure, the wavelength of light used is `lambda`. The distance between slits `S_(1)` and `S_(2)` is d (« D). The distance between `S_(3)` and `S_(4)` is `u=lambda D // 3d`. If the ratio of maximum to minimum intensity observed on screen is k. Find k.

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Consider the situation shown in figure-6.35. The two slits S_(1) and S_(2) placed symmetrically around the central line are illuminated by a monochromatic light of wavelength lambda . The separation between the slits is d. The light transmitted by the slits falls on a screen E_(1) placed at a distance D from the slits. The slit S_(3) is at the central line and the slit S_(4) is at a distance z from S_(3) . Another screen E_(2) is placed a further distance D away from E_(1) . Find the ratio of the maximum to minimum intensity observed on E_(2) if z is equal is : (a) (lambdaD)/(2d) " "(b) (lambdaD)/(d)" " (c )(lambdaD)/(4d)

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Knowledge Check

  • Three coherent point sources S_(1),S_(2) and S_(3) are placed on a line perpendicular to the screen as shown in the figure. The wavelength of the light emitted by the sources is lambda. The distance between adjacent sources is d=3lambda The distance of S_(2) from the screen is D (gtgt lambda) . Find the minimum (non zero) distance x of a point P on the screen at which complete darkness is obtained.

    A
    `(2sqrt(2)D)/7`
    B
    `sqrt((17))D/8`
    C
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    D
    `(4sqrt(2)D)/7`
  • In young's double slit experiment, distance between the slit S_1 and S_2 is d and the distance between slit and screen is D . Then longest wavelength that will be missing on the screen in front of S_1 is

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    C
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    D
    None of these
  • In YDSE apparatus shown in figure wavlength of light used is lambda . The screen is moved away form the source with a constant speed v. Initial distance between screen and plane of slits was D. At a point P on the screen the order of fringe will

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    B
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    remain constant
    D
    first increases and then decreases
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