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Two slits in YDSE are placed 2 millimete...

Two slits in YDSE are placed 2 millimeter from each other. Interference pattern is observed on a screen placed 2m from the plane of slits. What is the fringe width for a light of wavelength 400 nm ?

Text Solution

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`beta=(lambdaD)/d=0.4 mm`
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Two slits in YDSE are placed 2 millimetre from each other. Interference pattern is observedon a screen placed 2 m from the plane of slits. What is the fringe width for a light of wavelength 400 nm?

Two slits in YDSE are placed 1 mm from each other. Interference pattern is observed on a screen placed 1m from the plane of slits. What is the angular fringe width for a light of wavelength 400 nm

Knowledge Check

  • In Young's double slit experiment the two slits are d distance apart. Interference pattern is observed on a screen at a distance D from the slits. A dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light is

    A
    `(D^(2))/(2d)`
    B
    `(d^(2))/(2D)`
    C
    `(D^(2))/(d)`
    D
    `(d^(2))/(D)`

  • In Young's double slit experiment, the slits are 0.5mm apart and interference pattern is observed on a screen placed at a distance of 1.0m from the plane containg the slits. If wavelength of the incident light is 6000Å , then the separation between the third bright fringe and the central maxima is

    A
    (a) `4.0mm`
    B
    (b) `3.5mm`
    C
    (c) `3.0mm`
    D
    (d) `2.5mm`

  • Two slits, 4 mm apart, are illuminated by light of wavelength 6000 Å . What will be the fringe width on a screen placed 2 m from the slits

    A
    `0.12 mm `
    B
    `0.3 mm`
    C
    `3.0 mm`
    D
    `4.0 mm`

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