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In Young's experiment interference bands...

In Young's experiment interference bands were produced on a screen placed at 150 cm from two slits, `0.015` mm apart and illuminated by the light of wavelength `6500Å` . Calculate the fringe width.

Text Solution

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Given ` D=150 cm 150xx10^(2-) m , d=0.15 mm =0.15xx10^(-3)m, lambda =6500 Å6500xx10^(-10)m, beta=?`
`:. Beta =(lambdaD)/(d)`
`:. Beta =(6500xx10^(-10)xx150xx10^(-2))/(0.15xx10^(-3))`
`beta =(65xx15)/(15)xx10^(-12)xx10^(5)xx10^(3)`
`beta =65xx10^(7+3)`
`beta =65xx10^(-4),`
`beta=6.5xx10^(-3)m`
`beta =6.5m`
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In Young's experiment, interference bands are produced on the screen placed at 1.5 m from the two slits 0.15 mm apart and illuminated by light of wavelength 6599 Å . Find (a) the fringe width and (b) the change in the fringe width if the screen is taken away from the slit by 50 cm.

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Knowledge Check

  • In Young's experiment inteference bands are produced on the screen placed at 1.5m from the slits 0.15mm apart and illuminated by light of wavelength 6000 Å . If the scren is now taken away from the slit by 50 cm the change in the fringe width will be

    A
    `2xx10^(-4)m`
    B
    `2xx10^(-3)m`
    C
    `6xx10^(-3)m`
    D
    `7xx10^(-3)m`

  • 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`

  • In Young's experiment, interference fringes are obtained on a screen placed at some distance from the slits. When the screen is moved towards the slits by 5xx10^(-2)m , the fringe width is changed by 3.5xx10^(-5)m . If the separation between the slits is 1 mm, then th wavelength of light used is

    A
    4000 Å
    B
    5000 Å
    C
    6000 Å
    D
    7000 Å

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