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Light of wavelength 6328 Å is incident n...

Light of wavelength 6328 Å is incident normally on a slit of width 0.2 mm. Calculate the angular width of central maximum on a screen distance 9 m?

Text Solution

Verified by Experts

The correct Answer is:
`3 xx 10^(-3)m`
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Knowledge Check

  • Light of wavelength 6328 Å is incident normally on a slit of width 0.2 mm. Angular width of the central maximum on the screen will be :

    A
    `0.9^(@)`
    B
    `0.18^(@)`
    C
    `0.54^(@)`
    D
    `0.36^(@)`

  • The light of wavelength 6328Å is incident on a slit of width 0.2mm perpendicularly, the angular width of central maxima will be

    A
    (a) `0.36^@`
    B
    (b) `0.18^@`
    C
    (c) `0.72^@`
    D
    (d) `0.09^@`

  • Light of wavelength 600 nm is incident normally on a slit of width 0.2 mm. The angular width of central maxima in the diffraction pattern is

    A
    `6xx10^(-3)` rad
    B
    `4xx10^(-3)` rad
    C
    `2.4xx10^(-3)` rad
    D
    `4.5xx10^(-3)` rad

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    Light of wavelength 600 nm is incident normally on a slit of width 3 mm . Calculate linear width of central maximum on a screen kept 3 m away from the slit.

    The light of wavelength 6000 overset(*)A falls normally on a slit of width 2 mm. Calculate the linear width of central maxima on a screen kept 4 m away from the slit.

    Light of wavelength 6000 Å is incident normally on a single slit of width 0.03 mm. Find the width of the central maxima on the screen which is at a distance 1.5 m away from the slit. What will be the width if the apparatus is immersed in water of refractive index 1.33?

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