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In young's double slit experiment the di...

In young's double slit experiment the distance between two sources is`0.1//pimm`.the distance of the screen from the source is `25cm`.wavelength of light used is `5000Å`.Then the angular position of the first dark fringe is-

A

`0.10^(@)`

B

`0.15^(@)`

C

`0.30^(@)`

D

`0.45^(@)`

Text Solution

Verified by Experts

The angular position
`theta =(beta)/(D)=(lambda)/(d) (therefore beta=(lambdaD)/(d))`
The first dark frigue will be at half the fringe width from the mid point of central maximum.thsu the angular position of first dark fringe will be -
`alpha=(theta)/(2)=(1)/(2)[(lambda)/(d)] alpha=(1)/(2)[(5000xxpi)/(.1xx10^(-3))xx10^(-10)](180)/(pi)`
`alpha=0.45^(@)`
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Knowledge Check

  • In Young double slit interference experiment, the distance between two sources is 0.1 mm. The distance of the screen fr0m the sources is 20cm. Wavelength of light used is 5460Å . Then the angular position of the first dark fringe is

    A
    `0.08^(@)`
    B
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    D
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  • In Young's double slit experiment, distance between two sources is 0.1mm. The distance of screen from the sources is 20cm. Wavelength of light used is 5460 Å . Then, angular position of first dark fringe is approximately

    A
    `0.08^@`
    B
    `0.16^@`
    C
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    D
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  • In Young's double slit experiment, the distance between the two slits is 0.1 mm, the distance between the slits and the screen is 1 m and the wavelength of the light used is 600 nm. The intensity at a point on the screen is 75% of the maximum intensity. What is the smallest distance of this point from the central fringe ?

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