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In Young's double slit experiment, dista...

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

`0.20^@`

D

`0.32^@`

Text Solution

Verified by Experts

The correct Answer is:
B
`dsin theta = lambda/2`
:. `Theta = sin^-1(lambda/(2d))= sin^-1((5460 xx 10^-10)/(2xx 0.1 xx 10^-3))`
`= 0.16^@` .
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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
    `0.16^(@)`
    C
    `0.20^(@)`
    D
    `0.32^(@)`

  • In Young's double slit experiment, the distance between sources is 1 mm and distance between the screen and source is 1 m . If the fringe width on the screen is 0.06 cm , then lambda =

    A
    `6000 Å`
    B
    `4000 Å`
    C
    `1200 Å`
    D
    `2400 Å`

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    B
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