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In the figure is shown Young's double sl...

In the figure is shown Young's double slit experiment. Q is the position of the first bright fringe on the right side of O. P is the `11^(th)` bright fringe on the other side, as measured from Q. If the wavelength of the light used is `600 nm`. Then `S_(1)B` will be equal to

A

`6xx10^(-6) m`

B

`6.6xx10^(-6) m`

C

`3.138xx10^(-7) m`

D

`3.144xx10^(-7) m`

Text Solution

Verified by Experts

The correct Answer is:
A
`P` is `11^(th)` fringe from `Q` i.e. `10^(th)` fringe from `O`.
For `10^(th)` bright fringe, `Deltax=nlambda=10lambda=S_(1)B`
`S_(1)B=10lambda=10xx600xx10^(-9)=6xx10^(-6)m`
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Knowledge Check

  • In Young’s double slit experiment, the central bright fringe can be identified

    A
    By using white light instead of monochromatic light
    B
    As it is narrower than other bright fringes
    C
    As it is wider than other bright fringes
    D
    As it has a greater intensity than the other bright fringes

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    A
    `10^(-2)`
    B
    `3xx10^(-2)`
    C
    `2xx10^(-2)`
    D
    `1.5xx10^(-2)`

  • In a Young's double slit experiment, fringe width equal to 1 mm is observed. Then the distance of the nearest bright fringe from the central fringe will be :

    A
    1 mm
    B
    `0.5 mm`
    C
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    D
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