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In the Young's double slit experiment, t...

In the Young's double slit experiment, the intensities at two points `P_(1)` and `P_(2)` on the screen are respectively `I_(1)` and `I_(2)` If `P_(1)` is located at the centre of a bright fringe and `P_(2)` is located at a distance equal to a quarter of fringe width from `P_(1)` then `I_(1)/I_(2)` is
a.2 b.3 c.4 d.None of these

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Knowledge Check

  • In a Young's double slit experiment, let S_(1) and S_(2) be the two slits, and C be the centre of the screen. If angle S_(1)CS_(2)=theta and lambda is the wavelength, the fringe width will be

    A
    `(lambda)/(theta)`
    B
    `lambda theta`
    C
    `(2lambda)/(theta)`
    D
    `(lambda)/(2theta)`

  • In Young's double slit experiment, the slits are horizontal. The intensity at a point P as shown in figure is (3)/(4) I_(0) ,where I_(0) is the maximum intensity. Then the value of theta is, (Given the distance between the two slits S_(1) and S_(2) is 2 lambda )

    A
    `"cos"^(-1)((1)/(12))`
    B
    `"sin"^(-1)((1)/(12))`
    C
    `"tan"^(-1)((1)/(12))`
    D
    `"sin"^(-1)((3)/(5))`

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