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In YDSE, the slits have different widths...

In YDSE, the slits have different widths. As a result, amplitude of waves from slits are A and 2A respectively. If `I_(0)` be the maximum intensity of the intensity of the interference pattern then the intensity of the pattern at a point where the phase difference between waves is `phi` is given by `(I_(0))/(P)(5+4cos phi)` . Where P is in in integer. Find the value of P ?

Text Solution

Verified by Experts

The correct Answer is:
9
As the amplitude are A and 2A then the ratio of intensities is `1:4`
`I_(max) = I_(0) = I_(1)+I_(2) +2sqrt(I_(1)+I_(2)) = I+4I+2xx2I`
`I_(0) = 9I implies I = (I_(0))/(9)`
Intensity at any point :
`I^(1) = I_(1)+I_(2)+2sqrt(I_(1)+I_(2)) cos phi`
`I^(1) = I + 4I + 2sqrt(Ixx4I) cos phi`
`I^(1) = 5I + 4I cos phi , I^(1) = I [5+4cos phi]`
`I^(1) = (I_(0))/(9) [5+4 cos phi] = (I_(0))/(P) [5+4cos phi]`
`P = 9`
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Knowledge Check

  • Consider an YDSE that has different slit width. As a result, amplitude of waves from two slits are A and 2A, respectively. If I_(0) be the maximum intensity of the interference pattern, then intensity of the pattern at a point where phase difference between waves is phi is

    A
    `(I_(0))/(9) cos^(2) phi`
    B
    `(I_(0))/(3) sin^(2).(phi)/(2)`
    C
    `(I_(0))/(9) [5 + 4 cos phi]`
    D
    `(I_(0))/(9) [ 5 + 8 cos phi]`

  • Two coherent light beams of intensities I and 4I produce interference pattern. The intensity at a point where the phase difference is zero, will b:

    A
    I
    B
    4I
    C
    5I
    D
    9I

  • The intensity of each coherent source is I_(0) . Which of the following gives the intensity at a point where the phase difference between the superposing waves is phi

    A
    `I_(0) (1 cos phi)`
    B
    `2I_(0) cos^(2) (phi //2)`
    C
    `2I_(0) (1 + cos phi)`
    D
    `I_(0) cos^(2) (phi //2)`

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