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In a young's double slit experiment, the first maxima is observed at a fixed point P on the screen. Now the screen is continuously moved away from the plane of slits. The ratio of intensity at point P to the intensity at point O (center of the screen) is-



A
Remains constant
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B
Keeps on decreasing
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C
First decreases and then increases
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D
First decreases and then becomes constant
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Solution

The correct option is B Keeps on decreasing
Let, IO and IP be the intensity at central maxima and intensity at point P respectively.

I=4I0cos2(ϕ2) ......(1)

Where, ϕ=Phase difference

At central maxima, ϕ=0

IO=4I0

Intensity at point P ;

The phase difference ϕ can be written as,

ϕ=2πλ×δ

Where, δ=Path difference

δ=dsinθ

ϕ=2πλ×dsinθ=Dβ×2πsinθ [ β=λDd]

From (1) we get,

IP=4I0cos2(Dπsinθβ)

Now IPIO=4I0cos2(Dπsinθβ)4I0

=cos2(Dπsinθβ)

As D increases, cos2 value decreases, Hence IPIO Keeps on decreasing.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (B) is the correct answer.

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