    Question

# 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.  Suggest Corrections  2      Explore more