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Q.

In YDSE of equal width slits, if intensity at the centre of screen is $I_{0}$ , then intensity at a point on the screen which is at a distance of $\frac{β}{4}$ from central maxima is( $β = f r i n g e w i d t h)$

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a

$\frac{I_{0}}{2}$

b

$\frac{I_{0}}{8}$

c

Zero

d

$\frac{I_{0}}{4}$

answer is C.

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Detailed Solution

$I = I_{0} \cos^{2} \frac{ϕ}{2}$

Path difference $= dsinθ ≃ dtanθ = d . \frac{β / 4}{D} = d . \frac{λD}{4 . D . d} = \frac{λ}{4}$

$\begin{array}{l} ∴ ϕ = \frac{2 π}{λ} \times \frac{λ}{4} = \frac{π}{2} \\ ∴ intensity= \frac{I_{o}}{2} \end{array}$

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In YDSE of equal width slits, if intensity at the centre of screen is I0, then intensity at a point on the screen which is at a distance of β4 from central maxima is(β= fringe width)