Compare the intensities of two points located at respective distance $\frac{β}{4}$ and $\frac{β}{3}$ from the central maxima in a interference of YDSE ( $β$ is the fringe width)

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4:1

1:4

2:1

1:2

answer is B.

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Detailed Solution

$Δ θ = \frac{2 π}{λ} Δ x = \frac{2 π}{λ} (\frac{d}{D} \frac{β}{4}) = \frac{2 π}{λ} (\frac{d}{D} \frac{λ D}{4 d})$

$∴ Δ θ = \frac{2 π}{4} = \frac{π}{2} \Rightarrow I = 4 I_{0} \cos^{2} (\frac{π}{4})$

Similarly

$Δ θ = \frac{2 π}{3} \Rightarrow I = 4 I_{0} \cos^{2} (\frac{2 π}{2 \times 3}) = I_{0}$

$∴$ required ratio $= 2 : 1$

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