In the arrangement shown, wavelength of light used is $λ$ .The distance between slits S₁ and S₂ is d(<<D). S₄ is on the axis of a symmetry as shown. The distance between S₃ and S₄ is $u = \frac{λD}{3 d}$ . If the ratio of maximum to minimum intensity observed on screen is k. Find k.

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answer is 9.

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Detailed Solution

If intensity of light incident on slits S₁ and S₂ is I₀
The intensity through S₄ will be I₁ = 4I₀

Path difference at S₃ : $∆ x = \frac{du}{D} = \frac{λ}{3}$

$ϕ = \frac{2 π}{λ} \times (∆ x)$
The intensity through S₃ will be
$\begin{array}{l} I_{2} = 4 I_{0} \cos^{2} (ϕ / 2) = I_{0} \\ \Rightarrow k = \frac{I_{max}}{I_{min}} = {(\frac{\sqrt{4 I_{0}} + \sqrt{I_{0}}}{\sqrt{4 I_{0}} - \sqrt{I_{0}}})}^{2} = {(\frac{3}{1})}^{2} \\ \Rightarrow k = \frac{9}{1} \end{array}$