$P_{1}$ and $P_{2}$ are transparent plates having equal thickness $20 μ m$ and refractive indices $μ_{1} = 1.7 μ_{2} = 1.6$ . $P_{1}$ transmits 75% whereas $P_{2}$ transmits 50% of energy incident. [Given $S_{1}$ and $S_{2}$ are identical slits and the intensity at ‘O’ without plates is $l = 4 l_{0}$ ]

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The phase difference between waves coming from $S_{1}$ and $S_{2}$ at ‘O’ is $Δ ϕ = 10 π$

The optical path difference between waves coming from $S_{1}$ and $S_{2}$ at ‘O’ is $Δ x = 4 \times 10^{- 6} m$

The intensity at ‘O’ after placing $P_{1}$ and $P_{2}$ is $l = \frac{5 l_{0}}{4} + \sqrt{\frac{3}{2}} l_{0}$

The intensity at ‘O’ does not change even after placing plates $P_{1}$ and $P_{2}$

answer is B, C.

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Detailed Solution

$Δ x = (μ_{1} - μ_{2}) = 0.1 \times 20 \times 10^{- 6} = 2 \times 10^{- 6} m; Δ ϕ = \frac{2 π}{4000} \times \frac{2 \times 10^{- 6}}{10^{- 10}} = 10 π l' = 0.75 l_{0} + 0.5 l_{0} + \sqrt[2]{0.75 \times 0.5} l_{0} \cos 10 π = \frac{3}{4} l_{0} + \frac{l_{0}}{2} + 2 \sqrt{\frac{3}{8}} l_{0} = [\frac{5}{4} + \sqrt{\frac{3}{2}}] l_{0}$