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Q.

In a standard YDSE the region between screen and slits is immersed in a liquid whose refractive index varies with time T as $μ_{l} = \frac{5}{2} - \frac{T}{4}$ until it reaches a steady state value of $\frac{5}{4}$ . A glass plate of thickness 36 $μ$ m and refractive index 3/2 is introduced in front of one of the slits. Find the speed of central maxima when it is at ‘O’? (take d = 2mm and D = 1m)

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a

$2 \times 10^{- 3} m s^{- 1}$

b

$3 \times 10^{- 3} m s^{- 1}$

c

$4 \times 10^{- 3} m s^{- 1}$

d

$5 \times 10^{- 3} m s^{- 1}$

answer is B.

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Detailed Solution

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path difference $(Δ x) = [{(S_{2} P)}_{l i q} + (μ_{g} - μ_{l}) t - {(S_{1} P)}_{l i q u i d}]$

$(Δ x) = [{(S_{2} P - S_{1} P)}_{l i q} + (μ_{g} - μ_{l}) t] = μ_{1} {(S_{2} P - S_{1} P)}_{a i r} + (μ_{g} - μ_{l}) t$

$Δ x = μ_{l} (\frac{y d}{D}) + (μ_{g} μ_{l}) t$

For central maxima $Δ x = 0.$

$0 = μ_{l} (\frac{y d}{D}) + (μ_{g} - μ_{l}) t y = - \frac{D [\frac{3}{2} - (\frac{5}{2} - \frac{T}{4})] t}{d [\frac{5}{2} - \frac{T}{4}]}$

The time when y become zero is $0 = \frac{D (4 - T) t}{d (10 - T)}$

$D (4 - T) t = 0 \Rightarrow 4 - T = 0 \Rightarrow T = 4 \sec$ .

Speed of central maxima

$V = \frac{d y}{d x} = \frac{D t}{d} \frac{d}{d T} [\frac{4 - T}{10 - T}] V = \frac{t D}{d} \frac{d}{d T} [(4 - T) {(10 - T)}^{- 1}]$

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