In a standard YDSE the region between screen and slits is immersed in a liquid whose refractvie index varies with time as μ′=52−T4until it reaches a steady state value of 54. A glass plate of thickness 36μm and refractive index 3/2 is introduced in front of one of the slits. Find the velocity of central maximuma when it is at ‘O’?(take d=2mm and D=1m)b

Path difference(Δx)=S2Pliq +μg−μ2t−S1Pliquid (Δx)=S2P−S1Pliq +μg−μlt=μ′S2P−S1Pair +μg−μltΔx=μlydD+μg−μltFor central maxima Δx=0O=μlydD+μg−μlt y=−μg−μltDdμly=−D32−52−T4td52−T4y=−D1−T4td52−T4=D[4−T]td(10−T)The time when y becomes zero isO=D(4−T)td(10−T)D(4−T)t=0⇒4−T=0⇒T=4secSpeed of central maximaV=dydt=DtdddT4−T10−TV=tDd=(4−T)(10−T)−1=tDd−(4−T)+(10−T)(10−T)2=tDd−4+T+10−T(10−T)2 V=6Dtd(10−T)2Central maxima is at ‘O’ at T = 4secV=6×1×36×10−62×10−3(10−4)2=3×36×10−336V=3×10−3ms−1

In a standard YDSE the region between screen and slits is immersed in a liquid whose refractvie index varies with time as μ′=52−T4until it reaches a steady state value of 54. A glass plate of thickness 36μm and refractive index 3/2 is introduced in front of one of the slits. Find the velocity of central maximuma when it is at ‘O’?(take d=2mm and D=1m)b

Path difference(Δx)=S2Pliq +μg−μ2t−S1Pliquid (Δx)=S2P−S1Pliq +μg−μlt=μ′S2P−S1Pair +μg−μltΔx=μlydD+μg−μltFor central maxima Δx=0O=μlydD+μg−μlt y=−μg−μltDdμly=−D32−52−T4td52−T4y=−D1−T4td52−T4=D[4−T]td(10−T)The time when y becomes zero isO=D(4−T)td(10−T)D(4−T)t=0⇒4−T=0⇒T=4secSpeed of central maximaV=dydt=DtdddT4−T10−TV=tDd=(4−T)(10−T)−1=tDd−(4−T)+(10−T)(10−T)2=tDd−4+T+10−T(10−T)2 V=6Dtd(10−T)2Central maxima is at ‘O’ at T = 4secV=6×1×36×10−62×10−3(10−4)2=3×36×10−336V=3×10−3ms−1

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

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$2 \times 10^{- 3} {ms}^{- 1}$

$3 \times 10^{- 3} {ms}^{- 1}$

$4 \times 10^{- 3} {ms}^{- 1}$

$5 \times 10^{- 3} {ms}^{- 1}$

answer is B.

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Detailed Solution

Path difference
$(Δx) = [{(S_{2} P)}_{liq} + (μ_{g} - μ_{2}) t - {(S_{1} P)}_{liquid}]$
$\begin{array}{l} (Δx) = {(S_{2} P - S_{1} P)}_{liq} + (μ_{g} - μ_{l}) t \\ = μ_{'} {(S_{2} P - S_{1} P)}_{air} + (μ_{g} - μ_{l}) t \\ Δx = μ_{l} (\frac{yd}{D}) + (μ_{g} - μ_{l}) t \end{array}$
For central maxima $Δx = 0$
$\begin{array}{l} O = μ_{l} (\frac{yd}{D}) + (μ_{g} - μ_{l}) t y = \frac{- (μ_{g} - μ_{l}) tD}{{dμ}_{l}} \\ y = - \frac{D [\frac{3}{2} - (\frac{5}{2} - \frac{T}{4})] t}{d [\frac{5}{2} - \frac{T}{4}]} \\ y = - \frac{D [1 - \frac{T}{4}] t}{d [\frac{5}{2} - \frac{T}{4}]} = \frac{D [4 - T] t}{d (10 - T)} \end{array}$
The time when y becomes zero is
$\begin{array}{l} O = \frac{D (4 - T) t}{d (10 - T)} \\ D (4 - T) t = 0 \Rightarrow 4 - T = 0 \Rightarrow T = 4 \sec \end{array}$
Speed of central maxima
$\begin{array}{l} V = \frac{dy}{dt} = \frac{Dt}{d} \frac{d}{dT} [\frac{4 - T}{10 - T}] \\ V = \frac{tD}{d} = [(4 - T) (10 - T)^{- 1}] \\ = \frac{tD}{d} [\frac{- (4 - T) + (10 - T)}{(10 - T)^{2}}] \\ = \frac{tD}{d} [\frac{- 4 + T + 10 - T}{(10 - T)^{2}}] V = \frac{6 Dt}{d (10 - T)^{2}} \end{array}$
Central maxima is at ‘O’ at T = 4sec
$\begin{array}{l} V = \frac{6 \times 1 \times 36 \times 10^{- 6}}{2 \times 10^{- 3} (10 - 4)^{2}} = \frac{3 \times 36 \times 10^{- 3}}{36} \\ V = 3 \times 10^{- 3} {ms}^{- 1} \end{array}$