In a modified YDSE, the region between the screen and slits is immersed in a liquid whose refractive index varies with time as `mu_(1) = (5 // 2) - (T // 4)` until it reaches s steady state value of `5 // 4`. A glass plate of thickness 36 `mu m` and refractive index `3//2` is introduced in front of one of the slits.

Find the time when central maxima is at point O. located symmetrically on the x-asix.

Options

A. `2xx10^(-3)ms^(-1)`
B. `3xx10^(-3)ms^(-1)`
C. `4xx10^(-3)ms^(-1)`
D. `5xx10^(-3)ms^(-1)`

Answer

B

Explanation

Path different
`(Delta x) = [(S_(2)P)_(liq) + (mu_(g) - mu_(l))t- (S_(1)P)_("liquid")]`
`(Delta x) = (S_(2)P-S_(1)P)_(liq)+(mu_(g)-mu_(l))t`
`= mu_(l)(S_(2)P-S_(1)P)_(air)+(mu_(g)-mu_(l))t`
`Delta x = mu_(l) ((yd)/(D))+(mu_(g) - mu_(l))t`
For central maxima `Delta x =0`
`0 = mu_(l)((yd)/(D))+(mu_(g)-mu_(l))t y = (-(mu_(g)-mu_(l))tD)/(d mu_(l))`
`y = -(D[(3)/(2)-((5)/(2)-(T)/(4))]t)/(d[(5)/(2)-(T)/(4)])`
`y= - (D[1-(T)/(4)]t)/(d[(5)/(2)-(T)/(4)]) = (D[4-T]t)/(d(10-T))`
The time when y becomes zero is
`0 = (D(4-T)t)/(d(10-T))`
`D(4-T)t = 0 implies 4- T = 0 implies T = 4 sec`
Speed of central maxima
`V = (dy)/(dt) = (Dt)/(d)(d)/(dT) [(4-T)/(10-T)]`
`V = (tD)/(d) = [(4-T)(10-T)^(-1)]`
`= (tD)/(d)[(-(4-T)+(10-T))/((10-T)^(2))]`
`= (tD)/(d) [(-4+T+10-T)/((10-T)^(2))] V = (6Dt)/(d(10-T)^(2))`
Central maxima is at 'O' at `T = 4 sec`
`V = (6xx1xx36xx10^(-6))/(2xx10^(-3)(10-4)^(2)) = (3xx36xx10^(-3))/(36)`
`V = 3xx10^(-3)ms^(-1)`