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.

What is the speed of the central maxima when it is at

Options

A. `3 xx 10^(-3) m s_(-1)`
B. `4 xx 10^(-3) m s_(-1)`
C. `6 xx 10^(-3) m s_(-1)`
D. `1 xx 10^(-3) m s_(-1)`

Answer

a

Explanation

We consider a point P on the screen.
Optical path length,
`[S_(1) P]_("liquid") = [mu_(1) S_(1)P]_("air")`
`[S_(1) P - t]_("liquid") + t_("glass") = mu [S_(1) P - t]_("air") + [mu_(g) t]_("air")`
` = [mu_(1) S_(2) P + (mu_(g) - mu_(1)) t]_("air")`
Hence, optical path difference at P,
`Delta x = [mu_(1) S_(2) P + (mu_(g) - mu_(1)) t] - mu_(1) S_(1) P`
`= mu_(1) (S_(1) P - S_(1) P) + (mu_(g) + mu_(1))t`
For a point P at the screen in the absence of liquid,
`S_(2) P = S_(1) P =(yd)/(D)`
If a liquid is filled,
`[S_(2) P - S_(1) P]_("liquid") = (mu_(1) yd)/(D)`
Thus,
`Delta x = (mu_(1) yd)/(D) + [mu_(g) - mu_(t)] t`

a. For central maxima,
`Delta x = 0`
`y = [(mu_(g) - mu_(1))/(mu_(1))] (tD)/(d) = ((4 - T) tD)/((10 - T) d)`
At Point O, `y = 0`
Thus, `T = 4`
b. Speed of central maxima,
`v = |(dy)/(dt)| = (6 D t)/((10 - T)^(2) d)`
Central maxima is at O at time `t = 4` s.
`:. v = (6 D t)/(36 D) = ( 1 xx 36 xx 10^(-6))/(6 xx 2 xx 10^(-3)) = 3xx 10^(-3) m s^(-1)`