In fig., light of wavelength `lambda = 5000 Å` is incident on the slits (in a horizontally fixed place).
Here, `d = 1 mm` and `D = 1 m`
Take origin at O and XY plane as shown in the figure. The screen is released from rest from the initial position as shown

Velocity of 2nd maixma w.r.t central maxima at `t = 2`s is

Options

A. `(8 cm s^(-1)) hati + 20 m s^(-1) hat j`
B. `8 cm s^(-1) hat i`
C. `2 cm s^(-1) hat i`
D. `86 m s^(-1) hat i`

Answer

c

Explanation

Path difference corresponding to point P,
`Delta x = d sin theta = d sin theta`
`Delta x = (dx)/(D')`
For 2nd maxima, `Delta x = lambda`
`dx = 2 lambda D' implies x = (2 lambda D')/(d)`
Location of central maxima is
`(0, D + (gt^(2))/(2))` or `(0, D')`
Location of 2nd maxima is
`((2 lambda D')/(d) , D')`
Velocity of 2nd maxima w.r.t. central maxima is
`(vec v_(2nd))/(-3rd) = (3 lambda)/(D) = (2 lambda)/(d) [0 + g t] hat i = (2 lambda)/(d) xx g t hat i = 2 cm s^(-1) hat i`