A Young's double slit apparatus is immersed in a liquid of refractive index `mu_(1).` The slit plame touches the liquid surface. A parallel beam of monochromatic light of wavelength `lambda` (in air) is incident normally on the slits.
a. Find the fringe width.
b. If one of the slits (say `S_(2)`) is covered by a transparent slab of refrative index `mu_(2)` and thickless t as shown, find the new position of central maxima.
c. Now, the other slit, `S_(1)` is also covered by a slab of same thickness and refractive index `mu_(3)` as shown in Fig. 2.49 due to which the central maxima recovers it positon, find the value of `mu_(3)`
Find the ratio of intensities at O in the three conditions (a), (b), and (c).

Explanation

Fringe width, `w = (lambda_(mu) D)/(d) = (lambda D)/(mu_(1) d)`
Position of central maximum is shifted upward by a distance `((mu_(2) - 1) tD)/(d)`
c. `((mu_(2) - 1) tD)/(d) = (((mu_(3))/(mu_(1)) - 1) tD )/(d) implies (mu_(2))/(mu_(1))` or `mu_(3) = mu_(1) mu_(2)`
b. `I = I_(max) cos^(2) ((phi)/(2)),` where `phi = ((2 pi )/(lambda)) Delta x` or `(phi)/(2) = ((pi)/(lambda)) Delta x`
`implies I prop cos^(2) ((phi)/(2))`
The the first and third case, `Delta x = 0` while in the second case, `Delta x = (mu_(2) - 1)t`. Therefore, the desired ratio is
`I_(1) : I_(2) : I_(3) = 1 : cos^(2) {(pi (mu_(2) -1) t)/(lambda)} : 1`
.