In figure S is a monochromatic source of light emitting light of wavelength of wavelength `lambda` (in air). Light on slits `S_(1)` from S and then reaches in the slit `S_(2)` and `S_(3)` through a medium of refractive index `mu_(1)`. Light from slit `S_(2)` and `S_(3)` reaches the screen through a medium of refractive index `mu_(3)`. A thin transparent film of refractive index `mu_(2)` and thickness t is used placed in front of `S_(2)`. Point P is symmetrical w.r.t. `S_(2)` and `S_(3)`. Using the values `d = 1 mm, D = 1 m, mu_(1) = 4//3`,
`mu_(2) = 3//2, mu_(3) = 9//5`, and `t = (4)/(9) xx 10^(-5) m`,
a. find distance of central maxima from P,
b. If the film in front of `S_(2)` is removed, then by what distance and in which direction will be central maxima shift ?
.

Answer

a. the central order bright id formed at P only
b. `(40)/(27) mm`

Explanation

a. For the central order bright to be formed at Q,
`(SS_(1)) mu_(1) + (S_(1) Q) mu_(3) = (SS_(2)) mu_(1) + (S_(2) Q - t) mu_(3) + mu_(2) t`
`(S_(1) Q - S_(2) Q) mu_(3) = (SS_(2) - SS_(1)) mu_(1) + (mu_(2) - mu 3) t`
`(d sin phi) mu_(3) = (d sin theta) mu_(1) + (mu_(2) - mu_(3)) t`
` = (d^(2))/(D) mu_(1) + (mu_(2) - mu_(3) t`
`= (1^(2))/(10^(3)) xx (4)/(3) + ((3)/(2) - (9)/(5)) (4)/(9) xx 10^(-2) = 0`
Hence, `phi = 0` or the central order bright is formed at P only.
b. In absence of slab, `t = 0` and Eq. (i) gives
`(d sin phi) mu_(3) = (d sin theta) mu_(1)`
`(x)/(2D) mu_(3) = (d)/(D) mu_(1)` or `x = (2 mu_(1) d)/(mu_(3)) = (40)/(27) mm`.