In figure S is a monochromatic point source emitting light of wavelength `lambda=500 nm`. A thin lens of circular shape and focal length `0.10 m` is cut into two identical halves `L_(1)` and `L_(2)` by a plane passing through a doameter. The two halves are placed symmetrically about the central axis `SO` with a gap of `0.5 mm`. The distance along the axis from `A` to `L_(1)` and `L_(2)` is `0.15 m`, while that from `L_(1)` and `L_(2)` to `O` is `1.30 m`. The screen at `O` is normal to `SO`.
(a) If the `3^(rd)` intensity maximum occurs at point `P` on screen, find distance `OP`.
(b) If the gap between `L_(1)` and `L_(2)` is reduced from its original value of `0.5 mm`, will the distance `OP` increases, devreases or remain the same?

Explanation

The images of S due to both parts of lenses are fromed at `S_(1) ` and `S_(2)` whose distance `nu` from the lens is given by
`(1)/(f) = (1)/(nu) - (1)/(u)` Hence, `u = - 0.15 m. f = 0.10 m`.
`:. (1)/(0.10) = (1)/(nu) - (1)/(-0.15)`
`implies (1)/(nu) = (1)/(0.10) - (1)/(0.15) = (3 - 2)/(0.30) = (1)/(0.30)`
`:. nu = 0.30 m`
As the ray passing through optical center passes undeviated, therefore, angle `theta` subtended by lens gap and images `S_(1)` and ` S_(2)` must be same. We have,
`theta = (O_(1) O_(2))/(u) = (S_(1) S_(2))/(u + nu)`
`:. S_(1) S_(2) = (u + nu)/(u) O_(1) O_(2) = (0.30 + 0.15)/(0.15) xx 0.5mm`
`= 1.5 mm`
i.e, `S_(1) S_(2) = d = 1.5 mm = 1.5 xx 10^(-3) m`
Also, `D = 1.30 - 0.30 = 1.00 m.`
`S_(1)` and ` S_(2)` act coherent sources and produce interference The positons of maxima are given by
`y_(n) = (n D lambda)/(d)`
As point A is third maxima,
`:. y_(n) = OA =(3 d lambda)/(d) = (3 xx 1.00 xx 500 xx 10^(-9))/(1.5 xx 10^(-3))`
`=1000 xx 10^(-6) m = 1.0 mm`
b. If gap between`L_(1)`and`L_(2)` is reduced, then `S_(1) S_(2)`or d decrease and hence OA increases.