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?

Answer

(i)`1mm`,(ii)increase

Explanation

`S_(1)&S_(2)`are the image of `S`by upper&lower part
we have `m=(v)/(u)=(f)/(F+u) =(h_(I))/(h_(0))`
`(f)/(F+u)=(h_(I))/(h_(0))`
`h_(1)/(h_(0))=(0.10)/(0.10-0.15)`
for image by upper part
`(h_(0))/(h_(0))=-2`
`h_(1)=-2xx(-0.25)`
`=-0.5mm`
so `d ="distance between" S_(1)S_(2)`
`=2xx0.5mm`
`=1mm`
from equation `(1)(v)/(u)=(h_(1))/(h_(0))`
`v=-2xx(-0.15)m`
`v=0.3m,`
so `D=(1.3-0.3)m=1m`
`x=OA`
`=3beta`
`=3xx(lambdaD)/(d)`
`=(3xx500xx10^(-9)xx1)/(1.5xx10^(-3))m`
`=1mm`
(ii) `d` will decreases so`beta`will increases ,so`OA` will increase