Question

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 1

Image of `S` by lens:
`u=-0.15 m, f=0.10 m, v=?`
`1/v-1/u=1/f`
`1/v-1/-0.15=1/0.10 implies 1/v=1/0.10-1/0.15=(3-2)/0.30=1/0.30`
`v=0.30 m`

In similar triangle, `SA` `B` and `SS_(1)S_(2)`
`(AB)/0.15=(S_(1)S_(2))/0.45 implies0.5/0.15=d/0.45`
`d=1.5 mm`
Distance between sources `S_(1)S_(2), d=1.5 mm`
`D=1.0 m, lambda=500 nm`
(a) For `3^(rd)` maximum,
`y_(3)=(3Dlambda)/d=(3xx1xx500xx10^(-9))/(1.5xx10^(-3))=10^(-3) m=1 mm`
If `AB` is decreased, `S_(1)S_(2)` i.e. d decreases and hence `y` increases.