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?

Options

A. `1mm`
B. `2mm`
C. `1.5mm`
D. `2.5mm`

Answer

A

Explanation

" The images of S due to both parts of lenses are formed at " `S_(1) " and " S_(2)` " whose distance 'v' from the lens is give n by "
`(1)/(f) = (1)/(v) - (1)/(u) implies (1)/(0.1) = (1)/(v) - (1)/((-0.15))`
`(1)/(v) = (1)/(0.1) - (1)/(0.15) = (10)/(1) - (100)/(15) = (150-100)/(15) = (50)/(15) = (10)/(3)`
`v = (3)/(10) = 0.3m`
" From similar " `Delta_(les)SS_(1)S_(2) "and " SP_(1)P_(2)`
`(d)/(0.5xx10^(-3)) = (0.3+0.15)/(0.15)`
`d = (0.45)/(0.15)xx0.5xx10^(-3) = 1.5xx10^(-3)m`
" For third maximum "
`y_(n) = (3lambda D)/(d) = (3xx500xx10^(-9)xx1)/(1.5xx10^(-3)) = 10^(-9)xx10^(6)`
`y_(n) = 10^(-3)m implies y_(3) = 1mm`