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

Correct Answer - (i). 1 mm
(ii). Increases
(i). For the lens `u=-0.15m,f=+0.10m`
therefore using `(1)/(v)-(1)/(u)=(1)/(f)` we have
`(1)/(v)=(1)/(u)+(1)/(f)=(1)/((-0.15))+(1)/((0.10))` or `v=0.3m`
Linear magnitfication `m=(v)/(u)=(0.3)/(-0.15)=-2` ltbr. Hence two images `S_(1)` and `S_(2)` of S will be formed at 0.3m from the lens as shown in figure. Image `S_(1)` due to part 1 will be formed at 0.5 mm above its optic axis (m=-2). Similarly, `S_(2)` due to part 2 is formed 0.5 mm below the optic axis of this part as shown.
Hence `d=` distance between `S_(1)` and `S_(2)=1.5mm`
`Delta=1.30-0.30=1.0m=10^(3)mm`
`lamda=500nm=5xx10^(-4)mm`
Therefore, fringe width,
`omega=(lamdaD)/(d)=((5xx10^(-4))(10^(3)))/((1.5))mm=(1)/(3)mm`
now, as the point A is at the third maxima
`OA=3omega=3(1//3)mm` or `OA=1mm`
(ii). if the gap between `I_(1)` and `L_(2)` is reduced, d will decrease. Hence, the fringe width `omega` will increase or distance OA will increase