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

B

Explanation

(i) In this case, the two identical halves of convex lens will create two separate images `S_(1)` and `S_(2)` of the source S. These images `(S_9(1) and S_(2))` will behave as two coherent sources and the further dealing will be in accordance to Young's double slit experiment.
For lens `L_(1)`
The object is S
`u=-0.15m`, `v=?`, `f=+0.1m`
`(1)/(v)-(1)/(u)=(1)/(f)implies(1)/(v)=(1)/(f)+(1)/(u)=(1)/(0.1)+(1)/(-0.15)` `implies` `v=0.3m`
`triangleSO_(1)O_(2)` and `triangleSS_(1)S_(2)` are similar. Also the placement of `O_(1)` and `O_(2)` are symmetrical to S
`:'``(S_(1)S_(2))/(O_(1)O_(2))=(u+v)/(u)`
`implies``S_(1)S_(2)=((u+v)(O_(1)O_(2)))/(u)=((0.15+0.3))/((0.15))xx0.5xx10^(-3)`
`S_(1)S_(2)=d=1.5xx10^(-3)m` `:'``D=1.3-0.3=1m`
The fringe width
`beta=(lamdaD)/(d)=(500xx10^(-9)xx1)/(1.5xx10^(-3))=(1)/(3)xx10^(-3)`
`:'` Therefore, `OA=3beta=3xx(1)/(3)xx10^(-3)m=10^(-3)`
(ii) If the gap between `L_(1)` and `L_(2)` i.e., `O_(1)O_(2)` is reduced. Then d will be reduced. Then the fringe width will increase and hence OA will increase.