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Consider the situation shown in figure. ...

Consider the situation shown in figure. The two slits `S_1 and S_2` placed symmetrically around the central line are illuminated by a monochromatic light of wavelength `lamda`. The separation between the slits is d. The light transmitted by the slits falls on a screen `Sigma_1` placed at a distance D from the slits. The slit `S_3` is at the placed central line and the slit `S_4`, is at a distance z from `S_3`. Another screen`Sigma_2` is placed a further distance D away from 1,1. Find the ratio of the maximum to minimum intensity observed on `Sigma_2` if z is equal to a. `z=(lamdaD)/(2d)` b. `(lamdaD)/d` c. `(lamdaD)/(4d)`

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Verified by Experts

The correct Answer is:
A, B, C, D

i. when `z=(Dlamda)/(2d), at S_4`
minimum intensilty occurs, (dark fringe)
`rarr Amplitude =0`
At `S_3` path differenc e=0`
rarr maximum intensity occurs.
rarr Amplitude =2r
So on `Sigma^2` screen,
`l_(max)/l_(min)=((2r+0)^2)/((2r-0)^2)=1`
ii. When z=(lamdaD)/d `
As `S_4` maximum intensity occurs
`rarr Amplitude =sqrt2r`
At `S_3` also maximum intensity occurs.
`l_(max)/l_(min)=((2r+2r)^2)/((2r-2r)^2)=oo`
iii When `z=(lamdaD)/(4d), at S_4`
intensity =l_(max)/2`
`rarr Amplitude=sqrt2r`
`:. At S_3` Intensilty is minimum
`rarr Amplitude =2r
`l_(max)/l_(min)=((2r+sqrt(2r))^2)/((2r-sqrt(2r))^2)=34`
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