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In Young's double slit experiment the sl...


In Young's double slit experiment the slits, `S_(1) & S_(2)` are illuminated by a parallel beam of light of wavelength 4000 Å from the medium of refractive index `n_(1)=1.2` A thin film of thickness 1.2 `mum` and refractive index n=1.5 is placed infrom of `S_(1)` perpedicular to path of light. the refractive index of medium between plane of slits & screen is `n_(2)=1.4` if the light coming from the film and `S_(1)&S_(2)` have equal intensities I then intensity at geometrical centre of the screen O is

A

0

B

2I

C

4I

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
Path difference at O: `(mu_(rel)-1)t=((n)/(n_(2))-1)t`
`therefore `phase difference at O: `t((n)/(n_(2))-1)(pi)/(lamda_(2))` where `n_(1)lamda_(1)-n_(2)lamda_(2)`
`implies` phase difference `=(pi)/(2)implies"Resultant intensity "=2I`
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