Question

Figure shows a YDSE setup having identical slits `S_(1)` and `S_(2)` with d =5 mm and D = 1 m. A monochromatic light of wavelength `lamda = 6000 Å` is incident on the plane of slit due to which at screen centre O, an intensity `I_(0)` is produced with fringe pattern on both sides Now a thin transparent film of `11 mu m` thickness and refractive index `mu = 2.1` is placed in front of slit `S_(1)` and now interference patten is observed again on screen.

After placing the film of slit `S_(1)`, the intensity at point O screen is :
A. 0.1 mm
B. 0.06 mm
C. 0.02 mm
D. None of th

Answer 1

Correct Answer - C
After introducing the film, path difference at point O is
`Delta = t(mu-1)=11 xx10^(-6)xx1.1 = 12.1 xx10^(-6)m`
Phase difference at point O is
`phi=(2 pi)/(lamda)xx Delta`
`=(2pi)/(6 xx 10^(-7))12.1 xx 10^(-6)`
`=(12.1pi)/(3)=(pi)/(3)`
Thus intensity at point O is
`I=I_(0) "cos"^(2)(phi)/(2)=I_(0)"cos"^(2)(pi)/(6)= 3 (I_(0))/(4)`
Now path difference at point in terms of `lamda` is
`Delta _(0)=(12.1 xx10^(-6))/(6xx 10^(-7))lamda = 20.167 lamda`
Thus due to placement of thin film 20 bright fringes cross the point O.