A screen is at distance `D = 80` cm form a diaphragm having two narrow slits `S_(1)` and `S_(2)` which are `d = 2` mm apart.
Slit `S_(1)` is covered by a transparent sheet of thickness
`t_(1) = 2.5 mu m` slit `S_(2)` is covered by another sheet of thikness
`t_(2) = 1.25 mu m` as shown if Fig. 2.52.
Both sheets are made of same material having refractive index `mu = 1.40`
Water is filled in the space between diaphragm and screen. Amondichromatic light beam of wavelength `lambda = 5000 Å` is incident normally on the diaphragm.
Assuming intensity of beam to be uniform, calculate ratio of intensity of C to maximum intensity of interference pattern obtained on the screen `(mu_(w) = 4//3)`

Answer

`3//4`

Explanation

Path difference `Deltax=mu_(m)t_(1)+mu_(l)(S_(1)C-t_(1))-[mu_(m)t_(2)+mu_(l)(S_(2)C-t_(2))]`
`Deltax=mu_(m)(t_(1)-t_(2))-mu_(l)t_(1)+mu_(1)t_(2)`
`Deltax=mu_(m)(t_(1)-t_(2))-mu_(l)(t_(1)-t_(2))` ltbr `Deltax=(mu_(m)-mu_(l))(t_(1)-t_(2))`
`Deltax=(1.4-(4)/(3))(2.5xx10^(-6)-1.25xx10^(-6))`
`Deltax=(0.2)/(3)xx1.25xx10^(-6)`
`Deltax=(2.5)/(3)xx10^(-7)m`
`Deltax=(2500)/(3)Å=(5000)/(6)Å`
`Delta=(lambda)/(6)`
phase difference `phi=(pi)/(3)`
`phi=60^(@)`
Resultant intensity `I_(C)=I+I+2sqrt(IxxI)cos60^(@)`
`I_(C)=3I`
`I_(max)=I+I+2I`
`I_(max)=4I`
`(I_(C))/(I_(max))=(3)/(4)`