In the given set up the light of wavelength 5000A is used incident normally on the slits. The screen is at a distance D = 80 cm from 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_(2) = 1.25 mum` and slit `S_2` is covered by another sheet of the same material of the thickness `t_2 = 1.25 mum`. (as in the figure). Water is filled in the space between diaphragm and screen. Assuming intensity of beam to be uniform, calculate the ratio of intensity of C to maximum intensity of interference pattern obtained on the screen `(mu_(w) = 4/3, mu_("sheet") = 1.40)`

Answer

`00.75`

Explanation

Path difference at C
`Deltax = t_(1)(mu-1) - t_(2)(mu-1)`
`= (t_(1)-t_(2)) (mu_(s)/mu_(w)-1)`
`= (2.5 -1.25) ((14 xx 3)/(4 xx 10) -1) mum`
`= 25/400 mu m`
`= 1/16 mum`
`rArr phi = (2pi)/lambda xx Deltax`
`= (2pi xx 4)/(5000 xx 10^(-10) xx 3) xx 1/16 xx 10^(-6) (x^(2) - x/4)`
`= pi/3`
`rArr I_("max") = 4I_(0)`
Intensity at `C, I_( C) = 2I_(0) (1 + cos pi/3) = 3I_(0)`
Required ratio `= 1/L_(1) = 3/4 = 0.75`