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Light of wavelengthlambda = 500 nm falls...

Light of wavelength`lambda = 500 nm` falls on two narrow slits placed a distance d = ` 50 xx 10^-4` cm
apart, at an angle `phi= 30^@` relative to the slits as shown in figure. On the lower slit a transparent slab of thickness 0.1 mm and refractive index `3/2` is placed. The interference pattern is observed at a distance D=2m from the slits. Then, calculate

(a) position of the central maxima.
(b) the order of maxima at point C of screen .
(c)how many fringes will pass C, if we remove the transparent slab from the lower slit?

Text Solution

Verified by Experts

Phase difference `Deltaphi=kd sinphi+kd sinphi+kb(1-n)`, where `k=(2pi)/(lambda)`
Central maxima is obtained when `Deltaphi=0`
`kd sin alpha+kd sin theta+kb(1-n)=0`
`sintheta=(b(n-1))/(d)-sinalpha=(1)/(2)rArr` therefore `theta=30^(@)`
At `C`, `theta=0`
`Deltaphi=kd sin alpha-k b (n-1)=(kd)/(2)[1-(2b)/(d)(n-1)]`
`=(2pid)/(2lambda)[1-(2b)/(d)(n-1)]`
Substituting the values, we get, `Deltaphi= -100 pi=2m pi rArr m= -50`
As `I prop cos^(2)((Deltaphi)/(2))`
Hence at `C` there will be maxima.
For minima
`Deltaphi=(2m-1)pi`
for `m= -50` and `-49`, we get `Deltaphi= -101pi` and `-99pi`, respectively.
Hence the minima closest to `theta=0` are of order `m=-49` and `-50` on both side
The phase difference at `C` when we remove the slab
`Deltaphi=100pi` i.e. `m=50`
Hence number of frings passed over `=150`
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