In a Young's experiment, the upper slit is covered by a thin glass plate of refractive index `1.4` while the lower slit is covered by another glass plate having the same thickness as the first ont but having refractive index `1.7.` Interference pattern is observed using light of wavelength `5400 A.` It is found that the point `P` on the screen where the central maximum fell berore the glass were inserted now has `(3//4)^(th)` the original intensity. It is further observed that what used to be the `5th` maximum earlier, Iies below the point `O` while the `6th` minimum lies above `O.` The thickness of the glass plate in `nxx10^(-7)m.` Find the value of `n`
(Absorption of light by glass plate may by neglected)

Answer

93

Explanation

`mu_(1)=1.4, mu_(2)=1.7` and let `t` be the thickness of each glass plates.
Path difference `O,` due to insertion of glass plates will be -

`Deltax=(mu_(2)-mu_(1)) t=(1.7-1.4) t= 0.3 t ………..(1)`
Now since `5^(th)` maxima (earlir) lies below `O` and `6^(th)` minima above `O.`
This path difference should lie between `5lambda` and `5lambda+lambda//2`
So let `Deltax=5lambda+Delta ...........(2)`
where `Delta lt lambda//2`
Due to the path difference `Delta x,` the phase difference at `O` will be
`phi=(2pi)/(lambda)Delta x=(2pi)/(lambda)(5lambda+Delta)=10 pi+(2pi)/(lambda).Delta .......(3)`
Intensity at `O` is given `(3)/(4) I_(max)=I_(max) cos^(2)((phi)/(2))`
`therefore (3)/(4)I_(max)=I_(max) cos^(2) ((phi)/(2))` or `(3)/(4)=cos^(2) ((phi)/(2)) .......(4)`
For equations `(3)` and `(4)`, we find that
`Delta=lambda//6`
`i.e. Deltax = 5lambda+lambda//6=(31lambda//6)lambda=0.3 t therefore t=(31lambda)/(6(0.3))=((31)(5400xx10^(-10)))/(1.8)m`
or `t=9.3xx10^(-6) m 9.3 mu m`