In 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 one but having refractive index `1.7` interference pattern is observed using light of wavelength `5400 Å`
It is found that point P on the screen where the central maximum `(n = 0)` fell before the glass plates were inserted now has `3//4` the original intensity. It is further observed that what used to be the fourth maximum earlier, lies below point P while the fifth minimum lies above P.
Calculate the thickness of glass plate. (Absorption of light by glass plate may be neglected.
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Options

A. `9.3mu m`
B. `6.2 mu m`
C. `8.5 mu m`
D. `5.8mu m`

Answer

A

Explanation

`mu_(1)=1.4` and `mu_(2)=1.7` and let t be the thickness of each glass plates. Path difference at O, due to insertion of glass plates will be
`mu = 1.8}t`
`mu = 1.5`
`Delta x = (mu_(2)-mu_(1))t`
`= (1.7-1.4)t = 0.3 t` ….(1)
Now, since `5^(th)` maxima (earlier) lies below O and `6^(th)` minima lies above O. This path difference should lie between 51 and `51 + (lambda)/(2)`
So, let`Delta x = 51 = +D` .....(2)
where `Delta x lt (lambda)/(2)`
Due to the path difference Dx, the phase difference at O will be
`Delta phi = (2pi)/(lambda) Delta x = (2pi)/(lambda) (5lambda + Delta x) = (10 pi + (2pi)/(lambda). Deltax)` .....(3)
Intensity at O is given `(3)/(4)I_(max)` and since
`I(phi) = I_(max)cos^(2) ((phi)/(2)) (3)/(4)I_(max) = I_(max) cos^(2) ((phi)/(2))`
or `(3)/(4) = cos^(2) ((phi)/(2))` ....(4)
from Eqs. (3) and (4), we find that `D = (lambda)/(6)`
i.e., `Delta x = 5lambda + (lambda)/(6) = (31)/(6) 1 = 0.3 t (Delta x = 0.3 t)`
`t = (31lambda)/(6(0.3)) = ((31)(5400xx10^(-10)))/(1.8)m`
or `t = 9.3xx10^(-6) m= 9.3 mm`