In young’s experiment, the upper slits is covered a thin glass plate of refractive index $\frac{4}{3}$ and of thickness $9 λ,$ where $λ$ is the wavelength of light used in the experiment. The lower slits is also covered by another glass plate of thickness $2 λ$ and refractive index $\frac{3}{2}$ , as shown in figure if $I_{0}$ is the intensity at point P due to slits $S_{1} & S_{2}$ each then;

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Optical path difference between the waves from $S_{1} a n d S_{2}$ at point P is $2 λ$

intensity at point P is $4 I_{0}$

Optical path difference between the waves from $S_{1} a n d S_{2}$ at point P is $3 λ$

Two fringes have been shifted in upward direction after insertion of both the glass plates together

answer is A, B, C.

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Detailed Solution

Optical path difference due to $S_{1} i s = t (μ_{1} - 1) = 9 ƛ (\frac{4}{3} - 1) = 3 ƛ$
Optical path diff due to $S_{2} i s = t (μ_{2} - 1) = 2 ƛ (\frac{3}{2} - 1) = ƛ$
Path diff. at point $P P = 3 ƛ - ƛ = 2 ƛ$
Phase diff. $\frac{2 π}{ƛ} \times 2 ƛ = 4 π$
So net intensity at point P is $I_{n e t} = I_{0} + I_{0} + 2 I_{0} \cos 4 π$
$I_{n e t} = 4 I_{0}$
Optical path is conserved by S1 is more so fringe will shift at upper side.
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