A YDSE is performed in a medium of refractive index `4 // 3`, A light of 600 nm wavelength is falling on the slits having 0.45 nm separation . The lower slit `S_(2)` is covered b a thin glass plate of thickness 10.4 mm and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 m from the slits as shown in figure. (All the wavelengths in this problem are for the given medium of refractive index `4 // 3`, ignore absorption.)

Find the light intensity at point O relative t maximum fringe intensity.

Options

A. `(1)/(4) I_(max)`
B. `(5)/(4) I_(max)`
C. `(1)/(2) I_(max)`
D. `(3)/(4) I_(max)`

Answer

a

Explanation

Path difference at point P on the screen, `Delta x = (yd)/(D)`
At the position of central maxima, the optical path length `S_(2) P` and `S_(1) P` and equal.
`((S_(2) P - t))/(c // mu_(1)) = (1)/(c // mu_(2)) = (S_(1) P)/(c // mu_(1))`
where ` mu_(1) = 4 // 3, mu_(2) = 3 // 2`.
`mu_(1) (S_(1) P - S_(2) P) = (mu_(2) - mu) t`
`mu_(1) ((yd)/(D)) = (mu_(2) - mu_(1)) t`
`y = ((mu_(2) - mu_(1)) t D)/(mu_(1) d)`
`= ([(3 // 2) - (4 // 3)] xx 10.4 xx 1.5)/((4.3) xx 0.45 xx 10^(-3)) = 4.33 mm`

b. At point O, net path difference.
`Delta x = ((mu_(2))/(mu_(1)) - 1) t`
Net phase difference,
`Delta phi = (2 pi)/(lambda) Delta x`
` = (2 pi)/(6 xx 10^(-7)) ((1.5)/(4//3) - 1) (10.4 xx 10^(-6)) = ((13)/(3)) pi`
Thus, intensity
`I = I_(max) cos^(2) (phi // 2)`
`= I_(max) cos^(2) ((13 pi)/(6)) = (3)/(4) I_(max)`
c. For maximum intensity at point O,
`Delta x = n lambda`
Path difference at point O,
`Delta x = ((1.5)/(4//3) - 1) (10.4 xx 10^(-6)) = 1300 nm`
Thus, maximum intensity will correspond to `(1300)/(2) nm`,
`(1300)/(3) nm`,...
In the given range, required values are 650 nm and 433.33 nm.