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.)

Now, if 600 nm, find the wavelength of the ligth that forms maximum exactly at point O.

Options

A. 650 nm, 433.3 nm
B. 550 nm, 750 nm
C. 450 nm, 645.3 nm
D. 375 nm, 525.3 nm

Answer

b

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.