In the Young's double-slit experiment, the amplitude os source `S_(1)` is three times the amplitude of the source `S_(2)`. These sources are covered by perfectly transparent thin plates of same thickness but different refractive indices 1.6 and 1.5, respectively. Now, if the plates are interchanged and the amplitude of source `S_(1)` is made same as that of `S_(2)`, then find the amout by which the intensity is changed at a point where previous central maxima was formed. Take thickness of the plate equal to `(110//3) lambda`, where `lambda` is the wavelength of light used.

Answer

`(1)/(16)`

Explanation

Ratio of intensities of the sources `S_(1)` and `S_(2)`,
`(I_(1))/(I_(2)) = 9`
Let `I_(2) = I_(0)` and `I_(1) = 9 I_(0)`.
Intensity at any point on the screen.
`I = 9 I_(0) + I_(0) + 2 sqrt( 9 I_(0) I_(0)) cos gamma`
For maxima at P,
`I = 2 I_(0) [5 + 2] = 16 I_(0)`
If transparent plates are interchanged, the central maxima will shift downward by some amout j.
`j = (D t (mu_(1) - mu_(2)))/(d)`
`Delta x_(p) = (yd)/(D) + t (mu_(2) - mu_(1)) = 0`
`implies (yd)/(D) = t(mu_(1) - mu_(2))`
When `mu_(2)` and `mu_(1)` are interchanged
` Delta x_(p_(1)) = (yd)/(D) t(mu_(1) - mu_(2)) = 2t(mu_(1) - mu_(3))`
Phase diffenence at P,
`phi = (2pi)/(lambda) 2 (mu_(1) - mu_(2)) t = (44)/(3) pi`
Now, intensity at P,
`I' = I + I_(0) + 2 I_(0) cos ((44)/(3)) pi - I_(0)`
Hence, ratio of intensities at P,
`(I')/(I) = (I_(0))/(16 I_(0)) = (1)/(16)`