Young's double-slit experiment setup with ligth of wavelength `lambda = 6000 Å`, distance between two slit in 2 mm and distance between the plane of slits and the screen. Is 2 m. The slits are of equal intensity. When a sheet of glass of refractive index 1.5 (which permtis only a fraction `eta` of the incident light to pass through) and thickness `8000 Å` is placed in front of the lower slit, it is observed that the intensity at a point P, 0.15 mm above the central maxima, does not change.

The value of `eta` is

Options

A. 0.21
B. 0.42
C. 0.12
D. 0.5

Answer

a

Explanation

Without inserting the slab, path difference at P,
`Delta x =(yd)/(D) = (0.15 xx 10^(-3) xx 2 xx 10^(-3))/(2) = 1.5 xx 10^(-7) m`
Corresponding phase difference at P,
`phi = ((2 pi)/(lambda)) (Delta x)`
`= ((2 pi)/(6000 xx 10^(-10))) (1.5 xx 10^(-7)) = (pi)/(2)`
`(phi)/(2) = (pi)/(4)`
Therefore, intensity at P,
`I = 4 I_(0) cos^(2) (phi // 2) = 2 I_(0)`
Phase difference after placing to glass sheet,
`phi' = phi + (2 pi)/(lambda) (mu - 1)t`
`= (pi)/(2) + (2pi)/(6000 xx 10^(-10)) (1.5 - 1) (8000 xx 10^(-19))`
` = (11 pi)/(6)`
Now, the intensity at P is,
`I' = I_(0) + eta I_(0) + 2 sqrt(eta I_(0)^(2)) cos.(11 pi)/(6) = 2 I_(0)` (given)
solving this equation, we get `eta = 0.21`.