In a Young's double slit experiment, the slit separation d is 0.3 mm and the screen distance D is 1 m. A parallel beam of light of wavelength 600 nm is incident on the slits at angle  as shown in figure. On the screen, the point O is equidistant from the slits and distance PO is 11.0 mm. Which of the following statement(s) is/are correct ?

Options

A. For `alpha = (0.36)/pi` degree, there will be destructive interference at point O.
B. Fringe spacing depends on `alpha`
C. For `alpha = (0.36)/pi` degree, there will be destructive interference at point P.
D. For `alpha = theta` there will be constructive interference at point P.

Answer

C

Explanation

Option (a)
Angle, `alpha = (0.36)/180 = 2 xx 10^(-3)` rad
For rays at an angle `alpha`, path difference can be written as follows:
`Delta H = d sin alpha`
Here angle `alpha` is very small hence,
`Delta x = d alpha = (3 xx 10^(-4)) (2 xx 10^(-3)) = 6 xx 10^(-7)m`
`Deltax = nlambda implies 6 xx 10^(-7) = n xx 6 xx 10^(-7) implies n = 1`
Hence constructive interfernece will take place at O. Hence option (a) is wrong.
Option (b)
Fringe width is given by `beta = (lambda D)/(d)` , hence it is independent of `alpha`. So option (b) is wrong.
Option (c )
Path difference at point P is given by :
`Delta x_P = d alpha + (y d)/D = d(alpha + y/D)`
`Delta x_P = (3 xx 10^(-4))(2 xx 10^(-3) + 11 xx 10^(-3)) = 39 xx 10^(-7)m`
`Delta x_P = n lambda implies 39 xx 10^(7) = n xx 6 xx 10^(-7) implies n = 6.5`
Hence destructive interference will take place at P hence option (c ) is correct.
Option (d)
When `alpha = 0` then path difference at P is given by
`Delta x_P = (dy)/D`
`Delta x_P = (3 xx 10^(-4)) (11 xx 10^(-3))//1 = 33 xx 10^(-7) m`
`Delta x_p = n lambda implies 33 xx 10^(-7) = n xx 6 xx 10^(-7) implies n = 5.5`
Hence destructive interference will take palce at P hence option (d) is wrong.