In a Young's double slit set up, the wavelength of light used is 54nm. The distance of screen form slits is 1m. The slit separation is 0.3mm.
(a) Compare the intensity at a point P distant 10mm from the central fringe where the
intensity is `I_0`.
(b) Find the number of bright fringes between P and the central fringe.

Explanation

Given, `lambda=546nm = 5.46 xx 10^-7m, D = 0.1m and d=0.3mm = 0.3 xx 10^-3m `
(a) At a distance `y=10mm=10 xx 10^-3m,` from central fringe, the path difference will be
`Deltax = (yd)/D = ((10 xx10^-3)(0.3 xx 10^-3))/1.0 = 3.0 xx 10^(-6) m.
The corresponding phase difference between the two interfering beams will be
` phi = 2pi/lambda.Deltax `
` = (2pi/(5.46 xx (10^-7)) (3.0 xx (10^-6))` radian
`= 1978^@ `
`:. phi/2 = 989^@`
`I = I_0 cos^2 phi/2`
` =I_0 cos^2 (989) `
` = 3.0 xx (10^-4) I_0`
(b) Fringe width, `omega = (lambdaD)/d = ((5.46 xx (10^-7) (1.0))/(0.3 xx (10^-3)))m `
= 1.82 mm
Since, `y/omega = 10/1.82 = 5.49`
Therefore, number of bright fringes between P and central fringe will be 5 (excluding the
central fringe. )