Answer:
Phase difference = pi/4 radians
Explanation:
Given:
- The wavelength of incident light λ = 600 nm
- The split separation d = 0.85 mm
- Distance of screen from split plane L = 2.8 m
Find:
What is the phase difference between the two interfering waves on a screen, at a point 2.5 mm from the central bright fringe?
Solution:
- The phase difference can be evaluated by determining the type of interference that occurs at point y = 2.5 mm above central order. We will use the derived results from Young's double slit experiment.
sin ( Q ) = m*λ /d
m = d*sin(Q) / λ
- Where, m is the order number and angle Q is the angle for mth order of fringe from central bright fringe.
r = sqrt ( L^2 + 0.0025^ )
Where, r is the distance from split to the interference bright fringe.
r = sqrt(2.8^ + 0.0025^) = 2.8
sin(Q) = 0.0025 / 2.8
Hence. m = 0.00085*0.0025 / 2.8*(600*10^-9)
m = 1.26
- We know that constructive interference would occurred at m = 1 and destructive interference @ m = 1.5. They have a phase difference of pi/2 radians.
- The order number lies in between constructive and destructive interference i.e m ≈ 1.25 then the corresponding phase difference = 0.5*(pi/2).
Answer: Phase difference = pi/4 radians
