Two narrow, parallel slits separated by 0.85 mm are illuminated by 600 nm light, and the viewing screen is 2.8 m away from the slits. What is the phase difference between the two interfering waves on a screen, at a point 2.5 mm from the central bright fringe? Answer in radians.

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).

                 Phase difference = pi/4 radians

what was the answer?

dang that balls went far