A beam of light of wavelength 522nm passes through two slabs

A beam of light of wavelength 522nm passes through two slabs of material of identical thickness d= 1.20 micrometers, as shown in the figure. The slabs have different indices of refraction: n1= 1.39 and n2= 1.56. What is the phase difference (in radians, do not enter units) between the two parts of the beam after it passes through the slabs?

Solution

Here, d = 1.20 x 10^-6 m

Wavelength in 1 = (522 / 1.39) . 10^–9 m = 375.54x10^-9 m

Number of wavelengths in 1 = (1.2x10^–6) / (375.54x10^-9) = 3.195

Now, wavelength in 2 =   (522 / 1.56) . 10^–9 = 334.61x10^-9 m

Number of wavelengths in 2 = (1.2x10^–6) / (334.61x10^-9) = 3.586

Difference = 3.586 - 3.195 = 0.391 wavelength

Now, 1 wavelength = 2 radians

Therefore, the requisite phase difference = 2 x0.391 = 2.46 rad