Light wave interference on thin film

Let a point P exist just above the surface of a glass slab of thickness $$L$$. Light is incident vertically on this slab. The light at point P, at some instant has phase angle $$\phi$$. The refracted light will travel a total path of $$L + L$$ before again reaching the upper surface just below P. It takes time $$T = 2L/c$$ for the light to travel this path and its phase angle at this time is $$\phi + (2π/\lambda)\cdot2L = \phi + (4\pi L/\lambda)$$.

The new light passing through point P at this instant, has a phase angle of φ + ωT since the wave has propagated forward. The light ray which reflects at this instant has phase angle of φ + ωT + π since reflection adds π to phase angle.

inputting value of $$T = 2L/c$$ in $$\phi + \omega T + \pi$$ we get:

$$\phi + 2\omega L/c + \pi = \phi + 4πfL/c + \pi = \phi + 4πL/\lambda + \pi$$

Thus the reflected ray has phase angle $$\phi + 4πL/λ + π$$ and refracted ray which has just returned to the surface has phase angle $$\phi + (4πL/λ)$$

According to this, their phase different is π and hence the film should appear dark! I know this is wrong somewhere but where have i gone wrong?

Nothing wrong. That is how the calculations are done. A perfectly flat glass with parallel edges is an etalon. Under monochromatic light of just the right wavelength, it is dark.

You should consider that part of the wave is reflected off the top surface to the bottom and back up. And some makes another round trip. You get an infinite series. The sum of the series is the real answer.

A very small change in wavelength or thickness or angle changes it from destructive to constructive interference. Under normal light you have both happening and it doesn't look different from ordinary.

The same idea goes into thin film anti-reflection coatings. These are often a single layer of MgFl with a thickness of $$\lambda / 4$$. Because it is so thin, it takes a much bigger change in $$\lambda$$ to get constructive interference.

This is often used on glasses. They are designed for total destructive interference in the middle of the visible spectrum. This is why they have a faint purple reflected wave.

You can do a better job with two layers. You can choose two wavelengths with totally destructive interference.

The condition for constructive interference is if the reflected wave comes back in phase with the original wave. The reflected wave must have a phase change of $$2 \pi N$$ for some integer N.

This means

$$\pi + (4 \pi L/ \lambda_{glass}) = 2 \pi N$$

or

$$L = (N - 1/2)\frac{\lambda_{glass}}{2}$$

For $$N = 1$$,

$$L = \lambda_{glass}/4$$

You also get constructive interference in the reflected wave if $$L/\lambda_{glass}$$ is $$3/4$$ or $$5/4$$ or ...

Note that $$\lambda_{glass}$$ is the wavelength of light in the glass. Given an index of refraction $$n$$,

$$\lambda_{glass} = \frac{\lambda_{air}}{n}$$

• Thanks for helping out! Although I get this part: "You should consider that part of the wave is reflected off the top surface to the bottom and back up. And some makes another round trip. You get an infinite series. The sum of the series is the real answer." I am still unable to understand where the role of wavelength comes in mathematically since in my equations, the conditions for interference are independent of wavelength. Commented Mar 31, 2023 at 15:59
• Ahhh thanks! It finally clicked to me after a few days where i was going wrong! Commented Apr 8, 2023 at 8:49