Thin film interference question

Question

Suppose two microscope slide are held apart by a couple of hairs, so that there is a space between the slides of 0.80 micrometers. (Thus, we have an air space between the glass slides. The hairs merely hold the spacing constant.) When white light is shined on the slides at normal incidence and the reflected light is viewed, there are two visible wavelengths which are reflected constructively. What are these wavelengths?

Answer 1

Let

d = 0.8 μm = 800 nm be the thickness of the air gap,

λ (to be calculated) be the wavelength of light experiencing constructive reflection.

The light is reflected twice:

1) At the boundary between the first glass slide and the air gap

light gets reflected without change in phase because glass has higher

coefficient of refraction than air.

2) At the boundary between the air gap and the second glass slide

light gets reflected with reversal of phase, because the relationship

between the two media is the opposite.

The light reflected at the second interface then travels back to

the first glass slide.

There it interferes constructively with the first reflection,

provided the difference in phase is an integer multiple of 2π.

The light at the second interface has a phase difference (vis a vis the first interface)

equal to 2πd/λ.

The reflected light undergoes a phase reversal, which amounts to adding π to the phase.

Then the light travels back to the first interface undergoing another phase shift of 2πd/λ.

The total phase shift is then 4πd/λ + π.

And this is the quantity that must be an integer multiple of 2π.

That is,

4πd/λ + π = 2kπ

for some integer k.

From that,

λ = 4d/(2k-1) = 3200/(2k-1)

For this light to be visible, the wavelength must fall in the range 380 nm ≤ λ ≤ 780 nm.

By trying various values of k, we get two visible wavelengths

λ = 3200/5 = 640 nm

λ = 3200/7 = 457 nm