物理deffraction-thin film

1. Using 2nt = (m-0.5)λ for constructive interference to occur at the reflected ray.
where n is the refractive index of the film
t is its thickness
λ is the wavelength of the incident light
m is an integer

hence, λ = 2nt/(m-0.5) = 2 x 1.33 x 320/(m-0.5)
The value of λ that falls within the visible spectrum (between 400 nm ~ 700 nm) occurs when m = 2
i.e. λ = 2x1.33x320/(2-0.5) nm = 567 nm

[note: normal incident means the incident ray is perpendicular to the surface, i.e. the ray is along the normal, angle of incident = 0]

2. Using the above formula
λ = 2x 1.5 x 400/(m-0.5) nm
Again, the value of λ falls within the visible spectrum occurs when m = 3,
hence, λ = 2x1.5x400/(3-0.5) nm = 480 nm

3. Use the same formula,
2nt = (m-0.5)λ, here n = 1 for an air film
thus, m = 2t/λ + 0.5 = 2 x 48000/680 + 0.5 = 141.6
therefore, there will be 141 bright fringes be observed.

The interference occurs from the rays reflected on top and bottom of the AIR FILM, hence the wavelength of light in the AIR FILM needs to be used.

可唔可以解釋下點解又low-medium-high 時要用(m+1/2)λ系destructive interferance到,但當low-high-low, or high-low-high時,又系用(mλ)既?

Remember that whenever there is reflection from a less dense to a denser medium, there is a phase change of λ/2 on the reflected ray.
Hence, when, say a water film, is on a glass plate (refractive index 1.5), the reflected rays on top of the water suffers a phase change of λ/2 (from air to water), and also at the bottom (from water to glass). The two phase changes of λ/2 thus cancelled out. Therefore, use the formula 2nt = mλ for constructive interference.

However, if the water film is surrounded by air at the top and bottom, there is phase change at the top (air to water, less dense to denser medium) but NOT at the bottom (water to air, denser to less dense medium), hence the formula 2nt = (m-0.5)λ needs to be used.