A glass plate of index 1.59 is covered with a thin, uniform layer of oil of index 1.26.?

Question

A light beam of variable wavelength normally incident from air onto the oil surface. Observation of the reflected beam shows destructive interference at 503 nm and constructive interference at 709 nm with no intervening maxima or minima. Find the thickness of the oil layer. Answer in units of nm.

Answer 1

Let d be the thickness of oil.
All variables denoted by capital letters
are complex, i^2 = -1.

There are five waves involved in the problem:

in vacuum:
amplitude of incident forvard wave A_v = 1,
E = exp(k_v x)
B = exp(k_v x)
amplitude of reflected backward wave Cv,
E = Cv exp(-k_v x)
B = -Cv exp(-k_v x)

in oil:
amplitude of refracted forvard wave Ao,
E = Ao exp(k_v x)
B = n_o^2 Ao exp(k_v x)
amplitude of reflected backward wave Co,
E = Co exp(-k_v x)
B = - n_o^2 Co exp(-k_v x)

in glass:
amplitude of refracted forvard wave Ag = 1.
E = Ag exp(k_g (x-d))
B = n_g^2 Ag exp(k_g (x-d))

Both E and B are continuous on both surfaces,
these conditions can be written as a set of
four simultaneous linear eqations for four
variables:

1 + Bv = Ao + Co
1 - Bv = n_o^2 Ao - n_o^2 Co
Ao exp(k_v d) + Co exp(-k_v d) = Ag
n_o^2 Ao exp(k_v d) - n_o^2 Co exp(-k_v d) = n_g^2 Ag

Solution of these equations is straightford,
but real pain in the neck, so I think I'll
pass the batton here.

Answer 2

Destructive interference would be when waves are out of phase by 1/2 so they cancel and Contructive wavelengths are in phase and add.

The possible destructive thicknesses for the oil are 251, 754, 1257, 1760, etc or 503(n + .5) and the constructive thicknesses are 709, 1418, 2127, etc You are looking the the numbers to be equal. It looks like this happens around 4250-4270nm or so splitting the difference would be about 4260nm. However this is the distance through the oil and back so cut that in half to about 2130nm for the thickness of the oil.

I hope I did this right.