# Thinnest film that will give constructive interference for red light

## Homework Statement

When a thin soap ﬁlm is very thin, we see it as black. How thick (in nm) is the ﬁlm
in the region where we see the ﬁrst red band? Take the wavelength of red light to be 752
nm and the index of refraction of the soap to be n = 1.33.

## Homework Equations

[2nt/λ] - 1/2 = m

## The Attempt at a Solution

Solving for t:
First with m = 1 I moved the 1/2 over to get 3/2, Then I multiplied 3/2 by the wavelength 752e-9 and divided this result by 2n, n being 1.33.
The answer I attained after solving for t was: 424 nm.
The correct answer is 141 nm.

I still don't know what I'm doing wrong here. I have noticed that if I divide my answer by 3 I get the correct answer, though this may just be coincidence.
Also if anyone is using Walker edition 4, on page 988 the practice problem seems to be this exact same scenario and I am having the exact same problem.
Thanks!

So then the desired equation is 2ndcosθ=(m-.5)λ?
How would I figure out the angle? I thought it was the right equation.

If the thickness of the ﬁlm is t, the diﬀerence in phase between a ray reﬂecting oﬀ the
top surface of the ﬁlm and one reﬂecting oﬀ the bottom surface of the ﬁlm is (2πn/λ)2t + π
the factor π coming from the phase change on reﬂection of the ray in air reﬂecting oﬀ
the ﬁlm. Set this phase change to 2π for the ﬁrst constructive interference to get that
t = λ/4n = 752/(4 × 1.33) = 141nm.

I don't really understand how he got his answer either even though it is spelled out quite succinctly.

haruspex
Homework Helper
Gold Member
2020 Award
So then the desired equation is 2ndcosθ=(m-.5)λ?
How would I figure out the angle? I thought it was the right equation.
It's the angle of incidence, so for the OP it's 0.
the diﬀerence in phase between a ray reﬂecting oﬀ the
top surface of the ﬁlm and one reﬂecting oﬀ the bottom surface of the ﬁlm is (2πn/λ)2t + π
the factor π coming from the phase change on reﬂection of the ray in air reﬂecting oﬀ
the ﬁlm. Set this phase change to 2π for the ﬁrst constructive interference to get that
t = λ/4n = 752/(4 × 1.33) = 141nm.
That's the same equation as I referenced. For the mth constructive interference phase, (2πn/λ)2t + π = 2πm. Rearranging: (2nt/λ) + 1/2 = m. You just had the sign wrong on the 1/2.