Thinnest film that will give constructive interference for red light

[prettykitty]

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!

 

[haruspex]

[2nt/λ] - 1/2 = m

Are you sure that's the right equation? http://en.wikipedia.org/wiki/Thin-film_interference#Soap_bubble gives a different one (and the desired answer).

 

[prettykitty]

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

Here is my instructors answer:
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]

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.

 

[Nickie]

Hello,

The main issue with your solution is that you are using the wrong equation. The equation you are using, [2nt/λ] – 1/2 = m, is for the general case of constructive interference in a thin film. However, in this problem, we are specifically looking for the thinnest film that will give constructive interference for red light.

In this case, we can use the equation t = (m + 1/2)λ/2n, where m is the order of the interference (in this case, m = 1), λ is the wavelength of the light, and n is the index of refraction of the soap. Plugging in the given values, we get:

t = (1 + 1/2)(752e-9)/2(1.33) = 141 nm

So, the thinnest film that will give constructive interference for red light is 141 nm, which is the correct answer. I hope this helps clarify things for you.