Optical path difference in a soap film interference

[lorenz0]

Homework Statement

A soap water film $n = 1.46$ with a thickness of $d = 10^{-6} m$ is placed on a glass plate with a refractive index of $n = 1.6.$ The film is perpendicularly illuminated by green light with a wavelength of $566 nm.$ What is the optical path difference between the rays reflected from the two sides of the film, expressed in terms of the wavelength of the light inside the film?

Relevant Equations

$\Delta=\sum_i n_i d_i$

The ray that is reflected at the air-soap interface has an optical path length of $lambda_{soap}/2$, while the one refracted in the soap has an optical path length of $2 n_{soap} d+\frac{lambda_{glass}}{2}$ due to its traversing the soap film and also reflecting on the soap-glass interface. Thus the optical path difference between the two rays is $\Delta=2 n_{soap} d+\frac{lambda_{glass}}{2}-\frac{lambda_{soap}}{2}$ which can also be written as $\Delta=\frac{2 n_{soap} d+\frac{lambda_{glass}}{2}-\frac{lambda_{soap}}{2}}{lambda_{soap}}lambda_{soap}\approx 7.49 lambda_{soap}$, which according to the stated solution is wrong, since it should be $5.16 lambda_{soap}.$ I would be grateful if someone could tell me where I am going wrong. Thanks.

 

[kuruman]

There is phase change of 180° when the light reflects at the vacuum-soap interface and at the soap-glass interface so these cancel when you subtract. What's left?

 

[lorenz0]

That was my initial guess (that the phase shifts cancel out), and in that case it seems to me that $\Delta=\frac{2n_{soap}d}{lambda_{soap}} lambda_{soap}=\frac{2 n^2_{soap}d}{lambda_0} lambda_{soap}$, so it still is not in agreement with the stated solution

 

[kuruman]

That was my initial guess (that the phase shifts cancel out), and in that case it seems to me that $\Delta=\frac{2n_{soap}d}{lambda_{soap}} lambda_{soap}=\frac{2 n^2_{soap}d}{lambda_0} lambda_{soap}$, so it still is not in agreement with the stated solution

That's because you counted the index of refraction of soap twice. The number of wavelengths you can fit in distance $d$ in vacuum is $$N=\frac{2d}{\lambda_{vac}}.$$ How many wavelengths can you fit in the same distance $d$ when the wavelength is $\lambda_{vac}/n_{soap}$?

 

[lorenz0]

@kuruman Got it. Thank you very much