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Imagine this problem: A light incide in a thin film with thickness $d$, it incides in such way that the angle it makes with the normal is theta. The film has refractive index $n_2$ and the initial medium has refractive index $n_1$.

Now, the reference image for my calculations is this:

enter image description here

I just can't understand the signs i should put on the equation of diference of path! See:

For the light propagating ABC, i think we can say the path is essentialy $2dn_1/\cos \theta_2$. Now, the problem is with the path of the light AD, the light that is reflected at the above surace. Shouldn't it be "$2n_2d\tan(\theta_2)\sin(\theta_1) + \lambda/2$"? Where i am considering there is change of phase in the reflection.

So,, in the end, the difference of the optical path would be "$2dn_1/\cos \theta_2-(2n_2d\tan(\theta_2)\sin(\theta_1) + \lambda/2) = 2n_2d\cos\big(\theta_2) - \lambda/2$". But, apparently, this is wrong.

I tried to apply that for a real question essentially equal to the question i posted here. The author gaves that the difference of path would be "$2n_2d\cos\big(\theta_2) + \lambda/2$" and I can't understand why.

What is the critery for the choice of the sign due to the refletion?

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As far as the phase is concerned, adding or subtracting $π$ is totally equivalent. You can't tell the difference between the two. And the phase difference is what matters in optics.

The passage to the optical path is a convention: what is the difference of optical path which would give the same phase difference. So, adding or subtracting $λ/2$ is the same thing!

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  • $\begingroup$ Yes. But let me add something: In the problem i said above, the author asks for the thickness d necessary to occurs 2nd order interference. SO, in my case, i would need to use "$2\lambda = 2ndcos(\theta) -\lambda/2$", but he uses "$2\lambda = 2ndcos(\theta) +\lambda/2$", so that, even so add or subtract half of wavelength is equivalent, our answers differs! So who would be right? $\endgroup$
    – LSS
    Commented Sep 18, 2021 at 17:13
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    $\begingroup$ This is because, in this case, the definition of the interference order is also a convention. We must speak of the interference fringe for which the difference of the true optical path is 0, or $λ$, or...... There would be no more ambiguity. $\endgroup$
    – Vincent Fraticelli
    Commented Sep 18, 2021 at 17:18

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