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:
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?