1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ 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
    Sep 18 '21 at 17:13
  • 2
    $\begingroup$ 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. $\endgroup$ Sep 18 '21 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.