Shouldn't a part of the ray have to travel $\frac{\lambda_2}{2}$ extra path in the medium $n_2$?

Basically, my teacher had given the class a question regarding thin film interference that, there is a medium with refractive index $$n_1$$ below that there is another medium with thickness $$t$$ with refractive index $$n_2$$ and below that there is another medium with refractive index $$n_3$$.

Also $$n_3 > n_2$$ and $$n_2 > n_1$$, as well as we know that the wavelength of light in medium with refractive index $$n_2$$ is $$\lambda_2$$.

Now if a ray from the top (from medium with refractive index $$n_1$$) is projected downwards and an observer's eye is in that medium then, we have to find the condition such that there is formation of maxima in the eye.

I know that for formation of maxima, the path difference between the part of the ray that gets reflected back by the medium $$n_2$$ and the part of the ray that gets reflected back by the medium $$n_3$$ after entering $$n_2$$ should be $$n \lambda$$ (where $$n$$ is a non negative integer), which can be written as an equation as follows: $$2\frac{n_2}{n_1}t + \frac{\lambda_2}{2} - \frac{\lambda_1}{2} = n\lambda_1$$ where $$\lambda_1$$ is the wavelength of light in medium $$n_1$$ .

But according to my teacher, the extra path that the ray would have to travel when it is reflected back by medium $$n_2$$ (since it will be going from a rarer to denser medium) and the extra path the ray has to travel when it is reflected back by medium $$n_3$$ would be same i.e. $$\frac{\lambda_1}{2}$$.

So as per my teacher the equation should be $$2\frac{n_2}{n_1} t = n\lambda_1$$

But shouldn't it be $$\frac{\lambda_2}{2}$$ for the part of the ray that gets reflected back by the medium $$n_3$$ since when light wave goes from a rarer to a denser medium its phase changes by $$\pi$$?

• What do you mean by "reflected by the medium n2" - I assume you're referring to the n1-n2 interface? Sep 8, 2021 at 9:39
• @catalogue_number yes I am referring to the n1-n2 interface and when I say reflected back by medium n3 after entering n2, I am referring to the n2-n3 interface Sep 8, 2021 at 9:58
• Regardless of the relative $n$ ratio at an interface, the wavelength remains $\lambda_2$ all the time it's inside the $n_2$ medium. Sep 8, 2021 at 17:04
• @CarlWitthoft What I am asking is since the ray goes from medium $n_2$ to medium $n_3$ it should experience a phase change of $\pi$, which can be rephrased as it would have to travel $\frac{\lambda}{2}$ distance extra. For that shouldn't the wavelength be $\lambda_2$ instead of $\lambda_1$ like my teacher has assumed ? Sep 9, 2021 at 2:40
• Ahhh, as opposed to $n_3 < n_2$ . But take a look at the ray it's interfering with that was reflected from the first interface, with similar phase flip rules. Sep 9, 2021 at 16:23