What's the condition for the intensity of the interference of light to be zero?

Question

I need to solve this question:

"Suppose light comes from a medium $n_1$, refract a part of it in a medium of index $n_2$. Below the $n_2$ material, there is a $n_3$ material with refractive index greater than $n_2$. So there occurs part of reflection and part of refraction. What is the condition that needs to be satisfied if we want that the intensity of the reflected light be zero?"
(The incidence is perpendicular, $n_3>n_2>n_1$.)

I got the right answer which is $n_2= \sqrt{n_1 n_3}$. The problem is that I am not sure why this is right.

To get this answer, intuitively I thought that the reflection coefficients of intensity should be equal, to cancel. That is, let $r_{ij}=(\frac{n_2- n_3}{n_i+ n_j})^2$. So the condition is $r_{12}=r_{23}$.

But, in fact, I have no idea why this is right.
See:

The first electric field incides at the intersection 12. So, $E_{ref} = -RE, E_{trans} = TE$. The transmitted electric fields incides at intersection 23, and the reflected field is $E_{ref} = -R'TE$. This reflected fields incides at 12 and got transmitted with amplitude $E = -R'TTE$.

So, summarising, we have, at the end:
$$E = -ER - ET^2R'e^{ikn_{2}d}$$

To get the intensity equal to zero, $|E|*|E| = 0$.

This is equivalent to $$R²+T^4R'^2 + 2RT^2R'cos(2kn_2d)$$

Now, if we substitute the answer here, it will not be zero. Honestly, I am not sure what is wrong.

I would appreciate if you help me to understand why $n_2= \sqrt{n_1 n_3}$ makes the intensity zero. In other words, how to get $r_{12}=r_{23}$ by some algebra, and not by mere intuition?

Also, where is the error in my second attempt? I think maybe the error is that it should, in fact, have a progression geometric in T. But not sure about that.

thx

Answer 1

You started correctly, but have to sum up over infinite times of reflections, transmissions etc., which leads to series expression. Then you have to find for the phase a destructive condition $d=(2m+1)\frac{\lambda_1}{4}$.

Next, for optimal performance (not to gain destructive intereference), you must demand the equality of the reflectance R12 = R23 = $\vert{\frac{n_1-n_2}{n_1+n_2}}\vert^2 =\vert{\frac{n_2-n_3}{n_2+n_3}}\vert^2$ which gives $n_2=\sqrt{n_1n_3}$ .

Remember light flux is proportional to the square of the electric field amplitude