# Double slit formula derivation. Why not $I_{\theta} = 4 I_m (\cos \beta)^2 \left( \frac{\sin{\alpha}}\alpha \right)^2$?

The intensity of the double slits is given by $$I_{\theta} = I_m (\cos \beta)^2 \left( \frac{\sin{\alpha}}\alpha \right)^2$$ where $$\alpha = \frac{\pi a}{\lambda}\sin \theta$$ $$\beta = \frac{\pi d}{\lambda} \sin \theta$$ where $$d$$ is the distance between the centerlines of the slits and $$a$$ is the width of each slits.

I understand that the intensity of a single slit is $$I_\theta = I_m \left( \frac{\sin{\alpha}}\alpha \right)^2$$, but if there are two slits shouldn't the intensity become four times as much: $$I_\theta =4 I_m \left( \frac{\sin{\alpha}}\alpha \right)^2$$? Because the two diffracted waves are coherent and thus they interfere and so the amplitude is twice as much as of a single slit and since the intensity is proportional to the square of the amplitude, the resultant intensity of the two slits should be multiplied by 4

Note: I know the proof of all of formulas written here.

• Is there a definition of Im provided? I'm guessing it is maximum intensity. In that case it would absorb the factor of 4 into the definition of Im. Feb 4, 2022 at 11:46

Here's one of many explanations (emphasis mine) .

If one of the slits in the double-slit experiment is covered, what will be the new intensity? Please explain.

If only one slit is open the intensity (= rate of arrival of photons somewhere) becomes evenly distributed (no fringes) and at one quarter of the peak intensity from when both slits are open. The peak probability amplitude is halved, and the intensity goes as the square of the modulus of the probability amplitude; the new even intensity is a quarter of the old peak intensity value.

You are correct that when you go from one slit to two slits, the center maximum is 4x greater in intensity.

I don't know which book you are going off of, but I am willing to bet that it is mainly concerned with the relative intensity rather than the absolute intensity. In other words, we aren't interested in calculating the exact value of $$I_{\theta}$$ based on all the variables (such as the distance from the slits to the target screen or the initial intensity of the laser, etc, etc), and instead we are interested in finding the overall look of the intensity pattern with respect to $$\theta$$.

Because we are talking about relative intensity, whether we put $$I_{m}$$ or $$4I_{m}$$ is going to be a matter of convention.

It seems like the book chose $$I_{m}$$ to be the intensity of the double-slit pattern at the center maximum. You're asking why can't it be the intensity of the pattern at the center maximum when one of the slits is covered.

The answer is that it can be that, but it ends up not mattering if all you care about is the relative intensity graph. The book simply chose a different convention.