# Why there is always a bright fringe at angle zero in light's double-slit experiment?

The explanation I read is that because at $$\theta = 0$$ the waves travel the same distance from both slits to A and thus have no phase difference. They add up constructively so there must be a bright fringe there.

But I am wondering if the waves arrive at A when their amplitude is zero (phase = $$n*180^o$$ for a sin wave), wouldn't the resultant amplitude also be zero and thus no maximum appears at $$\theta = 0$$?

And for a more analytical description, the diffraction intensity from two slits is $$I(\theta) = I_0 \mathrm{sinc}^2\left(\frac{L\pi}{\lambda}\sin\theta\right)\cos^2\left(\frac{\pi a}{\lambda}\sin\theta\right)$$ where $$L$$ is the width of the slit.

If $$I(0) = 0$$, wouldn't there be no pattern everywhere on the screen?

• Why would $I(0)=0$? Both $\operatorname{sinc}$ and $\cos$ are equal to $1$ at $0$... Aug 29, 2021 at 10:07
• Because $I_0 = 0$ Aug 29, 2021 at 10:11
• But $I_0$ is a constant. Aug 29, 2021 at 10:12
• OP, if $I_0=0$ you have no light going through the slots at all. Please clarify why you think having no light at all is the same thing as the typical double slit system. Aug 29, 2021 at 10:49
• @Winniebear, not sure if this is what you mean, but yes, you can set $I_0$ to $0$, to the make the intensity zero everywhere, and then there will be no bright spot in the middle. But that's also not a very interesting case, and not really representative of the double-slit experiment. So the "always" isn't absolute, but must be taken in the context. ("Bright" is also not absolute, as you could flood the screen with another light, swamping the diffraction pattern.) Aug 29, 2021 at 22:24

The instantaneous displacement (electric field vector) changes with both position and time according to ($$kx-\omega t$$). At a fixed point, it is still oscillating at high frequency (on the order of $$10^{15} \;\text{Hz}$$ for visible light). The intensity at a point on the screen is proportional to the average of the square of the total displacement $$\langle\left(E_0\sin\omega t +E_0\sin(\omega t +\phi)\right)^2\rangle$$ where $$\phi$$ represents the path difference. At point A, the waves interfere constructively ($$\phi=0$$). Therefore, the only way for the intensity to be zero is if the amplitude $$E_0=0$$.