Young's double slit gives bright fringes where $d\sin\theta = m\lambda$. But how bright is the screen between the fringes? And why does a grating, with thousands of slits, give such sharp lines? The phase-difference formula answers both.
Key idea
For two narrow slits a distance $d$ apart, the intensity at angle $\theta$ is $$I = I_{\max}\cos^2\!\left(\frac{\pi d\sin\theta}{\lambda}\right).$$ For $N$ equal slits the principal maxima stay at $d\sin\theta = m\lambda$, but they are $N^2$ times as bright as one slit alone and only about $\frac{1}{N}$ as wide, with $N - 1$ dark points and $N - 2$ small secondary maxima between neighboring principal maxima.
Why it works
Light from the two slits travels paths that differ by $\Delta r = d\sin\theta$, so the waves arrive with phase difference $\phi = \frac{2\pi d\sin\theta}{\lambda}$. Two equal waves with that phase difference add to amplitude $2E_0\cos\frac{\phi}{2}$. Intensity goes as amplitude squared: $I = 4E_0^2\cos^2\frac{\phi}{2}$. The largest value, $I_{\max} = 4I_0$, is four times one slit, because the amplitudes add before squaring. With $N$ slits, each next slit adds the same phase $\phi$. At a principal maximum all $N$ amplitudes line up: amplitude $NE_0$, intensity $N^2I_0$. Turn $\theta$ a little and the $N$ small phase steps soon add up to a full circle, so the waves cancel. That happens when $N\phi = 2\pi$, which is closer to the maximum the larger $N$ is.
Worked examples
Worked example: Between the fringes
Two slits 0.100 mm apart are lit with 600 nm light, and the screen is 2.00 m away. Find the fringe spacing and the intensity, as a fraction of $I_{\max}$, at 3.00 mm from the center.
- Small angles: $\sin\theta \approx y/L$, so bright fringes are spaced $\Delta y = \frac{\lambda L}{d} = \frac{(600 \times 10^{-9})(2.00)}{1.00 \times 10^{-4}} = 12.0$ mm.
- At $y = 3.00$ mm: $\frac{\pi d\sin\theta}{\lambda} = \frac{\pi d y}{\lambda L} = \frac{\pi(1.00 \times 10^{-4})(3.00 \times 10^{-3})}{(600 \times 10^{-9})(2.00)} = \frac{\pi}{4}$.
- $I/I_{\max} = \cos^2(\pi/4) = 0.500$.
Answer: Fringes 12.0 mm apart; half intensity at 3.00 mm, a quarter of the way to the next fringe
Worked example: Five slits
Five equal slits replace the two. Compared with one slit alone, how bright is a principal maximum? How many secondary maxima lie between neighboring principal maxima? Where is the first dark point next to the central maximum, for $d = 0.100$ mm and $\lambda = 600$ nm?
- At a principal maximum the five amplitudes add in step: $(5E_0)^2 = 25I_0$.
- Between principal maxima there are $N - 1 = 4$ dark points and $N - 2 = 3$ secondary maxima.
- First dark point: $N\phi = 2\pi$, so $d\sin\theta = \lambda/N$: $\sin\theta = \frac{600 \times 10^{-9}}{5(1.00 \times 10^{-4})} = 1.20 \times 10^{-3}$, $\theta = 1.20$ mrad.
- For two slits the first dark point is at $d\sin\theta = \lambda/2$, 2.5 times farther out.
Answer: $25I_0$; 3 secondary maxima; first dark point at 1.20 mrad
Common mistake
Thinking more slits make the bright lines wider (“more light, bigger spots”). They make them narrower and brighter. The positions are fixed by $d\sin\theta = m\lambda$; the width of each line shrinks like $1/N$, since $N$ phase steps of a small size already add to a full circle. That is why a grating with thousands of lines separates wavelengths so cleanly.
Common mistake
Taking $I_{\max}$ as twice one slit's intensity. Amplitudes add, then you square: two slits in step give $2^2 = 4$ times one slit's intensity, not 2.