In 1801 Thomas Young sent light through two close slits and saw bright and dark stripes, proof that light is a wave. Each slit acts as a source, the two are in step, and the waves from them interfere. Where they arrive in step, the screen is bright.
Key idea
For slits a distance $d$ apart, bright fringes appear where $$d\sin\theta = m\lambda, \quad m = 0, \pm 1, \pm 2, \dots$$ and dark ones where $d\sin\theta = (m + \frac{1}{2})\lambda$. For small angles on a screen at distance $L$: $y_m = \dfrac{m\lambda L}{d}$, so bright fringes are $\Delta y = \dfrac{\lambda L}{d}$ apart.
Why it works
From a far point on the screen, the rays from the two slits are almost parallel, and the lower ray travels farther by the short side of a small right triangle: $d\sin\theta$ (sine of the angle times the hypotenuse $d$). A path difference of a whole number of wavelengths puts crest on crest. For small angles, $\sin\theta \approx \tan\theta = y/L$, which gives $y = m\lambda L/d$.
Common mistake
Thinking that moving the slits closer together squeezes the fringes closer. It spreads them: $\Delta y = \lambda L/d$, so a smaller $d$ gives a bigger spacing. Check the extreme: as $d$ shrinks toward zero, the two slits become one source and the pattern spreads out into one wide blur.
Worked example: Fringe spacing
Light of $600\ \text{nm}$ passes through slits $0.20\ \text{mm}$ apart onto a screen $2.0\ \text{m}$ away. How far apart are the bright fringes?
- $$\Delta y = \frac{\lambda L}{d} = \frac{(6.00 \times 10^{-7})(2.0)}{2.0 \times 10^{-4}}.$$
- $\Delta y = 6.0 \times 10^{-3}\ \text{m} = 6.0\ \text{mm}$.
Answer: $6.0\ \text{mm}$
Worked example: Measuring a wavelength
With slits $0.10\ \text{mm}$ apart and a screen $1.5\ \text{m}$ away, the third bright fringe (not counting the center) is $2.7\ \text{cm}$ from the center. Find $\lambda$.
- $y_3 = \dfrac{3\lambda L}{d}$, so $\lambda = \dfrac{y_3 d}{3L}$.
- $$\lambda = \dfrac{(0.027)(1.0 \times 10^{-4})}{3(1.5)} = 6.0 \times 10^{-7}\ \text{m}.$$
- That is $600\ \text{nm}$, orange light.
Answer: $600\ \text{nm}$
Worked example: A larger angle
Slits $2.0\ \mu\text{m}$ apart are lit with $500\ \text{nm}$ light. Find the angle of the second-order bright fringe.
- The angle is not small, so use $d\sin\theta = m\lambda$ directly.
- $\sin\theta = \dfrac{2(500 \times 10^{-9})}{2.0 \times 10^{-6}} = 0.50$, so $\theta = 30^\circ$.
Answer: $30^\circ$