Young's double slit experiment

The experiment that settled it

In 1801 Thomas Young passed sunlight through a tiny hole and then through two more holes close together. On a screen he saw bright and dark bands. A particle picture of light could not explain it. A wave picture could, if light from the two holes added in the way of the last lesson.

A narrow slit $S_0$, lit by a monochromatic source, sends a wavefront to two narrow slits $S_1$ and $S_2$ a distance $d$ apart. The two slits are at the same distance from $S_0$, so they are always in phase: they are coherent. The screen is a distance $D$ away, with $D$ much larger than $d$. The point $O$ on the screen is on the line through the middle $M$ of the slits.

The path difference

Take a point $P$ at a height $y$ above $O$. The wave from $S_2$ travels farther than the wave from $S_1$. With the centre of $S_1S_2$ at the origin, the squares of the paths are

Subtracting, $S_2P^2 - S_1P^2 = 2yd$. So $(S_2P - S_1P)(S_2P + S_1P) = 2yd$. For $D \gg d$ and $D \gg y$, both paths are almost $D$, so $S_2P + S_1P \approx 2D$ and

In terms of the angle $\theta$ of the line $MP$, this is $\Delta = d\sin\theta \approx d\theta$ for small angles.

Bright and dark fringes

Put $\Delta$ into the conditions of the last lesson.

• Bright fringes: $\dfrac{yd}{D} = n\lambda$, so $y_n = \dfrac{n\lambda D}{d}$, with $n = 0, \pm1, \pm2, \dots$ The place $n = 0$ is the central bright fringe.
• Dark fringes: $y = \left(n + \tfrac12\right)\dfrac{\lambda D}{d}$. The first dark fringe is at $n = 0$, half a fringe width from the centre.

The distance between two neighbouring bright fringes is the same everywhere. It is called the fringe width:

The angular fringe width is $\beta/D = \lambda/d$. Dark fringes are also $\beta$ apart.

The intensity on the screen is $I = 4I_0\cos^2\!\left(\dfrac{\pi y d}{\lambda D}\right) = 4I_0\cos^2\!\left(\dfrac{\pi y}{\beta}\right)$, where $I_0$ is the intensity from one slit alone.

Which changes make the fringes wider: red light in place of blue, slits closer together, or a screen farther away? All three. The fringe width grows with $\lambda$ and $D$, and falls as $d$ grows.

A worked example. Sodium light of wavelength $589$ nm is used with slits $0.50$ mm apart. The screen is $1.20$ m away. Find the fringe width, the angular fringe width and the distance of the 4th bright fringe and the 3rd dark fringe from the centre.

1. Fringe width. $\beta = \dfrac{\lambda D}{d} = \dfrac{589 \times 10^{-9}\ \text{m} \times 1.20\ \text{m}}{0.50 \times 10^{-3}\ \text{m}} = 1.41\ \text{mm}$.
2. Angular fringe width. $\dfrac{\beta}{D} = \dfrac{1.41\ \text{mm}}{1.20\ \text{m}} = 1.18\ \text{mrad}$.
3. The 4th bright fringe. $y_4 = 4\beta = 5.65\ \text{mm}$.
4. The 3rd dark fringe. The first is at $\tfrac12\beta$, so the third is at $2.5\beta = 3.53\ \text{mm}$.
5. Check. The small angle is $1.18$ mrad, far below $1$ rad, so $\sin\theta \approx \theta$ is fine.

White light, water and the conditions

• White light. All colours have a bright central fringe, since the path difference is zero for every colour there. The centre is white. Next, the fringe width $\beta \propto \lambda$ is smaller for blue than for red. The first bright fringes are therefore coloured, with violet near the centre and red outside. With $d = 0.50$ mm and $D = 1.20$ m, a first fringe of $400$ nm light is at $0.96$ mm and one of $700$ nm light is at $1.68$ mm. After a few fringes the colours overlap and wash out.
• In water. The wavelength falls to $\lambda/n$. The fringe width becomes $\beta/n$. With the same setup, sodium light in water of index $1.33$ gives $1.06$ mm in place of $1.41$ mm.
• One slit closed. The fringes vanish. A wide patch from one slit remains (course two of this chapter).

The conditions for a clear pattern:

• The light is monochromatic, and the two slits are lit by the same wavefront, so they are coherent.
• The slits are narrow, so each spreads light over the whole screen.
• $d \ll D$, and $d$ is small enough that $\beta$ is large enough to see. A distance $d$ of a few tenths of a millimetre is typical.
• The two slits give nearly equal intensities. Then $I_{\min}$ is close to zero and the contrast is best.

Other books. Some books call $\beta$ the fringe separation or fringe spacing. The word "order" for $n$ is common in A-Level and IB. The result is the same.

Practise with new numbers

Try it in Python

The approximation $\Delta = yd/D$ is good when the angles are small. The program finds where the bright fringes really are from the exact path difference, and compares them with $n\lambda D/d$.

Common mistakes

• Mixing up the number of the dark fringe. The first dark fringe is at $\tfrac12\beta$, the second at $\tfrac32\beta$, and so on.
• Mixing the units. The wavelength is in nm, the slit distance in mm and the screen distance in m. Change all to metre before using $\lambda D/d$.
• Thinking that fringe width depends on the slit width. It depends on $\lambda$, $D$ and $d$ only. The slit width changes how bright the pattern is.
• Forgetting the medium. The wavelength in water or glass is $\lambda_0/n$.

Remember these

• $\Delta \approx yd/D$ for $D \gg d$ and $D \gg y$.
• Bright: $y_n = n\beta$. Dark: $y = (n + \tfrac12)\beta$. All fringes are $\beta = \lambda D/d$ apart.
• The intensity is $4I_0\cos^2(\pi y/\beta)$.
• In white light, the centre is white and the first fringes have violet inside and red outside.
• In a liquid, $\beta$ is divided by $n$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Young's experiment is done in air. What happens to the fringe width in each case? (a) The screen moves to twice the distance. (b) The slits move to twice their separation. (c) The whole setup goes into water of index $1.33$.

2. Why is the centre of the pattern white when white light is used, but the fringes next to it are coloured?

3. What is the path difference at the 5th bright fringe from the centre?

4. One of the two slits is closed. What do you see on the screen?

References

• OpenStax, University Physics Volume 3, Chapter 3 (Interference): Young's double slit.
• NCERT Physics Class 12, Chapter 10 (Wave Optics): interference and Young's double slit experiment.