Young's slits and diffraction gratings

Young's double-slit experiment

In 1801 Thomas Young shone light through two narrow slits very close together and saw a pattern of bright and dark fringes on a screen. This was strong evidence that light is a wave. Two particles hitting a screen would give two bright patches. Two overlapping waves can interfere.

Each slit acts as a new source of waves (this is diffraction). The two sets of waves overlap on the screen. For the pattern to stay steady, the sources must be coherent: they have the same frequency and a constant phase difference. Two slits lit by the same laser beam are coherent, because both are fed by the same wavefront.

At any point on the screen, what matters is the path difference, the extra distance the light from one slit travels compared with the other.

For small angles, the geometry gives the fringe spacing (the distance between neighbouring bright fringes):

where $\lambda$ is the wavelength, $D$ the distance from slits to screen and $d$ the slit separation. Why does it make sense? A larger $\lambda$ means the waves must travel further to get out of step, so the fringes spread out. A larger $d$ means the two paths differ quickly as you move across the screen, so the fringes squeeze together.

The diffraction grating

A diffraction grating is a piece of glass or plastic with a very large number of equally spaced parallel slits, often hundreds per millimetre. The slit spacing $d$ is the distance between neighbouring slits, and it is found from the number of lines per unit length:

where $N$ is the number of lines per metre. For example, $250$ lines per millimetre is $2.50 \times 10^{5}$ lines per metre, so $d = 4.0 \times 10^{-6}\ \text{m}$.

Light leaving neighbouring slits at an angle $\theta$ to the straight-through direction has a path difference of $d \sin\theta$. All the slits reinforce one another when that path difference is a whole number of wavelengths.

Because $\sin\theta$ cannot be bigger than $1$, there is a limit to the number of orders: $n \le \dfrac{d}{\lambda}$. Always round down to a whole number.

White light and sharper maxima

Look at $\sin\theta = n\lambda/d$ again. For a fixed order, a longer wavelength gives a bigger angle. So when white light is used, each order is spread out into a spectrum. Violet is nearest the centre and red is furthest out. This is what you see on the back of a CD, whose closely spaced tracks act like a grating. That answers the question at the start of the page.

The central maximum ($n = 0$) is different. Every wavelength has zero path difference there, so all colours arrive in phase. The central maximum is white.

Higher-order spectra are wider, so they can overlap. Orders $n$ and $n + 1$ overlap when the red end of order $n$ is at a larger angle than the violet end of order $n + 1$.

Why a grating beats a double slit

With only two slits the bright fringes are wide and fuzzy. With a grating, light from the many slits cancels almost perfectly at every angle except those where $d\sin\theta = n\lambda$. The result is narrow, sharp and bright maxima. More slits are lit, so more energy goes into each maximum.

This also lets a grating separate two very close wavelengths, such as the two yellow lines of a sodium lamp. The angle between two wavelengths $\lambda_1$ and $\lambda_2$ in the same order is larger when $d$ is smaller, that is, when the grating has more lines per millimetre. When angles are small enough that $\sin\theta \approx \tan\theta \approx \theta$, the separation on a screen at distance $D$ is approximately $D\,\Delta\theta$. State clearly in your answer where you use this approximation.

Measuring the wavelength of light

In the student experiment you can find $\lambda$ with either set-up.

Double slit. Measure $D$ with a metre stick. Measure the distance across several fringe spacings and divide by the number of spacings to get $\Delta y$. Then use $\lambda = \Delta y\, d / D$. Measuring across, say, $6$ fringes reduces the percentage error, because the same ruler uncertainty of about $\pm 0.5\ \text{mm}$ is shared over a larger distance.

Grating. Measure $\theta$ for several orders on both sides of the centre and use $\lambda = d\sin\theta / n$. Taking both sides and averaging reduces the error from the beam not being exactly normal to the grating.

A graph is better than a single calculation. Since $\sin\theta = \dfrac{\lambda}{d}\, n$, a graph of $\sin\theta$ against $n$ is a straight line through the origin. Its gradient is $\dfrac{\lambda}{d}$, so $\lambda = \text{gradient} \times d$.

When you compare two methods, use the percentage difference $= \dfrac{|\lambda_1 - \lambda_2|}{\text{mean}} \times 100\%$ and compare it with the percentage uncertainty in your measurements. The largest source of error is usually the measurement with the smallest quantity compared with its uncertainty. For a double slit that is often the tiny slit separation $d$ or the fringe width. For a grating it is usually the angle $\theta$, read from a protractor to about $\pm 0.5^\circ$ or worse.

Summary

• Interference needs coherent sources. Bright where the path difference is $n\lambda$, dark where it is $\left(n + \tfrac{1}{2}\right)\lambda$. The centre is always bright.

• Young's slits: $\Delta y = \dfrac{\lambda D}{d}$. Measure across several fringes to reduce error.

• Grating: $d = 1/N$, and $n\lambda = d\sin\theta$. The highest order is $n \le d/\lambda$, rounded down. Maxima appear on both sides of the centre.

• With white light the centre is white and each higher order is a spectrum, violet closest to the centre. Higher orders can overlap.

• A grating gives sharper, brighter maxima than a double slit, and more lines per millimetre separate close wavelengths better.

• A graph of $\sin\theta$ against $n$ has gradient $\lambda/d$.

Back to the start: the tracks on a CD act as a grating, so each wavelength of white light leaves at its own angle and you see a spectrum.