The double slit with diffraction and the grating

Fringes inside a lamp of light

In the last course we treated each slit of a double slit as a point. A real slit has a width $a$, and it spreads the light by diffraction. The fringes are therefore not equally bright. The pattern on the screen is the fine fringes of the two slits, shaped by the single-slit pattern of one slit:

with $\alpha = \pi a\sin\theta/\lambda$. The single-slit pattern is a wide envelope.

The central maximum of the envelope ends where $a\sin\theta = \lambda$. A bright fringe of order $m$ is at $d\sin\theta = m\lambda$. It lies inside the central maximum if $m < d/a$.

• If $d/a$ is not a whole number, the number of bright fringes inside is $2\lceil d/a\rceil - 1$ (the $\lceil\ \rceil$ means "round up").
• If $d/a$ is a whole number, the order $m = d/a$ falls exactly on the first minimum of the envelope. It is a missing order, and the number of fringes inside is $2(d/a) - 1$.
• In general, an order is missing when $d\sin\theta = m\lambda$ and $a\sin\theta = n\lambda$ are true at the same angle, which gives $m = n\,d/a$.

A worked example. A double slit has slits of width $0.080$ mm, a distance $0.400$ mm apart. How many bright fringes fall inside the central maximum, and which orders are missing?

1. The ratio. $\dfrac{d}{a} = \dfrac{0.400\ \text{mm}}{0.080\ \text{mm}} = 5$.
2. The orders inside. $m < 5$: the orders $0, \pm1, \pm2, \pm3, \pm4$.
3. The number. $2 \times 5 - 1 = 9$ bright fringes.
4. The missing orders. $m = n\,d/a = 5n$: the orders $\pm5$, $\pm10$ and so on.
5. Check. The central maximum is $\dfrac{2\lambda D}{a}$ wide and a fringe is $\dfrac{\lambda D}{d}$ wide, so it holds $\dfrac{2d}{a} = 10$ fringe widths. This agrees with $9$ bright fringes plus the two missing ones at the edges.

The diffraction grating

A diffraction grating is a plate with a very large number of narrow, equally spaced slits, often $100$ to $1000$ in each millimetre. The distance between neighbouring slits is $d$. If the plate has $N_1$ lines in each millimetre, then $d = \dfrac{1\ \text{mm}}{N_1}$.

Light falls on the grating straight on. The waves from neighbouring slits add up in step in the direction where the path difference is a whole number of wavelengths:

Each such direction is a principal maximum of order $m$. Since $\sin\theta$ cannot be more than $1$, the highest order that can be seen has $m < d/\lambda$.

Why is a grating better than a double slit? With only two slits the bright fringes are broad. With $N$ slits the principal maxima stay at the same angles but become narrow and bright. Between two principal maxima there are $N - 2$ very weak maxima. A bright, thin line can be measured much more exactly, so the wavelength of a spectral line is found accurately.

Raise the number of lines in each millimetre and see what happens to the orders. They move farther from the straight-on direction, and fewer of them fit into the half circle. Why? $d$ gets smaller, so $\sin\theta = m\lambda/d$ gets bigger.

A worked example. A grating has $5000$ lines in each centimetre. Light of wavelength $600$ nm falls on it. Find the distance between the lines, the angles of the first and second orders, and the highest order.

1. The distance. $d = \dfrac{1\ \text{cm}}{5000} = 2.00\ \mu\text{m}$.
2. The first order. $\sin\theta_1 = \dfrac{\lambda}{d} = \dfrac{600\ \text{nm}}{2000\ \text{nm}} = 0.30$, so $\theta_1 = 17.5^\circ$.
3. The second order. $\sin\theta_2 = 0.60$, so $\theta_2 = 36.9^\circ$.
4. The highest order. $m < \dfrac{d}{\lambda} = \dfrac{2000}{600} = 3.33$, so $m_{\max} = 3$, at $\theta = 64.2^\circ$.
5. Check. The next order, $m = 4$, would need $\sin\theta = 1.20$, which is more than $1$. So it does not exist.

White light on a grating

For white light, each colour has its own angle in each order. The zero order ($m = 0$, straight on) is white, since all the colours have $\theta = 0$. In every other order there is a spectrum, with violet nearest to the centre and red farthest. Red is bent more than violet, as $\sin\theta = m\lambda/d$ grows with $\lambda$. This is the opposite of a prism, where violet is bent more.

The spectra of different orders can overlap. The red end of order $m$ overlaps the violet end of order $m + 1$ when $m \times 700\ \text{nm} > (m + 1) \times 400\ \text{nm}$. This is true when $m > \tfrac43$. So the first and second orders never overlap, but the second and third always do.

Practise with new numbers

Try it in Python

The program counts the bright fringes inside the central diffraction maximum of a double slit, and finds the missing orders.

The second program finds where the spectra of white light fall for a grating of $500$ lines in each millimetre.

Common mistakes

• Counting a missing order as a fringe. When $d/a$ is a whole number, the order $m = d/a$ is missing. It falls on a dark place of the envelope.
• Taking $d$ as the number of lines. $d$ is the distance between two lines, so $d = 1/N_1$.
• Using an order that is too high. If $m\lambda/d$ is more than $1$, there is no such order.
• Thinking that violet is bent more in a grating. In a grating, red is bent more than violet.

Remember these

• A real double slit: fine fringes inside the single-slit envelope. Fringes inside the central maximum: $m < d/a$. A missing order has $m = n\,d/a$.
• A grating: $d\sin\theta = m\lambda$, with $d = 1/N_1$. The highest order has $m < d/\lambda$.
• More slits make the principal maxima narrower and brighter.
• In white light, each order beyond zero is a spectrum with violet inside and red outside.

Check yourself

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

1. A double slit has $d = 4a$. Which order is missing, and how many bright fringes lie in the central maximum?

2. A grating has $5000$ lines in each centimetre. What is the distance between the lines?

3. Why does the zero-order image of a grating look white when white light is used?

4. A grating gives a first-order maximum at $20^\circ$. At what $\sin\theta$ is the second order?

References

• OpenStax, University Physics Volume 3, Chapter 4 (Diffraction): double-slit diffraction and diffraction gratings.
• NCERT Physics Class 12, Chapter 10 (Wave Optics): diffraction; gratings are treated in A-Level and IB Physics.