Unit 14 · Topic 14.8

14.8 Double-Slit Interference and Diffraction Gratings

Light through two narrow slits makes a row of evenly spaced bright fringes, located where the path length difference from the two slits is a whole number of wavelengths: $d\sin\theta = m\lambda$ . Young's double-slit experiment was key evidence that light is a wave. A diffraction grating has many evenly spaced slits, which make sharp, bright maxima and spread white light into rainbows.

Key terms

double-slit interference
slit separation
bright fringe
order number
diffraction grating
Young's experiment

Young's double-slit experiment

In the early 1800s, Thomas Young sent light through two narrow, closely spaced slits and saw a pattern of bright and dark bands on a screen. Particles going through two slits would just make two bright stripes. Bands of light and dark mean waves from the two slits were interfering, so the experiment was strong evidence that light is a wave.

The two slits act as two sources of waves in step with each other, because both are lit by the same beam.

Where the bright fringes are

From each slit to a point on the screen, the light travels a slightly different distance. For slits a distance d apart (center to center), the path length difference at angle θ is $\Delta D = d\sin\theta$ .

If the difference is a whole number of wavelengths, crests arrive together and the fringe is bright: $d\sin\theta = m\lambda$ , with m = 0, 1, 2, …. The number m is the order. The central fringe, m = 0, sits straight ahead, where both paths are equal.

If the difference is an odd number of half wavelengths (½λ, 1½λ, 2½λ, …), a crest meets a trough and the fringe is dark.

For small angles, the sheet gives $d\left(\frac{y_{\max}}{L}\right) \approx m\lambda$ , where $y_{\max}$ is the distance from the center of the pattern to the mth bright fringe. The fringes are evenly spaced, with spacing $\Delta y = \frac{\lambda L}{d}$ .

Interference inside a diffraction envelope

Each slit has some width, so each one also makes its own single-slit diffraction pattern (14.7). The real double-slit pattern is the evenly spaced interference fringes, with their brightness set by the single-slit envelope: bright fringes in the middle, fading outward, and missing where the envelope has a dark band.

If you only consider interference, you get uniformly bright, evenly spaced maxima. Adding diffraction explains why the outer fringes are dimmer.

Diffraction gratings

A diffraction grating has hundreds or thousands of evenly spaced slits per millimeter. It uses the same condition, $d\sin\theta = m\lambda$ , where d is the spacing between neighboring slits. A grating with N lines per meter has $d = \frac{1}{N}$ .

With many slits, the light from all of them lines up only very close to the angles where $d\sin\theta = m\lambda$ , so the maxima are much sharper and brighter than with two slits. That makes gratings good for measuring wavelengths precisely.

Grating angles are usually large, so use $d\sin\theta = m\lambda$ directly. Don't use the small-angle version.

Because sin θ can't exceed 1, the highest order you can see is the largest whole number m that is less than $\frac{d}{\lambda}$ .

With white light, the central maximum (m = 0) is white, because every color has its m = 0 maximum straight ahead. Each higher order spreads into a rainbow, with violet closest to the center and red farthest out, since red has the longest wavelength.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Fringe spacing
Light with λ = 550 nm passes through two slits 0.25 mm apart onto a screen 1.2 m away. How far apart are neighboring bright fringes?
Show the solutionHide the solution
1. Step 1: Neighboring bright fringes differ by one order, so $\Delta y = \dfrac{\lambda L}{d} = \dfrac{(550 \times 10^{-9})(1.2)}{0.25 \times 10^{-3}} = 2.64 \times 10^{-3}$ m.
2. Step 2: That's about 2.6 mm between fringes.
Answer: About 2.6 mm
Example 2Calculator allowed
Measuring a wavelength
In a double-slit experiment with d = 0.20 mm and L = 1.50 m, the 4th-order bright fringe is 1.80 cm from the center of the pattern. Find the wavelength.
Show the solutionHide the solution
1. Step 1: Rearrange $d\left(\dfrac{y_{\max}}{L}\right) = m\lambda$ : $\lambda = \dfrac{d\,y_{\max}}{mL} = \dfrac{(0.20 \times 10^{-3})(0.0180)}{(4)(1.50)}$ .
2. Step 2: $\lambda = 6.0 \times 10^{-7}$ m = 600 nm, which is orange light.
3. Step 3: Measuring out to the 4th fringe and dividing by 4 reduces the error compared with measuring a single fringe spacing.
Answer: λ = 600 nm
Example 3Calculator allowed
Grating with large angles
A grating has 600 lines per millimeter. Find the angle of the first-order maximum for red light (650 nm) and blue light (450 nm), and the highest order of red light you can see.
Show the solutionHide the solution
1. Step 1: Slit spacing: $d = \dfrac{1\text{ mm}}{600} = 1.67 \times 10^{-6}$ m.
2. Step 2: Red, m = 1: $\sin\theta = \dfrac{\lambda}{d} = \dfrac{650 \times 10^{-9}}{1.67 \times 10^{-6}} = 0.390$ , so θ = 23.0°.
3. Step 3: Blue, m = 1: $\sin\theta = \dfrac{450 \times 10^{-9}}{1.67 \times 10^{-6}} = 0.270$ , so θ = 15.7°. Red is farther out, as expected.
4. Step 4: Highest red order: $\dfrac{d}{\lambda} = \dfrac{1.67 \times 10^{-6}}{650 \times 10^{-9}} = 2.56$ , so m = 2 is the highest (m = 3 would need sin θ = 1.17, which is impossible).
5. Step 5: Trap: at these angles you must use sin θ directly. The small-angle form would be noticeably off for the 2nd order, at 51°.
Answer: Red 23.0°, blue 15.7°; highest red order is m = 2.

Common mistakes

Using the small-angle approximation for a grating. Grating angles are often large; use $d\sin\theta = m\lambda$ .
Mixing up the double-slit and single-slit equations. For two slits, $d\sin\theta = m\lambda$ gives bright fringes; for one slit, $a\sin\theta = m\lambda$ gives dark bands.
Using lines per millimeter as d. Take the reciprocal and convert to meters.
Putting violet farthest from the center in a grating spectrum. Red, with the longest wavelength, is farthest out.

On the exam

Expect questions that ask how the fringe spacing changes with d, L or λ, and why. Use $\Delta y = \frac{\lambda L}{d}$ to justify.
You may be asked what the double-slit pattern shows about light. The answer is wave behavior, because interference needs waves. That sets up the particle evidence in Unit 15.

Connected topics

Videos

AP Physics 2: Double Slit Interference (Unit 14) - Interference Patterns Made Easy
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Check yourself

5 questions on 14.8 Double-Slit Interference and Diffraction Gratings. Pick an answer to see if you got it, and why.

Question 1 of 5Calculator allowed

Light with a wavelength of 600 nm passes through two slits 0.25 mm apart. What is the spacing between neighboring bright fringes on a screen 1.5 m away?

Question 2 of 5Calculator allowed

In a double-slit experiment, the distance between the slits is doubled while the wavelength and screen distance stay the same. What happens to the bright fringes on the screen?

Light with a wavelength of 500 nm shines straight onto a diffraction grating that has 600 lines per millimeter.

Described experiment

Question 3 of 5Calculator allowed

At what angle from the central maximum does the first-order bright maximum appear?

Question 4 of 5Calculator allowed

What is the highest order of bright maximum that can appear for this light and grating?

Question 5 of 5Calculator allowed

White light is sent through the same grating. Which describes the pattern?

0 of 5 answered