Interference and Diffraction — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Double-slit interference, single-slit diffraction, diffraction gratings, path difference conditions for constructive/destructive interference, fringe spacing calculations, order determination, and wavelength measurement techniques.

You should already know: Wave superposition principle, wave properties of light, and how refractive index changes the wavelength of light in a medium.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Interference and Diffraction?

Interference is the effect that occurs when two or more coherent light waves (waves with a constant phase difference) superpose, producing a resultant wave with an amplitude that depends on the phase difference of the incoming waves. Diffraction is the bending of light waves as they pass through a narrow aperture or around an obstacle, an effect that only exists because light acts as a wave. Diffraction inherently produces interference patterns, as different parts of the diffracted wave superpose on one another.

Per the AP Physics 2 Course and Exam Description (CED), this topic makes up approximately 10-15% of the total exam score, and it appears in both multiple choice (MCQ) and free response (FRQ) sections, often combined with other optics or wave topics. Standard notation used throughout this chapter: $d$ = separation between adjacent slits, $a$ = width of a single slit, $\lambda$ = wavelength of light, $\theta$ = angle from the central maximum, $L$ = distance from slits/grating to the screen, $m$ = order number of a fringe or maximum, $\Delta x$ = fringe spacing, $N$ = number of lines per unit length for diffraction gratings. Interference and diffraction are the foundational experimental evidence for the wave nature of light, correcting the purely particle model of geometric optics.

2. Double-Slit Interference

Young’s double-slit experiment was the first definitive proof that light acts as a wave. In this setup, coherent monochromatic light passes through two narrow slits separated by distance $d$, and the light from both slits projects onto a screen a distance $L$ away from the slits. For any point on the screen, the path difference between the light coming from the two slits is $\Delta r=d\sin \theta$, where $\theta$ is the angle between the central axis and the line from the slits to the point on the screen.

Constructive interference (bright fringe) occurs when the path difference is an integer multiple of the wavelength: $$d\sin \theta =m\lambda (m=0,\pm 1,\pm 2,...)$$ Destructive interference (dark fringe) occurs when the path difference is a half-integer multiple of the wavelength: $$d\sin \theta =\left(m+\frac{1}{2}\right)\lambda (m=0,\pm 1,\pm 2,...)$$ For small angles ($\theta <10^{\circ }$), $\sin \theta \approx \tan \theta =\frac{y}{L}$, where $y$ is the vertical distance from the central maximum to the fringe. This simplifies to the fringe spacing formula (distance between adjacent bright fringes): $$\Delta x=\frac{\lambda L}{d}$$

Worked Example

A student performs a double-slit experiment with 500 nm green light, a slit separation of 0.25 mm, and a screen placed 2.0 m from the slits. Calculate the spacing between adjacent bright fringes, and the total distance from the 0th order central maximum to the 3rd order bright fringe.

1. Convert all units to meters to ensure consistency: $\lambda =500\times 10^{-9}\text{m}$, $d=0.25\times 10^{-3}\text{m}$, $L=2.0\text{m}$.
2. Apply the small-angle fringe spacing formula: $\Delta x=\frac{\lambda L}{d}$.
3. Calculate: $\Delta x=\frac{(500\times 10^{-9})(2.0)}{0.25\times 10^{-3}}=4.0\times 10^{-3}\text{m}=4.0\text{mm}$.
4. The distance from the 0th to 3rd order bright fringe is $3\Delta x=3(4.0\text{mm})=12\text{mm}=0.012\text{m}$.

Exam tip: Always convert all length units to the same base unit (usually meters) before plugging values into formulas; unit mismatches are the most common avoidable mistake on this topic.

3. Single-Slit Diffraction

When light passes through a single narrow slit of width $a$, each point within the slit acts as a secondary wave source, and these secondary waves interfere with one another to produce a diffraction pattern. The pattern features a wide, very bright central maximum, followed by much dimmer, narrower maxima on either side, with dark minima between each maximum. Unlike double-slit interference, single-slit maxima are not equally spaced.

The condition for dark minima (points of zero intensity) in single-slit diffraction is: $$a\sin \theta =m\lambda (m=\pm 1,\pm 2,...)$$ Note that $m=0$ is not a minimum, as all waves interfere constructively at the central maximum. For small angles, the distance from the central maximum to the first minimum on either side is $y=\frac{\lambda L}{a}$. The full width of the central maximum (distance between the two first minima on opposite sides of center) is therefore: $$W=\frac{2\lambda L}{a}$$ A key trend: decreasing the slit width $a$ increases the width of the central maximum, meaning more diffraction occurs for smaller apertures.

Worked Example

A 633 nm red laser passes through a single slit of width 0.10 mm, producing a diffraction pattern on a screen 1.5 m away. What is the full width of the central bright maximum?

1. Convert all units to meters: $a=0.10\times 10^{-3}\text{m}$, $\lambda =633\times 10^{-9}\text{m}$, $L=1.5\text{m}$.
2. Use the full central maximum width formula for small angles: $W=\frac{2\lambda L}{a}$.
3. Plug in values: $W=\frac{2(633\times 10^{-9})(1.5)}{0.10\times 10^{-3}}=1.9\times 10^{-2}\text{m}=1.9\text{cm}$.
4. Verify the trend: A 0.10 mm slit is narrow, so a 1.9 cm central width matches the expected behavior of diffraction through a small aperture.

Exam tip: Never mix up $a$ (single slit width) and $d$ (slit separation for double-slit/grating); swapping these gives an answer off by multiple orders of magnitude that will be immediately marked wrong.

4. Diffraction Gratings

A diffraction grating is an optical device with hundreds or thousands of equally spaced parallel slits etched into a glass or metal surface. It produces much sharper, brighter, and better-separated principal maxima than a double-slit, making it ideal for measuring the wavelength of unknown light or splitting white light into its component wavelengths to produce a spectrum.

For a grating with $N$ lines per unit length, the separation between adjacent slits is $d=\frac{1}{N}$. The condition for principal (bright) maxima is identical to the double-slit constructive interference condition: $$d\sin \theta =m\lambda (m=0,\pm 1,\pm 2,...)$$ Since $\sin \theta$ can never be greater than 1, the maximum possible order $m$ is the largest integer less than $\frac{d}{\lambda }$. Any order with $m>\frac{d}{\lambda }$ does not exist, as it would require a diffraction angle larger than 90 degrees.

Worked Example

A diffraction grating has 300 lines per mm. It is illuminated with white light, which has a wavelength range of 400 nm (violet) to 700 nm (red). What is the highest order complete full spectrum produced by this grating?

1. Calculate $d$ in meters: 300 lines per mm = 300,000 lines per meter, so $d=\frac{1}{300000}\approx 3.33\times 10^{-6}\text{m}$.
2. The longest wavelength (red) diffracts the most, so if red for order $m$ is still visible (i.e., $\theta <90^{\circ }$), all shorter wavelengths in the spectrum will also be visible for that order.
3. Calculate the maximum possible $m$ for red light: $m<\frac{d}{{\lambda }_{\text{red}}}=\frac{3.33\times 10^{-6}}{700\times 10^{-9}}\approx 4.76$.
4. Since $m$ must be an integer, the highest order complete spectrum is $m=4$. (For $m=5$, $\sin \theta =5(700\times 10^{-9})/(3.33\times 10^{-6})\approx 1.05>1$, which is impossible.)

Exam tip: Always check if your calculated $\sin \theta$ is greater than 1; if it is, that order does not exist, and you need to pick the next smaller integer order.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Swapping $d$ (slit separation) and $a$ (slit width) when solving problems, calculating $\Delta x=aL/\lambda$ instead of $\lambda L/d$. Why: The symbols are both lowercase letters, and students confuse what each variable measures. Correct move: Label every variable on your diagram as you read the problem: write $d$ between two slits, write $a$ across a single slit width before starting any calculations.
• Wrong move: Forgetting that the full width of the central maximum in single-slit diffraction is $2\lambda L/a$, using $\lambda L/a$ instead. Why: Students memorize that the first minimum is at $y=\lambda L/a$ from center, so they stop there instead of doubling for full width from $-y$ to $+y$. Correct move: Always confirm: "full width of central maximum" means distance between the two first minima on either side, so multiply the distance from center to one minimum by two.
• Wrong move: Using the small angle approximation $\Delta x=\lambda L/d$ for large angles (e.g., diffraction grating problems where $\theta >10^{\circ }$). Why: Students get used to small angles in basic double-slit experiments and automatically use the approximation regardless of angle size. Correct move: Only use the small angle fringe spacing formula when $\theta <10^{\circ }$; for larger angles, use the full $d\sin \theta =m\lambda$ relation and calculate position from $\tan \theta$.
• Wrong move: Calculating $d$ for a diffraction grating as $N$ (lines per meter) instead of $1/N$. Why: Problems usually give lines per mm or lines per cm, so students just plug the given number in without inverting or converting units. Correct move: If you have $N$ lines per unit length, $d=1/N$; always convert $N$ to lines per meter first to get $d$ in meters.
• Wrong move: Mixing up constructive and destructive interference conditions, claiming constructive interference occurs at half-integer path differences. Why: Students memorize the conditions backwards or confuse them with thin film interference rules. Correct move: Start every interference problem by writing: Constructive = integer multiple of $\lambda$, destructive = half-integer multiple of $\lambda$ at the top of your work.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

A double-slit experiment is set up in air with 700 nm red light, producing a fringe spacing of 5.0 mm. If the entire apparatus is fully immersed in pure water (refractive index $n=1.33$), what is the new fringe spacing? A) 6.65 mm B) 3.76 mm C) 5.0 mm D) 2.83 mm

Worked Solution: The wavelength of light in a medium with refractive index $n$ is ${\lambda }_n={\lambda }_0/n$, where ${\lambda }_0$ is the wavelength in vacuum/air. Fringe spacing is proportional to wavelength: $\Delta x={\lambda }_{n}L/d$. The new fringe spacing is $\Delta x_{\text{new}}=(5.0\text{mm})/1.33\approx 3.76\text{mm}$. This eliminates options A, C, D. The correct answer is B.

Question 2 (Free Response)

A student uses a diffraction grating to measure the wavelength of an unknown laser. The grating has 500 lines per mm, and the two first-order bright spots are 1.28 m apart on a screen 2.00 m from the grating. (a) Calculate the wavelength of the laser. (b) The student claims a second-order maximum for this laser exists. Justify whether the student is correct. (c) If the laser is replaced with one of shorter wavelength, what happens to the distance between the first-order maxima? Explain your reasoning.

Worked Solution: (a) The distance from the central maximum to one first-order spot is half the total distance between the two spots: $y=1.28/2=0.64\text{m}$. Grating slit separation: $d=1/(500\times 1000)=2.00\times 10^{-6}\text{m}$. $\tan \theta =y/L=0.64/2.00=0.32$, so $\sin \theta \approx 0.304$. Rearrange $d\sin \theta =m\lambda$ for $m=1$: $\lambda =(2.00\times 10^{-6})(0.304)=608\times 10^{-9}\text{m}=608\text{nm}$. (b) Check the value of $\sin \theta$ for $m=2$: $\sin \theta =m\lambda /d=2(608\times 10^{-9})/(2.00\times 10^{-6})=0.608$. Since $0.608<1$, the angle $\theta$ is valid, so the student is correct: the second-order maximum exists. (c) The distance between the first-order maxima will decrease. From $d\sin \theta =m\lambda$, a shorter wavelength $\lambda$ gives a smaller $\sin \theta$, a smaller diffraction angle $\theta$, and a smaller distance from the central maximum to each first-order spot, so the total distance between the two spots decreases.

Question 3 (Application / Real-World Style)

Astronomers use a diffraction grating with 600 lines per mm to analyze light from a distant star. A first-order spectral line from hydrogen is measured to have a diffraction angle of 14.5° from the central maximum. Calculate the wavelength of this line, and confirm if it falls in the visible spectrum (400 nm to 700 nm).

Worked Solution: First, calculate the grating slit separation: $d=1/(600\times 1000)\approx 1.667\times 10^{-6}\text{m}$. For first order, $m=1$, so rearrange $d\sin \theta =m\lambda$ to get $\lambda =d\sin \theta$. $\sin (14.5^{\circ })\approx 0.250$, so $\lambda =(1.667\times 10^{-6})(0.250)\approx 417\times 10^{-9}\text{m}=417\text{nm}$. This wavelength falls between 400 nm and 700 nm, so it is a visible violet spectral line from hydrogen in the star's light.

7. Quick Reference Cheatsheet

8. What's Next

After mastering interference and diffraction, the next key topics in Unit 6 Geometric and Physical Optics are thin film interference and optical resolution, both of which rely directly on the path difference and interference rules you learned here. Without understanding this chapter, you cannot correctly calculate the conditions for constructive/destructive interference in thin films, or apply the Rayleigh criterion for resolving power of optical instruments. This topic also underpins the wave nature of matter, a core concept in modern physics, where electrons and other subatomic particles produce interference patterns identical to light. Interference and diffraction also enable many modern technologies from CD/DVD reading to astronomical spectroscopy.

• Thin Film Interference
• Optical Resolution and Rayleigh Criterion
• Image Formation by Thin Lenses
• Wave-Particle Duality