Superposition and Interference

Superposition and Interference

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Board Coverage AQA Paper 2 | Edexcel CP2 | OCR (A) Paper 2 | CIE P2

Wave Interference

Explore the simulation above to develop intuition for this topic.

1. The Principle of Superposition

Definition. The principle of superposition states that when two or more waves overlap at a point, the resultant displacement is the algebraic sum of the individual displacements of each wave at that point.

Principle of Superposition. When two or more waves overlap, the resultant displacement at any point is the algebraic sum of the individual displacements:

$y_{\mathrm{total}} = y_1 + y_2 + y_3 + \cdots$

This principle is valid for linear waves (small amplitudes). It is a direct consequence of the linearity of the wave equation.

Constructive and Destructive Interference

Definition. Constructive interference occurs when two waves meet in phase (phase difference $= 0, 2\pi, 4\pi, \ldots$), producing a resultant amplitude greater than either individual amplitude.

Definition. Destructive interference occurs when two waves meet in antiphase (phase difference $= \pi, 3\pi, 5\pi, \ldots$), producing a resultant amplitude less than either individual amplitude, or zero if the waves have equal amplitude.

Consider two coherent waves of the same amplitude $A$ arriving at a point with phase difference $\Delta\phi$:

$y_1 = A\sin(\omega t), \qquad y_2 = A\sin(\omega t + \Delta\phi)$

The resultant is:

$y_{\mathrm{total}} = A\sin(\omega t) + A\sin(\omega t + \Delta\phi)$

Using the trigonometric identity $\sin\alpha + \sin\beta = 2\sin\frac◆LB◆\alpha+\beta◆RB◆◆LB◆2◆RB◆\cos\frac◆LB◆\alpha-\beta◆RB◆◆LB◆2◆RB◆$:

$y_{\mathrm{total}} = 2A\cos\left(\frac◆LB◆\Delta\phi◆RB◆◆LB◆2◆RB◆\right)\sin\left(\omega t + \frac◆LB◆\Delta\phi◆RB◆◆LB◆2◆RB◆\right)$

The resultant amplitude is:

$A_{\mathrm{resultant}} = 2A\left|\cos\frac◆LB◆\Delta\phi◆RB◆◆LB◆2◆RB◆\right|$

• Constructive interference: $\Delta\phi = 0, 2\pi, 4\pi, \ldots \implies A_{\mathrm{resultant}} = 2A$
• Destructive interference: $\Delta\phi = \pi, 3\pi, 5\pi, \ldots \implies A_{\mathrm{resultant}} = 0$

2. Path Difference and Phase Difference

Definition. Path difference is the difference in distance travelled by two waves from their respective sources to a given point: $\Delta x = |x_1 - x_2|$.

Definition. Phase difference is the difference in phase between two waves at a given point, measured in radians. It is related to path difference by $\Delta\phi = (2\pi/\lambda)\Delta x$.

For two waves of wavelength $\lambda$ travelling to a point via paths of lengths $x_1$ and $x_2$:

Path difference: $\Delta x = |x_1 - x_2|$

Phase difference:

$\boxed{\Delta\phi = \frac◆LB◆2\pi◆RB◆◆LB◆\lambda◆RB◆\Delta x}$

Derivation. One extra wavelength $\lambda$ corresponds to a full cycle, i.e., a phase difference of $2\pi$. By proportionality:

$\frac◆LB◆\Delta\phi◆RB◆◆LB◆2\pi◆RB◆ = \frac◆LB◆\Delta x◆RB◆◆LB◆\lambda◆RB◆ \implies \Delta\phi = \frac◆LB◆2\pi◆RB◆◆LB◆\lambda◆RB◆\Delta x$

$\square$

Conditions for interference:

Condition	Path Difference	Phase Difference
Constructive (maximum)	$n\lambda$	$2n\pi$
Destructive (minimum)	$(n + \frac{1}{2})\lambda$	$(2n+1)\pi$

where $n = 0, 1, 2, \ldots$

Definition. Coherence is the property of two or more waves having a constant phase relationship over time, which is necessary to produce a stable interference pattern. Coherent waves must have the same frequency and a constant phase difference.

Coherence. For a stable interference pattern, the two waves must be coherent — they must have a constant phase relationship. This requires:

• Same frequency (and hence wavelength)
• Constant phase difference

warning

warning not coherent because the phase difference fluctuates randomly. Interference requires coherent sources, typically produced by splitting a single wave.

3. Young's Double Slit Experiment

Derivation of the Double-Slit Interference Condition

1. Two coherent slits $S_1$ and $S_2$ are separated by distance $s$.
2. Light of wavelength $\lambda$ illuminates both slits.
3. A point $P$ on a distant screen ($D \gg s$) is at angle $\theta$ from the central axis.
4. The path difference from the two slits to $P$ is $\Delta x = s\sin\theta$.
5. For constructive interference, the path difference must equal a whole number of wavelengths:

$\boxed{s\sin\theta = n\lambda, \qquad n = 0, 1, 2, \ldots}$

$\square$

Derivation of the Fringe Spacing Formula

Two narrow slits $S_1$ and $S_2$, separated by distance $s$, are illuminated by coherent light of wavelength $\lambda$. A screen is placed at distance $D \gg s$ from the slits.

Consider a point $P$ on the screen at angle $\theta$ from the central axis. The path difference from the two slits is:

$\Delta x = s\sin\theta$

For constructive interference (bright fringe): $s\sin\theta = n\lambda$.

For small angles ($\sin\theta \approx \tan\theta \approx \theta$):

$\tan\theta = \frac{w}{D}$

where $w$ is the distance from the central maximum to the $n$-th bright fringe. Therefore:

$s \cdot \frac{w}{D} = n\lambda$

$\boxed{w = \frac◆LB◆n\lambda D◆RB◆◆LB◆s◆RB◆}$

The fringe spacing (distance between adjacent bright fringes) is:

$\boxed{\Delta w = \frac◆LB◆\lambda D◆RB◆◆LB◆s◆RB◆}$

Intuition. Larger wavelength means wider fringes. Greater slit separation means narrower fringes. Greater screen distance means wider fringes. The pattern scales linearly with all three quantities.

Single Slit Diffraction Envelope

Each slit has width $a$ and produces a diffraction pattern. The double-slit fringes are modulated by a single-slit diffraction envelope. If $a$ is too small, the fringes are very broad and hard to resolve. If $a$ is too large, the diffraction envelope is narrow and few fringes are visible.

The first minimum of the single-slit pattern occurs at $\sin\theta = \lambda/a$.

tip

tip (single slit before the double slit), narrow slits, monochromatic light, and the small-angle approximation. Quote the fringe spacing formula and explain each variable.

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Example: Young's Double Slit

Light of wavelength 590 nm passes through double slits separated by 0.50 mm onto a screen 1.5 m away. Calculate the fringe spacing.

Answer. $\Delta w = \frac◆LB◆\lambda D◆RB◆◆LB◆s◆RB◆ = \frac◆LB◆590 \times 10^{-9} \times 1.5◆RB◆◆LB◆0.50 \times 10^{-3}◆RB◆ = \frac◆LB◆8.85 \times 10^{-7}◆RB◆◆LB◆5.0 \times 10^{-4}◆RB◆ = 1.77 \times 10^{-3}$ m $= 1.77$ mm.

4. Diffraction Gratings

A diffraction grating has $N$ slits per unit length, with slit spacing $d = 1/N$.

Derivation of the Grating Equation

Consider $N$ equally spaced slits. For light emerging at angle $\theta$, the path difference between adjacent slits is $d\sin\theta$. For constructive interference from all $N$ slits simultaneously:

$d\sin\theta = n\lambda, \qquad n = 0, \pm 1, \pm 2, \ldots$

$\boxed{d\sin\theta = n\lambda}$

where $n$ is the order of the maximum.

Derivation. For two adjacent slits, constructive interference requires $d\sin\theta = n\lambda$. Since all slits are equally spaced, if adjacent slits constructively interfere, then all pairs do. The condition is the same as for two slits, but with $N$ slits the maxima are much sharper (because destructive interference from non-adjacent pairs suppresses the background).

Maximum number of orders. Since $|\sin\theta| \leq 1$:

$|n| \leq \frac◆LB◆d◆RB◆◆LB◆\lambda◆RB◆$

The highest order visible is $n_{\max} = \lfloor d/\lambda \rfloor$ (the greatest integer less than or equal to $d/\lambda$).

Grating vs Double Slit

Property	Double Slit	Diffraction Grating
Number of sources	2	Hundreds to thousands
Maxima sharpness	Broad	Very sharp
Maxima brightness	Low	High
Missing orders	None (ideally)	Possible (if slit width effects considered)

tip

Exam Technique If asked to find the number of visible orders: calculate $d/\lambda$, take the integer part, then note that orders $+n$ and $-n$ are both visible (plus the zeroth order). Total visible maxima $= 2n_{\max} + 1$.

5. Stationary Waves

Definition. A stationary (standing) wave is a wave formed by the superposition of two progressive waves of the same frequency and amplitude travelling in opposite directions, characterised by fixed positions of maximum amplitude (antinodes) and zero amplitude (nodes) with no net energy transfer.

Derivation from Superposition of Two Progressive Waves

Consider two identical waves travelling in opposite directions:

$y_1 = A\sin(kx - \omega t) \quad \mathrm{(travelling in } +x\mathrm{)}$ $y_2 = A\sin(kx + \omega t) \quad \mathrm{(travelling in } -x\mathrm{)}$

By superposition:

$y = y_1 + y_2 = A[\sin(kx - \omega t) + \sin(kx + \omega t)]$

Using $\sin\alpha + \sin\beta = 2\sin\frac◆LB◆\alpha+\beta◆RB◆◆LB◆2◆RB◆\cos\frac◆LB◆\alpha-\beta◆RB◆◆LB◆2◆RB◆$:

$y = 2A\sin(kx)\cos(\omega t)$

$\boxed{y = 2A\sin(kx)\cos(\omega t)}$

This is the equation of a stationary wave. Key features:

• The amplitude varies with position: $A(x) = 2A|\sin(kx)|$
• All points oscillate at the same frequency $\omega$
• The spatial and temporal parts are separated (hence "stationary")

Nodes and Antinodes

Definition. A node is a point on a stationary wave where the amplitude is always zero, occurring at positions where the two constituent waves always cancel.

Definition. An antinode is a point on a stationary wave where the amplitude is maximum, occurring midway between adjacent nodes.

Nodes are points of zero amplitude at all times. From $A(x) = 0$:

$\sin(kx) = 0 \implies kx = n\pi \implies x = \frac◆LB◆n\lambda◆RB◆◆LB◆2◆RB◆, \quad n = 0, 1, 2, \ldots$

Antinodes are points of maximum amplitude. From $|\sin(kx)| = 1$:

$kx = (n + \tfrac{1}{2})\pi \implies x = \frac◆LB◆(2n+1)\lambda◆RB◆◆LB◆4◆RB◆, \quad n = 0, 1, 2, \ldots$

The distance between adjacent nodes is $\lambda/2$. The distance between a node and the nearest antinode is $\lambda/4$.

Intuition. In a stationary wave, energy is trapped between nodes — it oscillates between kinetic and potential forms but does not propagate. This is fundamentally different from a progressive wave, where energy flows continuously.

Harmonics on a String Fixed at Both Ends

Definition. The fundamental frequency $f_1$ is the lowest frequency at which a stationary wave can form on a system, corresponding to the simplest mode of vibration.

For a string of length $L$ fixed at both ends, nodes must exist at $x = 0$ and $x = L$.

• Fundamental mode (1st harmonic): one antinode in the middle. $L = \lambda_1/2$, so $\lambda_1 = 2L$ and $f_1 = v/(2L)$.
• 2nd harmonic: two antinodes. $L = \lambda_2$, so $\lambda_2 = L$ and $f_2 = v/L = 2f_1$.
• $n$-th harmonic: $L = n\lambda_n/2$, so $\lambda_n = 2L/n$ and $f_n = nv/(2L) = nf_1$.

Derivation of Standing Wave Frequencies on a String

1. A string of length $L$ is fixed at both ends, so nodes exist at $x = 0$ and $x = L$.
2. The standing wave condition requires $L = n\lambda_n/2$ for integer $n$.
3. Therefore $\lambda_n = 2L/n$.
4. Using the wave equation $v = f\lambda$: $f_n = v/\lambda_n = nv/(2L)$.

$\boxed{f_n = \frac{nv}{2L}, \quad n = 1, 2, 3, \ldots}$

$\square$

$\boxed{f_n = \frac{nv}{2L}, \quad n = 1, 2, 3, \ldots}$

Harmonics in a Pipe Closed at One End

A closed end is a displacement node (pressure antinode). An open end is a displacement antinode (pressure node).

• Fundamental: $L = \lambda_1/4$, so $f_1 = v/(4L)$.
• 3rd harmonic (first overtone): $L = 3\lambda_3/4$, so $f_3 = 3v/(4L) = 3f_1$.

$\boxed{f_n = \frac{nv}{4L}, \quad n = 1, 3, 5, \ldots \mathrm{ (odd harmonics only)}}$

Only odd harmonics are present because an even number of quarter-wavelengths would require a node at the open end, which contradicts the boundary condition.

Harmonics in a Pipe Open at Both Ends

Both ends are displacement antinodes.

$\boxed{f_n = \frac{nv}{2L}, \quad n = 1, 2, 3, \ldots \mathrm{ (all harmonics)}}$

Problem Set

Details

Problem 1

Two coherent sources emit waves of wavelength 0.80 m. At a point P, the path difference is 2.00 m. Is the interference at P constructive or destructive?

Answer. $\Delta x / \lambda = 2.00 / 0.80 = 2.5 = 5/2$. This is an odd half-integer, so the interference is destructive (minimum).

If you get this wrong, revise: Path Difference and Phase Difference

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Problem 2

In Young's double slit experiment, the slits are 0.40 mm apart and the screen is 2.0 m away. The fringe spacing is measured as 2.8 mm. Calculate the wavelength of the light.

Answer. $\lambda = \frac◆LB◆s \cdot \Delta w◆RB◆◆LB◆D◆RB◆ = \frac◆LB◆0.40 \times 10^{-3} \times 2.8 \times 10^{-3}◆RB◆◆LB◆2.0◆RB◆ = \frac◆LB◆1.12 \times 10^{-6}◆RB◆◆LB◆2.0◆RB◆ = 5.6 \times 10^{-7}$ m $= 560$ nm.

If you get this wrong, revise: Young's Double Slit Experiment

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Problem 3

A diffraction grating has 500 lines per mm. Light of wavelength 600 nm is incident normally. Calculate the angle of the second-order maximum.

Answer. $d = 1/500$ mm $= 2.0 \times 10^{-6}$ m. $d\sin\theta = 2\lambda$. $\sin\theta = \frac◆LB◆2 \times 600 \times 10^{-9}◆RB◆◆LB◆2.0 \times 10^{-6}◆RB◆ = \frac◆LB◆1.2 \times 10^{-6}◆RB◆◆LB◆2.0 \times 10^{-6}◆RB◆ = 0.60$. $\theta = \arcsin(0.60) = 36.9^\circ$.

If you get this wrong, revise: Diffraction Gratings

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Problem 4

A stationary wave on a string of length 1.2 m has a fundamental frequency of 120 Hz. Find: (a) the wave speed on the string, (b) the frequency of the third harmonic.

Answer. (a) $f_1 = v/(2L)$, so $v = 2Lf_1 = 2 \times 1.2 \times 120 = 288$ m s$^{-1}$.

(b) $f_3 = 3f_1 = 360$ Hz.

If you get this wrong, revise: Harmonics on a String Fixed at Both Ends

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Problem 5

A pipe of length 0.85 m is closed at one end. The speed of sound is 340 m s$^{-1}$. Calculate the frequency of the first two harmonics.

Answer. $f_1 = \frac{v}{4L} = \frac◆LB◆340◆RB◆◆LB◆4 \times 0.85◆RB◆ = \frac{340}{3.4} = 100$ Hz.

The next harmonic is the 3rd: $f_3 = 3f_1 = 300$ Hz. (No 2nd harmonic exists for a closed pipe.)

If you get this wrong, revise: Harmonics in a Pipe Closed at One End

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Problem 6

Two waves meet at a point. Wave 1 has amplitude 3.0 mm and wave 2 has amplitude 4.0 mm. If they are in phase, what is the resultant amplitude? If they are in antiphase?

Answer. In phase: $A_{\mathrm{resultant}} = 3.0 + 4.0 = 7.0$ mm.

In antiphase: $A_{\mathrm{resultant}} = |3.0 - 4.0| = 1.0$ mm.

If you get this wrong, revise: Constructive and Destructive Interference

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Problem 7

A diffraction grating with 300 lines per mm is used with light of wavelength 540 nm. How many orders of maximum can be seen on each side of the central maximum?

Answer. $d = 1/300$ mm $= 3.33 \times 10^{-6}$ m. $n_{\max} = d/\lambda = 3.33 \times 10^{-6} / 540 \times 10^{-9} = 6.17$. So the highest visible order is $n = 6$. Total visible maxima $= 2(6) + 1 = 13$.

If you get this wrong, revise: Maximum number of orders

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Problem 8

Derive the stationary wave equation from two progressive waves $y_1 = A\sin(kx - \omega t)$ and $y_2 = A\sin(kx + \omega t)$.

Answer. $y = y_1 + y_2 = A[\sin(kx - \omega t) + \sin(kx + \omega t)]$.

Using the identity $\sin\alpha + \sin\beta = 2\sin\frac◆LB◆\alpha+\beta◆RB◆◆LB◆2◆RB◆\cos\frac◆LB◆\alpha-\beta◆RB◆◆LB◆2◆RB◆$:

$\alpha = kx - \omega t$, $\beta = kx + \omega t$.

$\frac◆LB◆\alpha + \beta◆RB◆◆LB◆2◆RB◆ = kx$, $\frac◆LB◆\alpha - \beta◆RB◆◆LB◆2◆RB◆ = -\omega t$.

$y = 2A\sin(kx)\cos(-\omega t) = 2A\sin(kx)\cos(\omega t)$ (since $\cos$ is even).

If you get this wrong, revise: Derivation from Superposition of Two Progressive Waves

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Problem 9

In a Young's double slit experiment using light of wavelength 550 nm, the fringe spacing on a screen 1.5 m away is 1.1 mm. Calculate the slit separation.

Answer. $s = \frac◆LB◆\lambda D◆RB◆◆LB◆\Delta w◆RB◆ = \frac◆LB◆550 \times 10^{-9} \times 1.5◆RB◆◆LB◆1.1 \times 10^{-3}◆RB◆ = \frac◆LB◆8.25 \times 10^{-7}◆RB◆◆LB◆1.1 \times 10^{-3}◆RB◆ = 7.5 \times 10^{-4}$ m $= 0.75$ mm.

If you get this wrong, revise: Young's Double Slit Experiment

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Problem 10

A string of length 0.60 m vibrates at its fourth harmonic at 400 Hz. Find the wave speed and the wavelength of the fourth harmonic.

Answer. $f_4 = \frac{4v}{2L} = \frac{2v}{L}$. $v = \frac{f_4 L}{2} = \frac◆LB◆400 \times 0.60◆RB◆◆LB◆2◆RB◆ = 120$ m s$^{-1}$.

$\lambda_4 = 2L/4 = L/2 = 0.30$ m.

If you get this wrong, revise: Harmonics on a String Fixed at Both Ends

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Problem 11

Explain the difference between progressive and stationary waves in terms of (a) energy transfer, (b) amplitude variation, and (c) phase relationship between neighbouring points.

Answer. (a) Progressive waves transfer energy; stationary waves do not (energy is trapped).

(b) In a progressive wave, all points have the same amplitude (for an ideal wave). In a stationary wave, the amplitude varies from zero (nodes) to maximum (antinodes).

(c) In a progressive wave, all points have the same phase (the wave pattern translates). In a stationary wave, all points between two adjacent nodes are in phase, but points in adjacent segments are in antiphase.

If you get this wrong, revise: Stationary Waves

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Problem 12

White light (wavelengths 400–700 nm) is incident on a diffraction grating with 400 lines per mm. For the second order, calculate the angular range subtended by the visible spectrum.

Answer. $d = 1/400$ mm $= 2.5 \times 10^{-6}$ m. For $n = 2$:

For 400 nm: $\sin\theta = \frac◆LB◆2 \times 400 \times 10^{-9}◆RB◆◆LB◆2.5 \times 10^{-6}◆RB◆ = 0.320$, $\theta = 18.7^\circ$.

For 700 nm: $\sin\theta = \frac◆LB◆2 \times 700 \times 10^{-9}◆RB◆◆LB◆2.5 \times 10^{-6}◆RB◆ = 0.560$, $\theta = 34.1^\circ$.

Angular range $= 34.1° - 18.7° = 15.4^\circ$.

If you get this wrong, revise: Diffraction Gratings

6. Advanced Double Slit Calculations

6.1 Finding the Position of a Specific Maximum

Example. Light of wavelength $480$ nm passes through double slits separated by $0.80$ mm onto a screen $2.0$ m away. Find the distance from the central maximum to the fifth bright fringe.

Answer. $w = \frac◆LB◆n\lambda D◆RB◆◆LB◆s◆RB◆ = \frac◆LB◆5 \times 480 \times 10^{-9} \times 2.0◆RB◆◆LB◆0.80 \times 10^{-3}◆RB◆ = \frac◆LB◆4.80 \times 10^{-6}◆RB◆◆LB◆8.0 \times 10^{-4}◆RB◆ = 6.0 \times 10^{-3}$ m $= 6.0$ mm.

6.2 Finding the Position of a Minimum

Example. Using the same setup ($\lambda = 480$ nm, $s = 0.80$ mm, $D = 2.0$ m), find the distance from the central maximum to the first dark fringe.

Answer. For the first minimum, the path difference equals $\lambda/2$. For small angles: $w = D\sin\theta \approx D \cdot \lambda/(2s) = 2.0 \times 480 \times 10^{-9}/(2 \times 0.80 \times 10^{-3}) = 6.0 \times 10^{-4}$ m $= 0.60$ mm.

This is exactly half the fringe spacing: $\Delta w = \lambda D/s = 1.20$ mm, and $0.60 = \Delta w/2$. Confirmed.

6.3 Non-Integer Wavelength Ratios

Example. Two coherent sources emit waves of wavelength $2.5$ cm. At point $P$, the path difference is $9.0$ cm. Describe the interference at $P$.

Answer. $\Delta x / \lambda = 9.0 / 2.5 = 3.6 = 18/5$. This is neither an integer (constructive) nor a half-integer (destructive). The interference is partial. The phase difference is $\Delta\phi = (2\pi/\lambda)\Delta x = 2\pi \times 3.6 = 7.2\pi = 0.2\pi$ (modulo $2\pi$).

The resultant amplitude is $A_R = 2A|\cos(\Delta\phi/2)| = 2A|\cos(0.1\pi)| = 2A \times 0.951 = 1.90A$. This is close to maximum ($2A$) because the path difference is close to $4\lambda = 10.0$ cm.

7. Path Difference for Minima and Maxima: Extended Analysis

7.1 General Conditions

For two coherent sources separated by distance $s$, at a point $P$ on a screen at distance $D \gg s$:

Condition	Path Difference	Fringe Type	Order
Bright	$n\lambda$	Constructive	$n = 0, 1, 2, \ldots$
Dark	$(n + 0.5)\lambda$	Destructive	$n = 0, 1, 2, \ldots$

The distance from the central maximum to the $n$-th bright fringe is $w_n = n\lambda D/s$. The distance to the $n$-th dark fringe is $w_n = (n + 0.5)\lambda D/s$.

7.2 White Light Fringes

If white light is used, each wavelength produces its own fringe pattern. The central maximum ($n = 0$) is white (all wavelengths constructively interfere at $\Delta x = 0$). Higher-order fringes are spectrally dispersed: blue light ($\lambda \approx 450$ nm) produces narrower fringes than red light ($\lambda \approx 700$ nm).

8. Phase Difference and Coherence in Detail

8.1 Quantifying Coherence

Two sources are coherent if their phase difference $\Delta\phi$ is constant in time. The coherence length is the maximum path difference over which interference can be observed:

$L_c = \frac◆LB◆\lambda^2◆RB◆◆LB◆\Delta\lambda◆RB◆$

where $\Delta\lambda$ is the bandwidth of the source. A narrow-band source (small $\Delta\lambda$) has a long coherence length.

Example. A sodium lamp at $\lambda = 589$ nm with $\Delta\lambda = 0.6$ nm: $L_c = (589 \times 10^{-9})^2 / (0.6 \times 10^{-9}) = 5.8 \times 10^{-4}$ m $= 0.58$ mm. A laser at $\lambda = 633$ nm with $\Delta\lambda = 10^{-6}$ nm has $L_c \approx 400$ m.

8.2 Why Independent Sources Are Not Coherent

Light from two independent sources is not coherent because each emits photons with random phase. The phase difference fluctuates randomly on a timescale of about $10^{-8}$ s, washing out any interference pattern. Young's experiment requires a single source illuminating both slits so the slits act as secondary coherent sources.

9. Diffraction Grating: Extended Analysis

9.1 Angular Dispersion and Resolving Power

The angular dispersion of a grating is $d\theta/d\lambda = n/(d\cos\theta)$. The resolving power is:

$R = \frac◆LB◆\lambda◆RB◆◆LB◆\Delta\lambda◆RB◆ = nN$

where $N$ is the total number of illuminated slits.

Example. A grating has $500$ lines/mm and is $20$ mm wide. In second order at $\lambda = 600$ nm, the minimum resolvable wavelength difference is $\Delta\lambda = 600/(2 \times 10\,000) = 0.030$ nm.

9.2 Missing Orders

If grating slits have finite width $a$, an order $n$ is missing when it coincides with a single-slit minimum: $n = (d/a) \times m$ for $m = 1, 2, 3, \ldots$.

Example. A grating with $d = 2.0$ $\mu$m and slit width $a = 0.50$ $\mu$m: $d/a = 4$, so orders $n = 4, 8, 12, \ldots$ are missing.

10. Common Pitfalls

1. Forgetting the small-angle approximation. $\Delta w = \lambda D/s$ assumes $\sin\theta \approx \tan\theta$, valid only when $D \gg s$. For large angles, use $\sin\theta = n\lambda/s$ directly.

2. Confusing slit separation $s$ with slit width $a$. $s$ is the separation between slit centres; $a$ determines the diffraction envelope.

3. Counting the zeroth order. The central maximum ($n = 0$) counts as a maximum. Total visible maxima $= 2n_{\max} + 1$.

4. Assuming all harmonics exist for closed pipes. A pipe closed at one end only supports odd harmonics ($n = 1, 3, 5, \ldots$).

5. Confusing nodes and antinodes at boundaries. A fixed end is a displacement node. An open end is a displacement antinode.

11. Extension Problem Set

Details

Problem 1

In a Young's double slit experiment, light of wavelength $620$ nm produces fringes that are $2.5$ mm apart on a screen $1.2$ m from the slits. Calculate the slit separation. If the slit separation is halved, what happens to the fringe spacing?

Answer. $s = \lambda D / \Delta w = 620 \times 10^{-9} \times 1.2 / 2.5 \times 10^{-3} = 2.98 \times 10^{-4}$ m $= 0.298$ mm. If $s$ is halved: $\Delta w$ doubles to $5.0$ mm.

If you get this wrong, revise: Young's Double Slit Experiment

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Problem 2

Two coherent sources $S_1$ and $S_2$ are $3.0$ mm apart and emit light of wavelength $600$ nm. A screen is placed $5.0$ m away. Calculate the positions of the central maximum and the first three minima.

Answer. Fringe spacing: $\Delta w = 600 \times 10^{-9} \times 5.0 / 3.0 \times 10^{-3} = 1.0$ mm. Central maximum at $w = 0$. Minima at $w = 0.50$ mm, $1.5$ mm, $2.5$ mm.

If you get this wrong, revise: Path Difference for Minima and Maxima

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Problem 3

A diffraction grating with 400 lines/mm is used with light of wavelength $500$ nm. Calculate the angle of the third-order maximum and the total number of visible maxima.

Answer. $d = 2.5 \times 10^{-6}$ m. $\sin\theta = 3 \times 500 \times 10^{-9}/2.5 \times 10^{-6} = 0.600$. $\theta = 36.9^\circ$. $n_{\max} = d/\lambda = 5$. Total $= 2(5) + 1 = 11$.

If you get this wrong, revise: Diffraction Gratings

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Problem 4

A stationary wave is formed on a string of length $0.80$ m with wave speed $320$ m s$^{-1}$. Calculate the fundamental frequency and the number of nodes in the fourth harmonic.

Answer. $f_1 = 320/(2 \times 0.80) = 200$ Hz. Fourth harmonic has $n + 1 = 5$ nodes.

If you get this wrong, revise: Harmonics on a String Fixed at Both Ends

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Problem 5

Two waves of the same frequency meet at a point. Wave $A$ has amplitude $5.0$ mm and wave $B$ has amplitude $3.0$ mm. The phase difference is $\pi/3$ radians. Calculate the resultant amplitude.

Answer. $A_R = \sqrt◆LB◆A^2 + B^2 + 2AB\cos\Delta\phi◆RB◆ = \sqrt◆LB◆25 + 9 + 2 \times 5 \times 3 \times \cos(\pi/3)◆RB◆ = \sqrt{25 + 9 + 15} = \sqrt{49} = 7.0$ mm.

If you get this wrong, revise: Constructive and Destructive Interference

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Problem 6

Explain why a diffraction grating produces sharper maxima than a double slit.

Answer. With $N$ slits, destructive interference occurs between all pairs, not just adjacent slits. Between two adjacent maxima, there are $N - 2$ positions where $N - 1$ slits cancel pairwise, suppressing the background intensity much more effectively than with 2 slits. As $N$ increases, maxima become narrower and brighter.

If you get this wrong, revise: Diffraction Gratings

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tip

tip Ready to test your understanding of Superposition and Interference? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Superposition and Interference with other physics topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.