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Single-Slit Fraunhofer Diffraction

Single-slit diffraction is the bending and interference of light waves when passing through a single aperture of comparable dimensions to the light's wavelength. It produces a central bright maximum flanked by alternating secondary minima and smaller maxima.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Qualitative understanding of diffraction, conditions for minima and maxima, width of the central maximum, and distinction between interference and diffraction.

PGT extension

Quantitative derivation of intensity distribution using phasor addition, path and phase difference calculations, angular width, and resolving power applications.

Prerequisites

Huygens-Fresnel Principle
Wave Superposition and Interference
Fraunhofer vs Fresnel Diffraction conditions

Physical Origin of Single-Slit Diffraction

Every point on the wavefront within the slit of width a acts as a source of secondary wavelet disturbances.
The resultant displacement at any point on the screen is the superposition of all wavelets traveling in a given direction θ.
Destructive interference occurs when wavelets from different parts of the slit cancel each other out completely due to path differences.

Condition for Minima (Dark Fringes)

Minima are formed on the screen when the path difference between waves from the top and bottom edges of the slit equals an integral multiple of the wavelength.
Mathematically expressed as a sin θ = nλ, where n = ±1, ±2, ±3, ...
The integer n = 0 corresponds to the central maximum, not a minimum.

Condition for Secondary Maxima (Bright Fringes)

Secondary maxima occur approximately halfway between the diffraction minima.
The condition for secondary maxima is given by a sin θ = (n + 0.5)λ, where n = ±1, ±2, ±3, ...
The intensity of secondary maxima decreases rapidly as the diffraction angle θ increases away from the central axis.

Intensity Distribution

The intensity I at any angle θ is given by I = I₀ (sin β / β)², where $\beta = (\pia / \lambda) \sin \theta$.
The central maximum contains the vast majority of the total incident optical energy.
The angular half-width of the central maximum is defined by the angular position of the first minimum: θ ≈ λ / a.

Derivations

Derivation of Minima Condition

Conditions: Fraunhofer diffraction approximation, plane wavefront incident normally on a single slit of width a.

Divide the slit of width a into two equal halves of width a/2.
The path difference between secondary wavelets originating from corresponding points in the upper and lower halves separated by a/2 is (a/2) sin θ.
Complete destructive interference between these two halves occurs when their path difference is half a wavelength: (a/2) sin $\theta = \lambda / 2$.
Generalizing this subdivision for 2n zones leads to the condition a sin $\theta = n\lambda$ for $n = \pm 1$, ±2, ±3, ...

Result: a sin $\theta = n\lambda (n = \pm 1$, ±2, ±3, ...)

Derivation of Intensity Distribution

Conditions: Integration of infinitesimal phasor amplitudes across the slit width a.

Consider the slit divided into N infinitesimal strips, each of width dy. The amplitude contribution from each strip is dE = (E₀ / a) dy.
The phase difference between the wave from element y and the center of the slit is d$\phi = (2\pi / \lambda) y \sin \theta$.
Integrate the resultant electric field amplitude E(θ) across the slit from -a/2 to +a/2.
E(θ) = ∫_-a/2^a/2 (E₀ / a) exp[i(2π/λ)y sin θ] dy, which evaluates to E₀ (sin β / β) where $\beta = (\pia/\lambda) \sin \theta$.
Since intensity I is proportional to the square of the amplitude (E), I(θ) = I₀ (sin β / β)².

Result: I(θ) = I_0 left(\frac{sin \beta}{\beta}\right)^2

Formula sheet

Formula / law	Equation	Variables and units	Conditions
Condition for Diffraction Minima	$$a \sin \theta = n \lambda$$	a = slit width, $\theta = diffraction angle$, $n = order of minimum (\pm 1$, ±2, ...), $\lambda = wavelength of light$ m for a and λ; radians or degrees for θ; dimensionless for n	Fraunhofer diffraction regime
Condition for Secondary Maxima	$$a \sin \theta = \left(n + \frac{1}{2}\right) \lambda$$	a = slit width, $\theta = diffraction angle$, n = integer order, $\lambda = wavelength$ m for a and λ; radians for θ	Approximate positions of secondary maxima
Intensity Distribution Formula	$$I = I_0 \left(\frac{\sin \beta}{\beta}\right)^2$$	$I = intensity at angle \theta$, $I_0 = intensity at central maximum$, $\beta = phase parameter (\pia \sin \theta / \lambda)$ W/m² for I and I_0; dimensionless for β	Valid for single-slit Fraunhofer diffraction
Phase Parameter beta	$$\beta = \frac{\pi a \sin \theta}{\lambda}$$	a = slit width, $\theta = diffraction angle$, $\lambda = wavelength$ radians	General single-slit geometry
Angular Width of Central Maximum	$$\theta_{cent} = \frac{2lambda}{a}$$	θ_cent = total angular spread of central maximum, $\lambda = wavelength$, a = slit width radians	Small angle approximation (sin θ ≈ θ)
Linear Width of Central Maximum	$$W = \frac{2lambda D}{a}$$	W = linear width on screen, D = distance from slit to screen, $\lambda = wavelength$, a = slit width m	D >> a, small angle approximation
Path Difference for Angle theta	$$\delta x = a \sin \theta$$	Δx = path difference between edge rays, a = slit width, $\theta = angle$ m	Plane wavefront approximation
First Minimum Angular Position	$$\theta_1 \approx \frac{\lambda}{a}$$	θ₁ = angular position of first dark fringe, $\lambda = wavelength$, a = slit width radians	Small angle approximation

Graphs and experiments

Intensity vs. Diffraction Angle Graph

Draw: A central intense peak at $\theta = 0 with height I_0$, flanked symmetrically by diminishing secondary peaks situated between minima located at $\theta = \pm \lambda/a$, ±2λ/a, etc.

Interpret: Shows rapid drop in intensity away from the center; secondary maxima are much dimmer than the central peak (approx 4.5% for the first side maximum).

Exam use: Directly tested in questions asking for the sketch of the diffraction pattern or identification of secondary maxima positions.

Slit Width Variation Experiment

Draw: Comparative sketches showing how narrowing slit width 'a' widens the diffraction pattern, while broadening 'a' compresses the pattern.

Interpret: Inversely proportional relationship between slit width and angular spread (θ ∝ 1/a).

Exam use: Conceptual MCQs testing the effect of altering wavelength or slit width on fringe spacing and central maximum breadth.

Monochromatic vs White Light Diffraction Setup

Draw: A schematic diagram showing a laser or monochromatic source versus a white light source forming a central white fringe with colored diffraction spectra on sides.

Interpret: Different wavelengths diffract by different angles (since θ ∝ λ), causing red light to bend outward more than violet light.

Exam use: Explaining the appearance of colored bands in white light single-slit experiments.

Solved numericals

TGT core

Light of wavelength 500 nm falls normally on a single slit of width 0.2 mm. Calculate the angular position of the first secondary minimum in radians.

Identify given values: $\lambda = 500 nm = 500 \times 10^{-9} m$, $a = 0.2 mm = 0.2 \times 10^{-3} m$, n = 1.
Apply the diffraction minimum formula: a sin $\theta = n\lambda$.
Substitute values: (0.2 × 10^-3) sin $\theta = 1 \times 500 \times 10^{-9}$.
Solve for sin θ: sin $\theta = (500 \times 10^{-9}) / (2 \times 10^{-4}) = 2.5 \times 10^{-3}$.
For small angles, sin θ ≈ θ ≈ 2.5 × 10^-3 rad.

Answer: 2.5 × 10^-3 rad

PGT extension

A single slit of width 0.10 mm is illuminated by a parallel beam of light of wavelength 600 nm. The diffraction pattern is observed on a screen placed 2.0 m away. Calculate the linear width of the central maximum.

Identify given parameters: $a = 0.10 mm = 1.0 \times 10^{-4} m$, $\lambda = 600 nm = 6 \times 10^{-7} m$, D = 2.0 m.
Recall the formula for linear width of the central maximum: $W = 2\lambdaD / a$.
Substitute the given values into the equation: $W = (2 \times 6 \times 10^{-7} m \times 2.0 m) / (1.0 \times 10^{-4} m)$.
Perform the multiplication in the numerator: 2 × 6 × 10^-7 × $2.0 = 2.4 \times 10^{-6}$.
Divide by the denominator: $W = 2.4 \times 10^{-6} / 1.0 \times 10^{-4} = 2.4 \times 10^{-2} m$.

Answer: 2.4 cm (or 0.024 m)

PGT extension

At what angle does the first secondary maximum occur for a single slit of width 'a' illuminated by light of wavelength λ?

The secondary maxima occur approximately midway between the minima, governed by the condition: a sin $\theta = (n + 0.5)\lambda$.
For the first secondary maximum, take n = 1.
Substitute $n = 1 into the condition: a \sin \theta = (1 + 0.5)\lambda = 1.5\lambda = (3/2)\lambda$.
Rearrange to solve for sin θ: sin $\theta = 3\lambda / (2a)$.

Answer: sin $\theta = 1.5\lambda / a$

Assumptions and limitations

Fraunhofer diffraction assumes both the source and the screen are effectively at an infinite distance from the slit, achieved experimentally using lenses.
The scalar wave approximation is used, ignoring polarization vector effects which become significant for sub-wavelength slit sizes.
The incident wavefront is treated as perfectly plane and parallel across the entire aperture.

Common misconceptions

Common error: The central maximum in single-slit diffraction has the same width and intensity as secondary maxima.

Correct: The central maximum is twice as wide as any secondary maximum and contains significantly higher light intensity.

The central maximum spans from $n = -1$ to $n = +1 (width 2\lambda/a)$, whereas secondary maxima lie between adjacent higher-order minima.

Common error: Diffraction minima correspond to points where light rays block each other out due to reflection.

Correct: Minima occur due to complete destructive interference among secondary wavelets originating from different regions of the same single slit.

Destructive phase cancellation occurs when path differences cause phase opposition across the aperture wavefronts.

Common error: The formula a sin $\theta = n\lambda represents conditions$ for bright fringes in single-slit diffraction.

Correct: In single-slit diffraction, a sin $\theta = n\lambda gives the positions of dark fringes (minima)$.

Students often confuse single-slit diffraction formulas with double-slit interference formulas, where nλ gives bright fringes.

Quick revision

Single slit dark fringe condition: a sin $\theta = n\lambda (n = 1$, 2, 3...).
Angular width of central maximum: 2λ/a.
Linear width of central maximum: 2λD/a.
First secondary maximum position: a sin $\theta = 1.5\lambda$.
Intensity of central maximum is proportional to E₀², while side maxima drop off rapidly.

Must remember

The condition for minima (dark fringes) in single-slit Fraunhofer diffraction is given by $a \sin \theta = n \lambda$, where $n = \pm 1, \pm 2, \pm 3, \dots$.
The integer $n = 0$ corresponds to the central maximum, not a minimum.
The condition for secondary maxima is given approximately by $a \sin \theta = \left(n + \frac{1}{2}\right) \lambda$, where $n = \pm 1, \pm 2, \pm 3, \dots$.
The intensity distribution is given by $I = I_0 \left(\frac{\sin \beta}{\beta}\right)^2$, where $\beta = \frac{\pi a \sin \theta}{\lambda}$.
The total angular width of the central maximum is $\theta_{cent} = \frac{2\lambda}{a}$, and the linear width is $W = \frac{2\lambda D}{a}$.

Glossary

Fraunhofer DiffractionA form of diffraction where both the source of light and the screen are effectively at an infinite distance from the aperture, using plane wavefronts.

Slit Width ($a$)The aperture dimension through which wavefronts pass to produce diffraction patterns.

Phase Parameter ($\beta$)A dimensionless variable defined as $\beta = \frac{\pi a \sin \theta}{\lambda}$ used in the intensity distribution formula.

Secondary MaximumBright fringes of decreasing intensity located between adjacent diffraction minima, occurring at $a \sin \theta = \left(n + \frac{1}{2}\right) \lambda$.

Central Maximum vs Secondary Maxima

Feature	Central Maximum	Secondary Maxima
Position	At $\theta = 0$ ($n = 0$)	Between adjacent minima, roughly at $a \sin \theta = \left(n + \frac{1}{2}\right) \lambda$
Angular Width	$\frac{2\lambda}{a}$	$\frac{\lambda}{a}$ (approximate width between adjacent minima)
Intensity	Maximum ($I_0$), contains vast majority of energy	Decreases rapidly as angle $\theta$ increases

Practice Quiz

Choose an option to reveal the explanation.

1. What is the condition for obtaining diffraction minima (dark fringes) in a single-slit Fraunhofer diffraction experiment?

Explanation: Minima in single-slit diffraction occur when the path difference between wavelets from the top and bottom edges is an integral multiple of wavelength, given by $a \sin \theta = n \lambda$ where $n \neq 0$.

2. What does the integer $n = 0$ represent in the equation $a \sin \theta = n \lambda$ for single-slit diffraction?

Explanation: For $n = 0$, $\sin \theta = 0$, which corresponds to $\theta = 0$, representing the bright central maximum.

3. What is the total angular width of the central maximum in single-slit Fraunhofer diffraction?

Explanation: The angular half-width is defined by the first minimum at $\theta \approx \frac{\lambda}{a}$. Therefore, the total angular spread of the central maximum is $2\theta_1 = \frac{2\lambda}{a}$.

4. How does the linear width of the central maximum on a screen change if the slit width $a$ is halved?

Explanation: The linear width is given by $W = \frac{2\lambda D}{a}$. Since $W$ is inversely proportional to $a$, halving $a$ doubles the linear width.

5. Which of the following equations gives the approximate angular positions of secondary maxima in single-slit diffraction?

Explanation: Secondary maxima occur approximately halfway between the diffraction minima, satisfying the condition $a \sin \theta = \left(n + \frac{1}{2}\right) \lambda$ for $n = \pm 1, \pm 2, \dots$.

6. What is the expression for the phase parameter $\beta$ used in the intensity distribution formula of single-slit diffraction?

Explanation: The phase parameter representing total phase difference across the slit is defined as $\beta = \frac{\pi a \sin \theta}{\lambda}$.

7. The intensity $I$ at any angle $\theta$ in single-slit Fraunhofer diffraction is given by:

Explanation: The diffraction intensity distribution formula is $I = I_0 \left(\frac{\sin \beta}{\beta}\right)^2$, derived using phasor addition or integration over the slit aperture.

8. What is the path difference between light rays originating from the top and bottom edges of a slit of width $a$ at a diffraction angle $\theta$?

Explanation: For plane wavefronts incident normally on a slit of width $a$, the path difference between extreme edge rays is given by $\Delta x = a \sin \theta$.

9. Which of the following statements is true regarding the central maximum compared to secondary maxima in single-slit diffraction?

Explanation: The central maximum spans from $n = -1$ to $n = +1$, making it twice as wide as secondary maxima, and it contains the vast majority of the incident optical energy.

10. What happens to the diffraction pattern on the screen if monochromatic light is replaced by white light?

Explanation: At the center ($\theta = 0$), all wavelengths interfere constructively, producing a white central maximum. Higher-order fringes are dispersed into spectra because the position depends on wavelength $\lambda$.

11. If the distance from the slit to the screen $D$ is doubled, how does the linear width of the central maximum $W$ change?

Explanation: Since $W = \frac{2\lambda D}{a}$, the linear width is directly proportional to the screen distance $D$. Doubling $D$ doubles $W$.

12. What type of wavefront is incident on the slit in Fraunhofer diffraction?

Explanation: In Fraunhofer diffraction, both the source and the screen are effectively at infinity, meaning plane wavefronts fall on the aperture.

13. What is the angular position of the first diffraction minimum for a slit of width $a$ illuminated by wavelength $\lambda$ (using small angle approximation)?

Explanation: For $n = 1$ in $a \sin \theta = n \lambda$, using $\sin \theta \approx \theta$ for small angles yields $\theta_1 \approx \frac{\lambda}{a}$.

14. Which physical phenomenon is responsible for the dark fringes in single-slit diffraction?

Explanation: Dark fringes (minima) arise from complete destructive interference of secondary wavelets originating from different sections of the single slit aperture.

15. What is the unit of the phase parameter $\beta$?

Explanation: The phase parameter $\beta = \frac{\pi a \sin \theta}{\lambda}$ represents a phase angle, and its unit is radians.

PYQ Patterns

These are pattern-based practice questions. Authentic year labels are only used when verified from a source.

1. In a single-slit diffraction experiment, the width of the slit is $0.2\text{ mm}$ and it is illuminated by light of wavelength $500\text{ nm}$. The distance of the screen from the slit is $1\text{ m}$. What is the linear width of the central maximum?

Explanation: Using $W = \frac{2\lambda D}{a}$, substitute $\lambda = 500 \times 10^{-9}\text{ m}$, $D = 1\text{ m}$, and $a = 0.2 \times 10^{-3}\text{ m}$. Thus, $W = \frac{2 \times 500 \times 10^{-9} \times 1}{0.2 \times 10^{-3}} = 5 \times 10^{-3}\text{ m} = 5.0\text{ mm}$.

2. The angular width of the central maximum of a diffraction pattern of a single slit of width $a$ with incident light of wavelength $\lambda$ is:

Explanation: The angular spread extends from $\theta = -\frac{\lambda}{a}$ to $\theta = +\frac{\lambda}{a}$, giving a total angular width of $\frac{2\lambda}{a}$.

3. For a single-slit diffraction, the condition for the second minimum is given by:

Explanation: The condition for diffraction minima is $a \sin \theta = n \lambda$. For the second minimum, the order is $n = 2$, so $a \sin \theta = 2\lambda$.

4. If light of wavelength $600\text{ nm}$ is incident normally on a slit, and the first diffraction minimum is observed at an angle of $30^{\circ}$, what is the width of the slit?

Explanation: For the first minimum ($n = 1$), $a \sin \theta = 1 \cdot \lambda$. Here $\theta = 30^{\circ}$, so $\sin(30^{\circ}) = 0.5$. Thus, $a = \frac{\lambda}{\sin 30^{\circ}} = \frac{600 \times 10^{-9}}{0.5} = 1.2 \times 10^{-6}\text{ m}$.

5. In single-slit diffraction, what happens to the angular width of the central maximum if the entire apparatus is immersed in water (refractive index $\mu = 4/3$)?

Explanation: When immersed in water, the wavelength becomes $\lambda' = \frac{\lambda}{\mu}$. The angular width $\theta_{cent} = \frac{2\lambda'}{a}$ also decreases by a factor of $\frac{1}{\mu} = \frac{3}{4}$.

6. The intensity at the secondary maximum in single-slit diffraction is much less than that of the central maximum because:

Explanation: At secondary maxima, the wavefront cannot divide into complete canceling pairs as it does for minima; leftover portions contribute to a lower net intensity.

Timed Topic Test

Attempt these questions under a ten-minute limit using a separate timer.

1. What is the path difference between waves from the two edges of a slit of width $a$ for the third minimum at angle $\theta$?

Explanation: The condition for minima is $a \sin \theta = n \lambda$. For the third minimum, $n = 3$, so the path difference $\Delta x = 3\lambda$.

2. The ratio of the intensity of the first secondary maximum to the central maximum intensity $I_0$ in single-slit diffraction is approximately:

Explanation: The first secondary maximum occurs around $\beta \approx 1.43 \pi$. Evaluating $I = I_0 \left(\frac{\sin \beta}{\beta}\right)^2$ yields roughly $\frac{1}{22} I_0$ or about $4.5\%$ of $I_0$.

3. If the wavelength of light used in a single-slit diffraction setup is doubled while keeping the slit width constant, the angular width of the central maximum will:

Explanation: The angular width is $\theta_{cent} = \frac{2\lambda}{a}$, which is directly proportional to wavelength $\lambda$. Doubling $\lambda$ doubles the angular width.

4. In a Fraunhofer diffraction pattern by a single slit, the screen is placed at the focal plane of a convex lens. If the lens is close to the slit, what is the role of the lens?

Explanation: A convex lens placed after the slit focuses parallel rays traveling in a specific diffraction angle $\theta$ onto a specific point on the focal plane screen.

5. What is the condition for destructive interference (minima) when dividing a slit of width $a$ into $2n$ equal parts?

Explanation: When the slit is divided into $2n$ zones, adjacent zones have a path difference of $\frac{\lambda}{2}$ relative to each other, leading to complete cancellation.

6. Which of the following describes the correct order of decreasing intensity among maxima in single-slit diffraction?

Explanation: Intensity drops rapidly away from the center; the central maximum is the brightest, followed by rapidly diminishing secondary maxima.

7. A slit of width $a$ is illuminated by light of wavelength $\lambda$. What is the number of diffraction minima visible on either side of the central maximum if $\lambda < a$

Explanation: Since $a \sin \theta = n \lambda$ and $\sin \theta \le 1$, the maximum order $n$ is limited by $\frac{a}{\lambda}$. Thus, the number of minima on one side is $\left\lfloor \frac{a}{\lambda} \right\rfloor$.

8. If the slit width $a$ is made equal to the wavelength of light $\lambda$, what is the angular span of the central maximum?

Explanation: When $a = \lambda$, the first minimum occurs at $\sin \theta_1 = \frac{\lambda}{a} = 1$, so $\theta_1 = 90^{\circ}$. The central maximum spreads from $-90^{\circ}$ to $+90^{\circ}$, covering a total span of $180^{\circ}$.

9. What is the phase difference between wavelets originating from the top and bottom of the slit corresponding to the first diffraction minimum?

Explanation: The path difference for the first minimum is $\lambda$, which corresponds to a phase difference of $\frac{2\pi}{\lambda} \times \lambda = 2\pi$ radians between the extreme edges.

10. How does single-slit diffraction differ fundamentally from double-slit interference regarding the envelope of the pattern?

Explanation: In a double-slit experiment using finite slits, the interference fringes are constrained and modulated by the single-slit diffraction envelope of each individual slit.