Fraunhofer diffraction is a phenomenon of light bending around obstacles or apertures where both the source of light and the observation screen are effectively at an infinite distance from the aperture. Under these conditions, the wavefronts incident on the aperture and diffracted from it are planar.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Qualitative understanding of single-slit diffraction patterns, central maxima, secondary maxima and minima conditions, and relation with wavelength and slit width.
PGT extension
Rigorous mathematical treatment of single-slit and double-slit Fraunhofer diffraction, intensity distribution, angular width of central maxima, and diffraction grating spectra.
Prerequisites
Huygens-Fresnel principle
Superposition of waves and phasor addition
Plane wavefronts and coherence
Condition for Fraunhofer Diffraction
The source of light and the screen are at infinite distances from the diffracting aperture, achieved practically using two convex lenses.
The incident wavefronts on the obstacle or aperture are planar.
The diffracted rays arriving at any point on the screen are parallel to each other.
Single-Slit Intensity Distribution
The diffraction pattern consists of a central bright maximum flanked by alternate secondary minima and weaker secondary maxima.
The angular half-width of the central maximum is given by sin(θ) = λ/a, where a is the slit width.
Most of the incident light energy is concentrated within the central maximum.
Minima and Secondary Maxima Conditions
Minima are formed when the path difference between light rays originating from the extreme edges of the slit is an integral multiple of wavelength (λ).
Secondary maxima are approximately located midway between the secondary minima, where the path difference is odd multiples of λ/2.
As the slit width a increases, the central maximum becomes narrower and more intense.
Comparison with Fresnel Diffraction
In Fresnel diffraction, either the source, the screen, or both are at finite distances from the aperture, requiring spherical or cylindrical wavefront analysis.
Fraunhofer diffraction is a special limiting case of Fresnel diffraction where wavefront curvature is negligible.
Fraunhofer diffraction is mathematically simpler to handle using rectilinear propagation approximations and Fourier transforms.
Derivations
Fraunhofer Diffraction at a Single Slit
Conditions: Normal incidence of plane wavefront of wavelength λ on a slit of width a, observation on a focal plane screen.
Divide the slit of width a into n equal infinitesimal strips. The path difference between consecutive strips for a diffraction angle θ is dx sin(θ).
The phase difference between consecutive strips is d$\phi = (2\pi/\lambda) dx \sin(\theta)$. Integrating across the slit width from -a/2 to a/2 gives the resultant amplitude $R = A₀ (\sin \alpha)/\alpha$, where $\alpha = (\pi a \sin \theta)/\lambda$.
The intensity I is proportional to R², yielding $I = I₀ (\sin \alpha / \alpha)²$.
Result: $I = I₀ (\sin \alpha / \alpha)^2$ where $\alpha = \frac{\pi a \sin\theta}{\lambda}$
Positions of Minima for Single Slit
Conditions: Intensity vanishing condition in single-slit diffraction.
Minima occur when the amplitude factor sin $\alpha = 0$, provided α ≠ 0.
This implies $\alpha = m\pi$, where $m = \pm 1$, ±2, ±3, ...
Substituting α gives (π a sin\theta)/λ = mπ, leading to a sin $\theta = m\lambda$.
f = focal length of lens, lambda = wavelength, a = slit width Metres (m)
Screen placed at focal plane
Grating Equation for Multiple Slits
$$(a + b) \sin\theta = nlambda$$
(a + b) = grating element, \theta = diffraction angle, n = order, lambda = wavelength Metres (m) for (a+b)
Normal incidence on diffraction grating
Resolving Power of Grating
$$R = \frac{\lambda}{Deltalambda} = n N$$
n = spectral order, N = total number of rulings, lambda = wavelength Dimensionless
Rayleigh criterion for resolution
Graphs and experiments
Intensity vs Phase Parameter (α) for Single Slit
Draw: A symmetrical curve showing a high central peak at $\alpha = 0 with decreasing secondary peaks at \alpha \approx \pm 1.43\pi$, ±2.46π, with intensity vanishing at $\alpha = \pm \pi$, ±2π, ...
Interpret: Illustrates how diffracted intensity drops rapidly away from the center, concentrating about 90% of energy in the central maximum.
Exam use: Frequently asked in PGT to identify positions of secondary maxima and minima.
Formation of Fraunhofer Diffraction using Convex Lenses
Draw: A point or slit source at the focus of lens L₁, parallel beam falling on aperture, and lens L₂ focusing parallel diffracted rays onto a screen.
Interpret: Demonstrates the experimental realization of infinite source and screen distances in a laboratory setup.
Exam use: TGT conceptual questions regarding the necessity and placement of lenses.
Grating Diffraction Spectrum
Draw: Sharp principal maxima peaks separated by broad regions of darkness and faint subsidiary maxima.
Interpret: Shows the sharp dispersive power of diffraction gratings compared to double slits.
Exam use: PGT numericals and conceptual questions on resolving power and missing orders.
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.
Identify given values: lambd$a = 500 nm = 5 \times 10^{-7} m$, $a = 0.2 mm = 2 \times 10^{-4} m$, m = 1.
Since sin\theta is very small, \theta approx 2.5 \times 10^{-3} radians.
Answer: 2.5 \times 10^{-3} radians
PGT extension
A slit of width 0.5 mm is illuminated normally with light of wavelength 589 nm. Find the linear width of the central maximum on a screen placed 1.0 m away from the lens.
Identify given values: $a = 0.5 mm = 5 \times 10^{-4} m$, lambd$a = 589 nm = 5.89 \times 10^{-7} m$, f = 1.0 m.
Linear width of central maximum is given by $\beta = (2 f \lambda) / a$.
Use grating formula for maximum order: (a + b) sin\theta = nlambda. For maximum order, set sin\theta = 1.
Calculate $n_max = (a + b) / \lambda = (2 \times 10^{-6}) / (6 \times 10^{-7}) = 3.33$.
Since order must be an integer, round down to the nearest integer.
Answer: 3
Assumptions and limitations
Assumes scalar wave theory, ignoring polarization effects of light waves.
Valid strictly when the distance of source and screen from the aperture are large compared to the size of the aperture.
Common misconceptions
Common error: Fraunhofer diffraction requires spherical wavefronts.
Correct: Fraunhofer diffraction involves plane wavefronts incident and diffracted.
Wavefront curvature is assumed negligible due to infinite distance.
Common error: The central maximum in single-slit diffraction has the same width as secondary maxima.
Correct: The central maximum has double the width of any secondary maximum.
The central maximum spans from $m = -1$ to $m = +1$, occupying an angular range of 2lambda/a.
Quick revision
Central maximum width is 2lambda/a in angular terms and 2flambda/a in linear terms for a single slit.
Phase Parameter alpha: alpha = \frac{pi a sin\theta}{lambda}
Intensity Distribution for Single Slit: $I = I_0 \left(\frac{sinalpha}{\alpha}\right)^2$
Must remember
In Fraunhofer diffraction, both the source of light and the screen are effectively at infinite distances from the aperture, using plane wavefronts.
The condition for secondary minima in a single-slit Fraunhofer diffraction is given by $a \sin\theta = m\lambda$ where $m = \pm 1, \pm 2, \dots$.
The angular width of the central maximum is $2\theta = \frac{2\lambda}{a}$, while the linear width is $\beta = \frac{2f\lambda}{a}$.
The intensity distribution for a single slit is given by $I = I_0 \left(\frac{\sin\alpha}{\alpha}\right)^2$ where $\alpha = \frac{\pi a \sin\theta}{\lambda}$.
Resolving power of a diffraction grating is given by $R = \frac{\lambda}{\Delta\lambda} = n N$, depending on the spectral order $n$ and total rulings $N$.
Glossary
Fraunhofer DiffractionA type of diffraction where source and screen are effectively at infinite distances, characterized by plane wavefronts.
Phase Parameter ($\alpha$)A dimensionless angular parameter defined as $\alpha = \frac{\pi a \sin\theta}{\lambda}$ used in single-slit intensity equations.
Grating ElementThe sum of the width of a ruling and an opaque space, denoted by $(a + b)$.
Resolving PowerThe ability of an optical instrument or grating to distinguish between two closely spaced spectral lines, given by $R = nN$.
Fraunhofer vs Fresnel Diffraction
Feature
Fraunhofer Diffraction
Fresnel Diffraction
Distance of Source/Screen
Infinite distances (achieved via lenses)
Finite distances
Wavefront Type
Plane wavefronts
Spherical or cylindrical wavefronts
Mathematical Complexity
Simpler (algebraic & Fourier methods)
Complex (involves half-period zones and integrals)
Practice Quiz
Choose an option to reveal the explanation.
1. What type of wavefronts are incident on the aperture in Fraunhofer diffraction?
Explanation: In Fraunhofer diffraction, the source is at an infinite distance (or made parallel using a convex lens), so the incident wavefronts on the aperture are planar.
2. The condition for secondary minima in single-slit Fraunhofer diffraction is:
Explanation: Minima are formed when the path difference between light rays from extreme edges is an integral multiple of wavelength, expressed as $a \sin\theta = m\lambda$ where $m \neq 0$.
3. What is the angular half-width of the central maximum in single-slit diffraction?
Explanation: The angular half-width of the central maximum is given by $\sin\theta \approx \theta = \frac{\lambda}{a}$.
4. What is the expression for the intensity distribution in single-slit Fraunhofer diffraction?
Explanation: The intensity distribution for a single slit is given by $I = I_0 \left(\frac{\sin\alpha}{\alpha}\right)^2$, where $\alpha = \frac{\pi a \sin\theta}{\lambda}$.
5. If the slit width $a$ is increased in a single-slit diffraction experiment, what happens to the central maximum?
Explanation: Since the linear width of the central maximum is $\beta = \frac{2f\lambda}{a}$, increasing $a$ makes the central maximum narrower and concentrates the light energy more, making it more intense.
6. Which of the following represents the grating equation for normal incidence?
Explanation: The grating equation for multiple slits under normal incidence is $(a + b) \sin\theta = n\lambda$, where $(a+b)$ is the grating element.
7. The resolving power of a diffraction grating is given by which formula?
Explanation: According to Rayleigh's criterion, the resolving power of a grating is $R = \frac{\lambda}{\Delta\lambda} = n N$, where $n$ is the order and $N$ is the total number of rulings.
8. In Fresnel diffraction, what is the key difference regarding distances compared to Fraunhofer diffraction?
Explanation: In Fresnel diffraction, either the source, the screen, or both are at finite distances from the aperture, resulting in spherical or cylindrical wavefront curvature.
9. What is the phase parameter $\alpha$ for single-slit diffraction defined as?
Explanation: The phase parameter $\alpha$ is defined as $\alpha = \frac{\pi a \sin\theta}{\lambda}$.
10. What is the linear width of the central maximum when the screen is placed at the focal plane of a convex lens of focal length $f$?
Explanation: The linear width is twice the angular half-width multiplied by focal length $f$, yielding $\beta = 2\theta f = \frac{2f\lambda}{a}$.
11. Secondary maxima in single-slit diffraction are approximately located where the path difference is:
Explanation: Secondary maxima occur approximately midway between secondary minima, where the path difference is odd multiples of $\frac{\lambda}{2}$.
12. What is the condition for secondary maxima given by the transcendental equation?
Explanation: Differentiating the intensity expression with respect to $\alpha$ and setting it to zero yields the transcendental equation $\tan\alpha = \alpha$ for secondary maxima positions.
13. Which misconception about Fraunhofer diffraction is false?
Explanation: Fraunhofer diffraction does *not* require spherical wavefronts; it involves plane wavefronts incident and diffracted. Saying it requires spherical wavefronts is a misconception.
14. How does the width of the central maximum compare to the secondary maxima in single-slit diffraction?
Explanation: The central maximum spans from $m = -1$ to $m = +1$, occupying an angular range of $\frac{2\lambda}{a}$, which is double the width of any secondary maximum ($\frac{\lambda}{a}$).
15. Fraunhofer diffraction is mathematically simpler because it can be treated using:
Explanation: Fraunhofer diffraction is mathematically simpler as it can be handled using rectilinear propagation approximations and Fourier transforms due to negligible wavefront curvature.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. In a single-slit Fraunhofer diffraction setup, light of wavelength $500\text{ nm}$ is incident normally on a slit of width $0.2\text{ mm}$. What is the angular width of the central maximum?
3. The first secondary minimum in a single-slit Fraunhofer diffraction pattern is observed at an angle of $30^\circ$ for a light of wavelength $500\text{ nm}$. The slit width $a$ is:
Explanation: Condition for first minimum ($m=1$): $a \sin\theta = 1\lambda \implies a \sin(30^\circ) = 500\text{ nm} \implies a(0.5) = 500\text{ nm} \implies a = 1000\text{ nm}$.
4. What is the maximum resolving power of a diffraction grating having $2000$ lines per cm across a total width of $5\text{ cm}$ in the 2nd spectral order ($n=2$)?
Explanation: Total number of lines $N = 2000 \times 5 = 10,000$. Resolving power $R = n N = 2 \times 10,000 = 20,000$.
5. In single-slit diffraction, if the wavelength of light is doubled while keeping the slit width constant, the linear width of the central maximum on the screen will:
Explanation: Linear width $\beta = \frac{2f\lambda}{a}$. Since $\beta \propto \lambda$, doubling the wavelength doubles the linear width.
6. Which of the following conditions correctly identifies the location of secondary minima for a single slit of width $a$?
Explanation: Secondary minima occur when $a \sin\theta = m\lambda$ for $m = \pm 1, \pm 2, \dots$. $m=0$ corresponds to the central maximum center.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. Why are convex lenses used in laboratory setups to observe Fraunhofer diffraction?
Explanation: Convex lenses are used to make light rays parallel (incoming) and focus parallel diffracted rays onto a focal plane screen, simulating infinite source/screen distances.
2. What fraction of incident light energy is predominantly concentrated in single-slit diffraction?
Explanation: Most of the incident light energy is concentrated within the central maximum of a single-slit diffraction pattern.
3. If the slit width $a$ is equal to the wavelength $\lambda$ of the incident light, what is the angular half-width of the central maximum?
Explanation: Since $\sin\theta = \frac{\lambda}{a}$, if $a = \lambda$, then $\sin\theta = 1$, which gives $\theta = 90^\circ$.
4. Which parameter determines the phase difference between wavelets from the top and bottom of a single slit at diffraction angle $\theta$?
Explanation: The phase parameter $\alpha = \frac{\pi a \sin\theta}{\lambda}$ represents half the phase difference between wavelets originating from the extreme edges of the slit.
5. In a diffraction grating experiment, missing spectra can occur if:
Explanation: A missing order occurs when the condition for a single-slit minimum ($a \sin\theta = m\lambda$) coincides with the condition for a grating maximum ($(a+b)\sin\theta = n\lambda$).
6. What happens to the diffraction pattern when a white light source is used instead of monochromatic light in single-slit diffraction?
Explanation: With white light, the center is white (all wavelengths overlap at $m=0$), while secondary maxima are colored because the position of maxima depends on wavelength $\lambda$.
7. In Fraunhofer diffraction, the rectilinear propagation approximation is valid because:
Explanation: Fraunhofer diffraction treats wavefronts as planar since source and screen are at infinite distances, rendering wavefront curvature negligible.
8. What is the ratio of the width of the central maximum to that of the first secondary maximum in single-slit diffraction?
Explanation: Central maximum width is $\frac{2\lambda}{a}$, while any secondary maximum width is $\frac{\lambda}{a}$, giving a ratio of $2:1$.
9. If the total number of rulings on a grating is doubled while keeping the grating element constant, what happens to the resolving power in a given spectral order?
Explanation: Since resolving power $R = nN$, doubling the total number of rulings $N$ doubles the resolving power $R$.
10. Which of the following describes the fundamental distinction between interference and diffraction?
Explanation: Operationally, diffraction is often viewed as interference among a continuous infinity of coherent wavelets originating from different points of the same unobstructed portion of a wavefront.