Diffraction of Light: Single Slit, Circular Aperture & Resolving Power | HRK Physics

House of Physics October 05, 2025
Waves and Oscillations HRK: Diffraction - Complete Physics Guide
Mastering Single Slit Diffraction, Diffraction Patterns, Intensity Analysis, Circular Aperture Diffraction, and Resolving Power
Diffraction Single Slit Diffraction Wave Optics Huygens Principle Resolving Power Reading Time: 20 min

📋 Table of Contents

📜 Historical Background

The understanding of diffraction developed over centuries:

  • Francesco Grimaldi (1665): First documented observations of light diffraction
  • Christiaan Huygens (1678): Proposed the wave theory of light and Huygens principle
  • Thomas Young (1803): Demonstrated interference and provided evidence for wave theory
  • Augustin-Jean Fresnel (1818): Developed mathematical theory of diffraction
  • Lord Rayleigh (1879): Established the Rayleigh criterion for resolving power

These developments established diffraction as a fundamental wave phenomenon with practical applications in optics and imaging.

Introduction to Diffraction

🔬 What is Diffraction?

Diffraction is the bending of light waves around the edges of an obstacle or an opening and their spread into the geometrical shadow of the obstacle or opening.

This phenomenon is most prominent when the size of an opening or the edges of an obstacle is of the order of the wavelength of the waves. Since the wavelength of light is very small (about \(10^{-7}\) m), light waves are diffracted by very tiny particles, sharp edges, and very narrow openings.

📝 Key Characteristics of Diffraction

  • Diffraction occurs with all types of waves (light, sound, water waves)
  • The effect is most noticeable when the obstacle size is comparable to the wavelength
  • Diffraction patterns consist of alternating bright and dark regions
  • The central maximum is always the brightest and widest

Huygens Principle and Theoretical Explanation

🌊 Huygens Principle

According to Huygens principle:

"Every point on a wavefront can be considered as a secondary source of wavelets that spread out in the forward direction at the speed of the wave. The new wavefront is the envelope of these secondary wavelets."

⚙️ Diffraction Explanation Using Huygens Principle

Incident Wavefront
↓
SLIT
↓

Working Principle:

  1. When a wavefront of monochromatic light reaches an opening, part of it falls on the slit
  2. This part of the wavefront acts as secondary sources
  3. Wavelets originating from these sources reach into the shadow region where light doesn't directly reach
  4. This spreading into the shadow region is diffraction

In the slit or opening, there are infinite points that act as point sources, all phase coherent. Therefore, diffraction of light can be regarded as the superposition of infinite phase coherent waves.

Comparison of Interference and Diffraction

Aspect Interference Diffraction
Definition Superposition of two phase coherent waves Superposition of infinite number of phase coherent waves
Number of Sources Two coherent sources Infinite number of coherent sources from a single aperture
Fringe Pattern Large number of equally spaced, equally bright fringes Few fringes with central maximum brightest and widest
Fringe Width All fringes of equal width Fringes of unequal width
Intensity All maxima have equal intensity Intensity decreases rapidly with order

💡 Key Insight

Both interference and diffraction are wave phenomena resulting from the superposition of coherent waves. The main difference lies in the number of sources: interference involves a finite number of sources (typically two), while diffraction involves an infinite number of sources from a single aperture.

Single Slit Diffraction

🔍 Single Slit Diffraction Setup

Consider the diffraction pattern of plane waves of light of wavelength \(\lambda\) that are diffracted by a single long, narrow slit of width \(a\). When the diffracted light reaches a viewing screen, waves from different points within the slit undergo interference and produce a diffraction pattern of bright and dark fringes.

Central Maxima in Single Slit Diffraction

🧮 Formation of Central Maximum

Explanation

The central bright fringe appears because Huygens wavelets from all points in the slit travel approximately the same distance to reach the center of the pattern. Thus, they arrive in phase and interfere constructively, producing the central bright fringe.

At the center of the screen (\(\theta = 0\)), the path difference between waves from any two points in the slit is zero, resulting in constructive interference.

First and Second Minima in Single Slit Diffraction

🧮 First Minimum Condition

Step 1: Dividing the Slit

To find the first dark fringe (minimum), we divide the slit into two zones of equal width \(\frac{a}{2}\).

Step 2: Path Difference Analysis

Consider rays from the upper boundary of each zone reaching point \(P_1\) on the screen. The path difference between these rays is:

\[ \text{Path Difference} = \frac{a}{2} \sin \theta \]

Step 3: Condition for Destructive Interference

For destructive interference (dark fringe), the path difference must be \(\frac{\lambda}{2}\):

\[ \frac{a}{2} \sin \theta = \frac{\lambda}{2} \]
\[ \Rightarrow a \sin \theta = \lambda \]

🧮 Second Minimum Condition

Step 1: Dividing the Slit

To find the second dark fringe, we divide the slit into four zones of equal width \(\frac{a}{4}\).

Step 2: Path Difference Analysis

Consider rays from the upper boundary of each zone. The path difference between successive rays is:

\[ \text{Path Difference} = \frac{a}{4} \sin \theta \]

Step 3: Condition for Destructive Interference

For destructive interference, the path difference must be \(\frac{\lambda}{2}\):

\[ \frac{a}{4} \sin \theta = \frac{\lambda}{2} \]
\[ \Rightarrow a \sin \theta = 2\lambda \]

General Condition for Minima in Single Slit Diffraction

\[ a \sin \theta = m\lambda \quad \text{for } m = \pm 1, \pm 2, \pm 3, \ldots \]

Where:

  • \(a\) = width of the slit
  • \(\theta\) = angle from the central axis to the minimum
  • \(\lambda\) = wavelength of light
  • \(m\) = order of the minimum (1, 2, 3, ...)
Sample Problem 1: Single Slit Diffraction

A single slit of width 0.1 mm is illuminated by light of wavelength 600 nm. Calculate the angular width of the central maximum.

Given:
\[ a = 0.1 \, \text{mm} = 1 \times 10^{-4} \, \text{m} \]
\[ \lambda = 600 \, \text{nm} = 6 \times 10^{-7} \, \text{m} \]
Position of first minimum:
\[ a \sin \theta = \lambda \]
\[ \sin \theta = \frac{\lambda}{a} \]
\[ = \frac{6 \times 10^{-7}}{1 \times 10^{-4}} \]
\[ = 6 \times 10^{-3} \]
\[ \theta = \sin^{-1}(6 \times 10^{-3}) \]
\[ \approx 0.3439^\circ \]
Angular width of central maximum = \(2\theta\)
\[ = 2 \times 0.3439^\circ \]
\[ = 0.6878^\circ \]

Intensity in Single-Slit Diffraction

💡 Intensity Distribution

The intensity pattern in single-slit diffraction is not uniform. The central maximum is the brightest and widest, with intensity decreasing rapidly for higher-order maxima.

Qualitative Analysis of Intensity

📊 Intensity Characteristics

  • Central Maximum: Brightest and widest, intensity = \(I_0\)
  • First Secondary Maximum: About 4.5% of central maximum intensity
  • Second Secondary Maximum: About 1.6% of central maximum intensity
  • Third Secondary Maximum: About 0.8% of central maximum intensity

The intensity decreases rapidly with increasing order of the maximum.

Quantitative Analysis of Intensity

🧮 Intensity Formula Derivation

Step 1: Phase Difference

The phase difference \(\phi\) between waves from the top and bottom of the slit is:

\[ \phi = \frac{2\pi}{\lambda} a \sin \theta \]

Step 2: Resultant Amplitude

The resultant amplitude at angle \(\theta\) is:

\[ A = A_0 \frac{\sin(\beta)}{\beta} \]

where \(\beta = \frac{\phi}{2} = \frac{\pi a \sin \theta}{\lambda}\)

Step 3: Intensity Formula

Since intensity is proportional to the square of amplitude:

\[ I = I_0 \left[ \frac{\sin(\beta)}{\beta} \right]^2 \]

where \(I_0\) is the intensity at the central maximum (\(\theta = 0\)).

📈 Single Slit Diffraction Intensity Pattern

[Graph: Intensity vs. angle showing central maximum and secondary maxima]

The intensity pattern shows a bright central maximum flanked by weaker secondary maxima, with dark fringes (minima) at positions given by \(a \sin \theta = m\lambda\).

Diffraction at a Circular Aperture

🔵 Circular Aperture Diffraction

When light passes through a circular aperture (like a lens or pinhole), it produces a diffraction pattern consisting of a central bright disk (Airy disk) surrounded by concentric dark and bright rings.

🧮 First Minimum for Circular Aperture

Step 1: Condition for First Minimum

The angular position of the first minimum for a circular aperture of diameter \(D\) is given by:

\[ \sin \theta = 1.22 \frac{\lambda}{D} \]

Step 2: Small Angle Approximation

For small angles (\(\theta\) in radians):

\[ \theta \approx 1.22 \frac{\lambda}{D} \]

Step 3: Angular Radius of Airy Disk

The angular radius of the Airy disk (central bright spot) is:

\[ \theta = 1.22 \frac{\lambda}{D} \]
Sample Problem 2: Circular Aperture Diffraction

A telescope has an objective lens of diameter 10 cm. If light of wavelength 550 nm is used, calculate the angular radius of the Airy disk.

Given:
\[ D = 10 \, \text{cm} = 0.1 \, \text{m} \]
\[ \lambda = 550 \, \text{nm} = 5.5 \times 10^{-7} \, \text{m} \]
Angular radius of Airy disk:
\[ \theta = 1.22 \frac{\lambda}{D} \]
\[ = 1.22 \times \frac{5.5 \times 10^{-7}}{0.1} \]
\[ = 6.71 \times 10^{-6} \, \text{radians} \]
Convert to arcseconds:
\[ \theta = 6.71 \times 10^{-6} \times \frac{180}{\pi} \times 3600 \]
\[ \approx 1.38 \, \text{arcseconds} \]

Resolving Power and Rayleigh Criterion

🔭 Resolving Power

The resolving power of an optical instrument is its ability to produce separate images of two closely spaced objects.

📜 Rayleigh Criterion

According to the Rayleigh criterion, two point sources are just resolvable when the central maximum of the diffraction pattern of one source coincides with the first minimum of the diffraction pattern of the other source.

🧮 Minimum Resolvable Angle

Step 1: Angular Separation

For a circular aperture, the minimum angular separation \(\theta_R\) that can be resolved is:

\[ \theta_R = 1.22 \frac{\lambda}{D} \]

Step 2: Linear Separation

If the objects are at distance \(L\) from the aperture, the minimum resolvable linear separation is:

\[ d = L \theta_R = 1.22 \frac{\lambda L}{D} \]

🔭 Telescope Resolution

The resolving power of a telescope determines how close two stars can be while still appearing as separate points of light. Larger aperture telescopes have better resolution.

🔬 Microscope Resolution

In microscopes, resolution determines the smallest details that can be observed. The minimum resolvable distance is given by:

\[ d = \frac{1.22 \lambda}{2 \sin \alpha} \]

where \(\alpha\) is the half-angle of the cone of light entering the objective lens.

📡 Radar and Sonar

Diffraction limits the resolution of radar and sonar systems. Larger antennas provide better angular resolution for detecting and distinguishing between closely spaced targets.

Sample Problem 3: Resolving Power of Human Eye

Estimate the minimum separation between two points that can be resolved by the human eye. Assume pupil diameter = 3 mm, wavelength = 550 nm, and distance from eye to object = 25 cm.

Given:
\[ D = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m} \]
\[ \lambda = 550 \, \text{nm} = 5.5 \times 10^{-7} \, \text{m} \]
\[ L = 25 \, \text{cm} = 0.25 \, \text{m} \]
Minimum angular resolution:
\[ \theta_R = 1.22 \frac{\lambda}{D} \]
\[ = 1.22 \times \frac{5.5 \times 10^{-7}}{3 \times 10^{-3}} \]
\[ = 2.237 \times 10^{-4} \, \text{radians} \]
Minimum resolvable separation:
\[ d = L \theta_R \]
\[ = 0.25 \times 2.237 \times 10^{-4} \]
\[ = 5.59 \times 10^{-5} \, \text{m} \]
\[ \approx 0.056 \, \text{mm} \]

Double Slit Interference and Diffraction Combined

🔍 Combined Effect

In a double-slit experiment, the observed pattern is actually a combination of:

  1. Interference between waves from the two slits
  2. Diffraction from each individual slit

The overall intensity pattern is the product of the interference pattern and the diffraction envelope.

🧮 Combined Intensity Formula

Step 1: Interference Term

For two slits separated by distance \(d\), the interference term is:

\[ I_{interference} = \cos^2\left(\frac{\delta}{2}\right) \]

where \(\delta = \frac{2\pi}{\lambda} d \sin \theta\)

Step 2: Diffraction Term

For each slit of width \(a\), the diffraction term is:

\[ I_{diffraction} = \left[ \frac{\sin(\beta)}{\beta} \right]^2 \]

where \(\beta = \frac{\pi a \sin \theta}{\lambda}\)

Step 3: Combined Intensity

The overall intensity pattern is:

\[ I = I_0 \left[ \frac{\sin(\beta)}{\beta} \right]^2 \cos^2\left(\frac{\delta}{2}\right) \]

📈 Double Slit Pattern with Diffraction Envelope

[Graph: Combined pattern showing interference fringes modulated by diffraction envelope]

The diffraction envelope determines the overall intensity distribution, while the interference creates the fine fringe structure within this envelope. Some interference maxima may be missing if they coincide with diffraction minima.

💡 Missing Orders

In the combined double-slit pattern, certain interference maxima may be missing. This occurs when:

\[ \frac{d}{a} = \frac{m}{n} \]

where \(m\) is the order of interference maximum and \(n\) is the order of diffraction minimum. When this ratio is an integer, the \(m\)-th interference maximum coincides with the \(n\)-th diffraction minimum, resulting in a missing order.

Frequently Asked Questions

❓ Why is the central maximum in single-slit diffraction twice as wide as other maxima?

The central maximum spans from the first minimum on one side to the first minimum on the other side. Since the first minima occur at \(\theta = \pm \sin^{-1}(\lambda/a)\), the angular width of the central maximum is \(2\sin^{-1}(\lambda/a)\).

In contrast, the secondary maxima are bounded by two minima. For example, the first secondary maximum lies between the first minimum at \(\theta = \sin^{-1}(\lambda/a)\) and the second minimum at \(\theta = \sin^{-1}(2\lambda/a)\), giving it approximately half the angular width of the central maximum.

Additionally, the central maximum is much brighter because waves from all parts of the slit contribute constructively at the center, while for secondary maxima, destructive interference from parts of the slit reduces the overall intensity.

❓ How does diffraction limit the resolution of optical instruments?

Diffraction limits resolution because when light passes through an aperture (like a lens), it spreads out due to diffraction. This spreading causes point sources to appear as diffraction patterns (Airy patterns for circular apertures) rather than perfect points.

According to the Rayleigh criterion, two point sources are just resolvable when the central maximum of one diffraction pattern falls on the first minimum of the other. This gives the minimum resolvable angular separation as:

\[ \theta_R = 1.22 \frac{\lambda}{D} \]

where \(\lambda\) is the wavelength and \(D\) is the diameter of the aperture.

This limitation explains why:

  • Large telescopes have better resolution than small ones
  • Microscopes have limited resolution regardless of magnification
  • Radio telescopes need very large dishes to achieve good resolution
❓ What is the difference between Fraunhofer and Fresnel diffraction?

The distinction between Fraunhofer and Fresnel diffraction lies in the distance between the aperture and the observation screen:

Aspect Fraunhofer Diffraction Fresnel Diffraction
Distance Source and screen are effectively at infinite distance Source or screen at finite distance
Wavefront Incident wavefront is plane Incident wavefront is spherical or cylindrical
Mathematical Treatment Simpler, uses Fourier transforms More complex, uses Fresnel integrals
Experimental Setup Requires lenses or large distances Can be observed with small distances
Pattern Pattern size proportional to distance Pattern has more complex structure

In practice, most textbook examples and basic experiments deal with Fraunhofer diffraction because it's mathematically simpler and the patterns are easier to analyze. Fresnel diffraction becomes important when studying diffraction at small distances or with converging/diverging beams.

📚 Master Wave Optics

Understanding diffraction is fundamental to wave optics, with applications spanning from basic physics to advanced technologies like microscopy, astronomy, and optical communications. Continue your exploration of wave phenomena to unlock deeper insights into the nature of light and matter.

Read More: Physics HRK Notes on Interference

© House of Physics | HRK Physics: Waves and Oscillations - Diffraction

Based on Halliday, Resnick, and Krane's "Physics" with additional insights from university physics curriculum

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