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Wave Optics & Wave Diffraction

Single Slit Diffraction

Analyze the mechanics of single-slit wave diffraction. Adjust micrometer-driven metal jaws, change wavelengths, overlay intensity curves, and use calibration brackets to measure central maximum widths.

Adjustable Single-Slit Laboratory ⓘ

Vary width and color parameters. Measure the central maximum width using calibration sliders.

Lab active

Single Slit Telemetry

a sinθ = mλ
Wavelength (λ)
600 nm
Slit Width (a)
0.10 mm
Screen Distance (D)
2.00 m
Minima Angle (θ)
0.34°
Central Max Width
24.00 mm

Principles of Single Slit Diffraction

Single slit diffraction is the wave interference pattern formed by the superposition of secondary Huygens wavelets emerging from different points across a single aperture. Under the small-angle approximation, the pattern is characterized by:

  • Broad Central Peak: The central maximum is formed at θ = 0, where wavelets from all portions of the slit arrive at the screen in phase. Its width is double the size of secondary maximum fringes.
  • Dark Fringe Condition: Minima occur when the path difference between wavelets from opposite edges of the slit satisfies a sinθ = mλ.

Single Slit Pattern

A single slit diffraction pattern consists of a wide central bright maximum flanked by alternating dark fringes and progressively narrower, dimmer secondary bright maxima.

Unlike double-slit interference, the fringes do not have uniform brightness, and the central maximum contains over 90% of the total light intensity.

Condition for Minima

The positions of dark fringes (minima) are given by the equation:

Diffraction Minima Condition

a × sinθ = m × λ

Here, a is the slit width, θ is the angle, λ is the wavelength, and m is the non-zero integer order (m = ±1, ±2, ±3, ...).

Linear Central Width

The linear width of the central maximum on a screen D meters away is defined by:

W = 2 × λ × D / a

Narrower slits (smaller a) or longer wavelengths (larger λ) make the central maximum significantly wider.

Step-by-Step Solved Problems

Study these step-by-step calculations to master single slit diffraction.

Example 1 Problem Statement

A single slit of width 0.10 mm is illuminated by monochromatic light of wavelength 589 nm. Find the angular spread of the central maximum.

View Mathematical Solution Steps
  1. Identify given values: Slit width a = 0.10 mm = 1 × 10-4 m, Wavelength λ = 589 nm = 5.89 × 10-7 m.
  2. Recall the angular half-width formula of central maximum: sinθ = λ / a.
  3. For small angles, sinθ ≈ θ (in radians). Therefore, half angular width θ ≈ λ / a = 5.89 × 10-7 m / 1 × 10-4 m = 5.89 × 10-3 rad.
  4. Calculate full angular spread (width): 2θ = 2 × 5.89 × 10-3 rad = 1.178 × 10-2 rad = 0.675°.

Final Derived Answer: Full angular spread 2θ ≈ 0.0118 rad (0.68°).

Example 2 Problem Statement

Monochromatic light of wavelength 632.8 nm falls on a slit of unknown width. If the first minimum is observed at 1.5 cm from the center of the screen placed at a distance of 1.2 m, find the slit width.

View Mathematical Solution Steps
  1. Identify given values: Wavelength λ = 632.8 nm = 6.328 × 10-7 m, Screen distance D = 1.2 m, Minimum position y1 = 1.5 cm = 0.015 m, Order m = 1.
  2. Recall the position of minima formula: ym = mλD / a.
  3. Rearrange for slit width a: a = mλD / ym.
  4. Substitute values: a = (1 × 6.328 × 10-7 m × 1.2 m) / 0.015 m.
  5. Calculate: a = (7.5936 × 10-7) / 0.015 = 5.062 × 10-5 m = 0.051 mm.

Final Derived Answer: Slit Width a ≈ 0.051 mm (or 51 μm).

Example 3 Problem Statement

A parallel beam of light of wavelength 500 nm falls on a single slit. If the distance from the screen is 2.5 m and the linear width of the central maximum is 2.5 cm, calculate the width of the slit.

View Mathematical Solution Steps
  1. Identify given values: Wavelength λ = 500 nm = 5 × 10-7 m, Screen distance D = 2.5 m, Linear width W = 2.5 cm = 0.025 m.
  2. Recall the linear width of central maximum formula: W = 2λD / a.
  3. Rearrange for slit width a: a = 2λD / W.
  4. Substitute values: a = (2 × 5 × 10-7 m × 2.5 m) / 0.025 m.
  5. Calculate: a = (2.5 × 10-6) / 0.025 = 1.0 × 10-4 m = 0.10 mm.

Final Derived Answer: Slit Width a = 0.10 mm.

Self-Check Questions

Question 1

Describe the diffraction pattern of a single slit and explain what the central maximum is.

Show Answer & Explanation

A single slit diffraction pattern consists of a broad, highly intense central bright band (the central maximum) centered on the screen, flanked on either side by alternating dark fringes (minima) and progressively narrower, much weaker secondary bright bands (secondary maxima).

Question 2

Why does the intensity of secondary maxima decrease rapidly in a single slit diffraction pattern?

Show Answer & Explanation

Secondary maxima occur because wavelets from different portions of the single slit only partially cancel each other out. As the diffraction angle increases, the slit is effectively divided into a larger number of odd segments (e.g. 3, 5, 7) where only the light from one unpaired segment interferes constructively while others cancel out, leading to rapidly decaying brightness.

Question 3

How does changing the slit width a affect the spacing and angular width of the diffraction pattern?

Show Answer & Explanation

Slit width "a" is inversely proportional to the angular spread (θ ≈ λ/a). Narrowing the slit (smaller a) spreads the diffraction bands wider across the screen. Widening the slit (larger a) collapses the pattern, concentrating the light into a narrow geometric line.

Question 4

Explain the role of the micrometer screw in a laboratory single slit assembly.

Show Answer & Explanation

The micrometer screw provides a mechanical gear mechanism that drives one of the metal slit jaws forward or backward against a spring. This allows the user to adjust the slit aperture width "a" with high precision (down to micrometers) and observe the corresponding change in diffraction width.

Question 5

State the equation for the first dark minimum in single slit diffraction.

Show Answer & Explanation

The condition for the first dark minimum (m = 1) is: a sinθ = λ, where "a" is the slit width, θ is the diffraction angle, and λ is the wavelength of light.

Question 6

If the wavelength of light is increased, what happens to the size of the central maximum?

Show Answer & Explanation

Since the width of the central maximum is directly proportional to wavelength (W = 2λD/a), increasing the wavelength (e.g., switching from green to red light) causes the central maximum to stretch and become wider on the screen.