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Wave Optics & Interference

Fringe Width

Master the physics of fringe spacing (\u03b2). Learn how wavelength (λ), slit separation (d), and screen distance (D) interact according to the formula β = λD/d, using interactive measurement scales and color bands.

Fringe Width Measurement Lab ⓘ

Vary wave parameters to stretch or compress the bright and dark bands, measuring β directly on the ruler.

Lab active

Fringe Spacing Telemetry

β = λD/d = 4.00 mm
Wavelength (λ)
600 nm
Slit Separation (d)
0.30 mm
Screen Distance (D)
2.00 m
Fringe Spacing (β)
4.00 mm
Fringe Position (y_m)
8.00 mm (m=2)

Postulates of Fringe Width Spacing

Fringe width represents the physical distance between any two adjacent bright bands or two adjacent dark bands in a stable interference pattern. It is derived under the small-angle approximation and is governed by:

  • Wavelength Dependence: Fringe width is directly proportional to the wavelength of light. Since red light has a longer wavelength than blue, red fringes spread further apart.
  • Geometric Scaling: Spacing is directly proportional to the distance from slits to screen (D) and inversely proportional to the separation between the double slits (d).

What is Fringe Width?

In YDSE, the bright and dark bands are equally spaced. The distance between the centers of consecutive bright (or dark) fringes is called the fringe width, denoted by \(\beta\).

Because all fringes are of equal width, the spacing remains constant across the observation zone, which allows for precise millimeter measurements.

Fringe Spacing Formula

The fringe width \(\beta\) is mathematically expressed by the following equation:

Fringe Spacing Formula

β = λ × D / d

Here, \(\lambda\) is the wavelength, \(D\) is the screen distance, and \(d\) is the slit separation.

Factors Affecting Spacing

We can control the size of the fringes in three main ways:

  • Wavelength (λ): Increasing wavelength stretches the fringes.
  • Screen Distance (D): Moving the screen away increases the fringe width but reduces light brightness.
  • Slit Separation (d): Bringing the slits closer together spreads the pattern.

Step-by-Step Solved Problems

Examine these step-by-step solutions to master the fringe width equation and spatial scaling.

Example 1 Problem Statement

A light of wavelength 500 nm is used in a double-slit setup where the slits are separated by 0.25 mm. The interference fringes are observed on a screen placed 1.2 m away. Calculate the fringe width.

View Fringe Width Proof Steps
  1. Identify given values: Wavelength λ = 500 nm = 5 × 10-7 m, Slit separation d = 0.25 mm = 2.5 × 10-4 m, Screen distance D = 1.2 m.
  2. Recall the fringe width formula: β = λD / d.
  3. Substitute the values: β = (5 × 10-7 m × 1.2 m) / (2.5 × 10-4 m).
  4. Calculate: β = (6 × 10-7) / (2.5 × 10-4) = 2.4 × 10-3 m = 2.4 mm.

Final Derived Answer: Fringe Width β = 2.4 mm.

Example 2 Problem Statement

In a Young's double-slit experiment, the fringe width is measured to be 1.8 mm. If the distance to the screen is doubled and the slit separation is halved, find the new fringe width.

View Fringe Width Proof Steps
  1. Identify initial fringe width: β1 = 1.8 mm.
  2. Recall the proportional relationship: β = λD/d. Therefore, β is directly proportional to D and inversely proportional to d.
  3. Express the new parameters in terms of the initial ones: D' = 2D, and d' = d / 2.
  4. Write the ratio equation: β' = λ(2D) / (d/2) = 4 × (λD/d) = 4 × β1.
  5. Substitute values: β' = 4 × 1.8 mm = 7.2 mm.

Final Derived Answer: New Fringe Width β' = 7.2 mm.

Example 3 Problem Statement

A double-slit experiment has a fringe width of 3.0 mm for light of wavelength 600 nm. If it is replaced with light of wavelength 450 nm, find the position of the 4th bright fringe from the center.

View Fringe Width Proof Steps
  1. Identify given values: Wavelength 1 λ1 = 600 nm, Wavelength 2 λ2 = 450 nm, Initial fringe width β1 = 3.0 mm, Target fringe order m = 4.
  2. Recall that β is proportional to wavelength (β2 = β1 × λ2 / λ1).
  3. Calculate new fringe width: β2 = 3.0 mm × (450 / 600) = 2.25 mm.
  4. Recall that the position of the m-th bright fringe is: ym = m × β.
  5. Substitute: y4 = 4 × 2.25 mm = 9.0 mm.

Final Derived Answer: Position of the 4th Bright Fringe y4 = 9.0 mm.

Self-Check Questions

Question 1

Define fringe width and state its mathematical relationship to wavelength, screen distance, and slit separation.

Show Answer & Explanation

Fringe width (β) is the center-to-center distance between two consecutive bright or dark fringes. It is mathematically defined by the formula: β = λD/d, where λ is the wavelength of light, D is the distance from the slits to the screen, and d is the distance between the two slits.

Question 2

Why is it necessary to keep the slit separation d extremely small to see fringes clearly?

Show Answer & Explanation

Fringe width β is inversely proportional to the slit separation d (β ∝ 1/d). If the slits are separated by a large distance (e.g. several centimeters), the fringe width becomes extremely small, causing the bright and dark bands to overlap closely and merge, rendering them indistinguishable to the naked eye. Narrow spacing is essential to stretch the pattern.

Question 3

What physical parameter of the light wave determines the color distribution of fringes in YDSE?

Show Answer & Explanation

The wavelength of light (λ). Since different colors of the spectrum have different wavelengths (e.g., Red ≈ 700 nm, Blue ≈ 450 nm), they produce different fringe widths. Red light produces wider fringes than blue light under identical conditions.

Question 4

How does moving the screen closer to the double-slit affect the width and brightness of the fringes?

Show Answer & Explanation

Moving the screen closer (smaller D) reduces the fringe width (β ∝ D), making the fringes more closely packed. However, because the light energy is concentrated over a smaller area, the overall brightness of the fringes on the screen increases.

Question 5

If YDSE is conducted in a vacuum and then in water (n = 1.33), does the fringe width increase or decrease? Explain.

Show Answer & Explanation

The fringe width decreases. When light enters water, its speed slows down, causing the wavelength to shorten to λ' = λ/1.33. Since β = λD/d, the fringe width decreases proportionally to β' = β/1.33.

Question 6

In YDSE, are the bright fringes wider than the dark fringes, or are they equal?

Show Answer & Explanation

Under the small-angle approximation, the bright and dark fringes are of equal width. The spacing between consecutive bright fringes is exactly equal to the spacing between consecutive dark fringes (β = λD/d).