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Wave Optics & Wave Interference

Young's Double Slit Experiment

Explore Thomas Young's classic 1801 experiment that proved the wave nature of light. Observe coherent wavefront splitting, path difference lines, and live fringe width calculations on a school laboratory optical bench.

Thomas Young's Wave Interference Lab ⓘ

Double-slit coherent source division. Measures fringe spacing directly against a millimeter scale.

YDSE Active

YDSE Telemetry

β = λD/d = 3.60 mm
Wavelength (λ)
600 nm
Slit Separation (d)
0.30 mm
Screen Distance (D)
2.00 m
Fringe Spacing (β)
4.00 mm
Path Difference (Δx)
1200 nm (m=2)

Postulates and Setup of YDSE

Thomas Young's double-slit experiment provides a direct demonstration of light interference. By passing light through two slits, he resolved the issue of creating coherent sources:

  • Wavefront Splitting: A single primary wavefront falls on S1 and S2. Because they originate from the same wave, S1 and S2 act as coherent secondary sources.
  • Superposition: The diffracted waves emerging from S1 and S2 spread out and overlap in the space ahead.
  • Interference Pattern: At points where the path difference is a whole number of wavelengths (\(m\lambda\)), they reinforce constructively to form bright bands. Where it is a half-wavelength (\((m+1/2)\lambda\)), they cancel destructively to form dark bands.

Coherence Generation

To produce stable interference fringes, the light waves must maintain a constant relative phase. Young achieved this by using:

• A single pinhole/slit first: Ensuring spatial coherence across the width of the beam.
• Monochromatic light: Providing a single constant wavelength, which avoids different colors overlapping and washing out the pattern.

Path Difference Conditions

The phase relationship at any point on the screen is dictated by the path difference \(\Delta x = S_2P - S_1P\):

Bright Fringes (Constructive)

d × sin(θ) = m × λ

Dark Fringes (Destructive)

d × sin(θ) = (m + 1/2) × λ

For small angles, \(\sin(\theta) \approx y/D\), which simplifies the fringe position on the screen to \(y = m\lambda D / d\).

Fringe Spacing (\u03b2)

Fringe width is the distance between two consecutive bright or dark bands. It is given by:

Fringe spacing

β = λ × D / d

• Wavelength (λ): Longer wavelengths (red) produce wider fringes than shorter ones (blue).
• Slit separation (d): Moving the slits closer together spreads the fringes wider.

Step-by-Step Solved Problems

Examine these step-by-step solutions to master calculations for YDSE fringe spacings, indices, and path differences.

Example 1 Problem Statement

In a Young's double-slit experiment, the screen is placed 2.0 m away from the double slit. The slits are separated by 0.30 mm and are illuminated by a light of wavelength 600 nm. Find the distance of the second bright fringe from the central maximum.

View YDSE Proof Steps
  1. Identify the given values: Screen distance D = 2.0 m, Slit separation d = 0.30 mm = 3 × 10-4 m, Wavelength λ = 600 nm = 6 × 10-7 m, Fringe order m = 2.
  2. Recall the formula for the position of the m-th bright fringe: ym = m × λD / d.
  3. Substitute values: y2 = 2 × (6 × 10-7 m × 2.0 m) / (3 × 10-4 m).
  4. Calculate the result: y2 = 2 × (1.2 × 10-6) / (3 × 10-4) = 2 × 0.004 m = 8.0 × 10-3 m = 8.0 mm.

Final Derived Answer: Distance of 2nd Bright Fringe y2 = 8.0 mm.

Example 2 Problem Statement

Light of wavelength 589 nm produces interference fringes of width 4.0 mm in a double-slit experiment. If the entire apparatus is immersed in water (refractive index 1.33), calculate the new fringe width.

View YDSE Proof Steps
  1. Identify given values: Initial fringe width β1 = 4.0 mm, Refractive index of water n = 1.33.
  2. Recall that when the setup is immersed in a liquid, the wavelength of light decreases to λ' = λ / n.
  3. Since β is directly proportional to wavelength λ, the new fringe width β' will decrease proportionally: β' = β / n.
  4. Substitute the values: β' = 4.0 mm / 1.33 ≈ 3.0 mm.

Final Derived Answer: New Fringe Width β' ≈ 3.0 mm.

Example 3 Problem Statement

In a Young's double-slit experiment, the distance of the 5th dark fringe from the central maximum is 1.35 cm when a light source of wavelength 600 nm is used. Find the slit separation if the screen is at a distance of 1.5 m.

View YDSE Proof Steps
  1. Identify given parameters: Fringe order for 5th dark fringe is m = 4 (since the dark fringes are at half-integers: m=0 is 1st, m=1 is 2nd... m=4 is 5th dark fringe), position y = 1.35 cm = 0.0135 m, wavelength λ = 600 nm = 6 × 10-7 m, screen distance D = 1.5 m.
  2. Recall the position formula for dark fringes: y = (m + 0.5) × λD / d.
  3. Rearrange the equation to solve for slit separation d: d = (m + 0.5) × λD / y.
  4. Substitute the values: d = (4.5) × (6 × 10-7 m × 1.5 m) / 0.0135 m.
  5. Calculate the result: d = 4.5 × (9 × 10-7) / 0.0135 = 4.05 × 10-6 / 0.0135 = 3 × 10-4 m = 0.30 mm.

Final Derived Answer: Slit Separation d = 0.30 mm.

Self-Check Questions

Question 1

Why is the central fringe in Young's double-slit experiment always bright?

Show Answer & Explanation

The central point on the screen lies exactly on the perpendicular bisector of the two coherent slit sources. Waves starting from S1 and S2 travel equal distances to reach this point, resulting in a path difference of zero (Δx = 0). Since a path difference of zero corresponds to constructive interference, waves arrive in the same phase and reinforce each other, forming a central bright band.

Question 2

How does YDSE demonstrate that light is a wave rather than a stream of particles?

Show Answer & Explanation

If light consisted of particles (as suggested by Newton's corpuscular theory), passing it through two slits would produce two distinct, sharp bright lines on the screen directly opposite the slits. However, YDSE produces a pattern of multiple alternating bright and dark lines (fringes) spreading far beyond the slits. This pattern is characteristic of wave diffraction and constructive/destructive superposition, showing that light behaves as a wave.

Question 3

Describe the effect on the fringe pattern if the monochromatic light source is replaced by a white light source in YDSE.

Show Answer & Explanation

If white light is used: (1) The central maximum remains white because all colors arrive with zero path difference and interfere constructively. (2) Fringes on either side will be colored, with violet appearing closest to the center and red furthest away, because fringe spacing is proportional to wavelength (β = λD/d). (3) After a few colored fringes, the different color bands overlap completely, resulting in a uniform white glow.

Question 4

What happens to the fringe pattern if one of the two slits in a double-slit setup is completely covered?

Show Answer & Explanation

If one slit is covered, the interference pattern disappears. Instead, a single-slit diffraction pattern is formed on the screen. This pattern consists of a very broad central bright band flanked by much fainter and narrower secondary maximum bands, showing no fine double-slit interference bands.

Question 5

How does the width of the slits influence the visibility of the interference fringes?

Show Answer & Explanation

If the slits are very narrow, they cause wide diffraction spreading, which ensures that the light fields from both slits overlap extensively on the screen, creating high-contrast, visible fringes. If the slits are too wide, the light propagates as geometric rays with minimal diffraction. They do not overlap sufficiently, causing the fringes to wash out into two bright regions.

Question 6

State the phase difference and path difference conditions required for the formation of a dark fringe on the screen.

Show Answer & Explanation

For a dark fringe (destructive interference) to form, the path difference between the waves must be an odd half-integer multiple of the wavelength: Δx = (m + 0.5)λ, where m = 0, ±1, ±2... This corresponds to a phase difference of an odd multiple of π (i.e. φ = (2m + 1)π).