18.1.3 Interference of Light Waves

Young's Double-Slit Experiment

The most famous demonstration of light interference is Young's Double-Slit Experiment (1801). In this experiment:

1. Monochromatic light from a single source $S$ falls on a barrier with two narrow slits $S_1$ and $S_2$ separated by a distance $d$.
2. The two slits act as coherent sources — they emit light of the same frequency and maintain a constant phase difference (both derived from the same source).
3. Light from $S_1$ and $S_2$ overlaps on a screen placed at distance $L$ from the slits.
4. An alternating pattern of bright fringes (constructive interference) and dark fringes (destructive interference) is observed on the screen.

This pattern — called an interference fringe pattern — is direct evidence of the wave nature of light.

Conditions for a Stable Interference Pattern

For a clear, stable interference pattern to be observed, the two sources must be:

In Young's experiment, both slits are illuminated by the same source, so they are automatically coherent.

Path Difference and Fringe Conditions

Consider a point $P$ on the screen. Let the distances from $S_1$ and $S_2$ to $P$ be $r_1$ and $r_2$ respectively.

Path difference: $\Delta =r_2-r_1$

Constructive Interference (Bright Fringe)

When the path difference is a whole number of wavelengths, the waves arrive in phase:

$\Delta =m\lambda (m=0,\pm 1,\pm 2,\dots )$

Destructive Interference (Dark Fringe)

When the path difference is a half-integer number of wavelengths, the waves arrive out of phase (antiphase):

$\Delta =\left(m+\frac{1}{2}\right)\lambda (m=0,\pm 1,\pm 2,\dots )$

Phase Difference Relationship

Path difference and phase difference $\varphi$ are related by:

$\varphi =\frac{2\pi }{\lambda }\Delta$

So a path difference of $\lambda /2$ corresponds to a phase difference of $\pi$ radians (destructive interference).

The Double-Slit Formula (Fringe Spacing)

For a screen at distance $L$ from the slits (with $L\gg d$), the positions of bright fringes are equally spaced. The fringe spacing (distance between adjacent bright or dark fringes) is:

$\boxed{\Delta y=\frac{\lambda L}{d}}$

where:

• $\Delta y$ = fringe spacing (m)
• $\lambda$ = wavelength of light (m)
• $L$ = distance from slits to screen (m)
• $d$ = slit separation (m)

This can also be written using the small-angle approximation from $d\sin \theta =m\lambda$:

$d\sin \theta =m\lambda$

where $\theta$ is the angle to the $m$-th order maximum.

Effect of Changing Variables

Worked Example

Problem: In a Young's double-slit experiment, the slit separation is $d=0.50\text{ mm}$, the screen is $L=1.2\text{ m}$ away, and the fringe spacing is measured to be $\Delta y=1.44\text{ mm}$. Find the wavelength of the light used.

Solution:

$\lambda =\frac{\Delta y\cdot d}{L}=\frac{(1.44\times 10^{-3})(0.50\times 10^{-3})}{1.2}$

$\lambda =\frac{7.2\times 10^{-7}}{1.2}=6.0\times 10^{-7}\text{ m}=600\text{ nm}$

This corresponds to orange-red light.

Summary

• Young's double-slit experiment demonstrates interference of light, proving its wave nature.
• Coherent, monochromatic sources of equal amplitude produce the clearest fringes.
• Bright fringes: $\Delta =m\lambda$; Dark fringes: $\Delta =(m+\frac{1}{2})\lambda$.
• Fringe spacing: $\Delta y=\frac{\lambda L}{d}$.