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Topic 14 of 15

Wave Optics: Interference

When two or more coherent light waves overlap, they interfere constructively or destructively depending on their path length difference. Young's double-slit experiment demonstrated the wave nature of light, and thin-film interference explains the colors seen in soap bubbles and oil slicks.

Key Concepts

Coherence

Two wave sources are coherent if they maintain a constant phase relationship. Interference patterns are stable only with coherent sources.

Path Length Difference

The difference

\Delta L

in the distances traveled by two waves from their sources to a meeting point. Determines whether interference is constructive or destructive.

Young's Double-Slit

Two slits separated by distance

d

produce bright fringes at angles where

d\sin\theta = m\lambda

(

m = 0, \pm1, \pm2, \ldots

) and dark fringes where

d\sin\theta = (m+\tfrac{1}{2})\lambda

Fringe Spacing

On a screen a distance

L

away, consecutive bright fringes are separated by

\Delta y = \lambda L/d

(for small angles).

Thin-Film Interference

Light reflecting from the top and bottom surfaces of a thin film interferes. A phase shift of

\pi

(half-wavelength) occurs at each reflection from a slower medium. Net phase shifts determine constructive or destructive conditions.

Phase Shift on Reflection

When light reflects from a medium with higher index of refraction, the reflected wave acquires a phase shift of

180°

(equivalent to a path difference of

\lambda/2

Key Equations

Double-Slit Bright Fringes

d\sin\theta = m\lambda\quad(m = 0,\,\pm1,\,\pm2,\ldots)

Condition for constructive interference; $d$ = slit separation, $\lambda$ = wavelength.

Double-Slit Dark Fringes

d\sin\theta = \left(m + \tfrac{1}{2}\right)\lambda

Condition for destructive interference.

Fringe Spacing

\Delta y = \dfrac{\lambda L}{d}

Distance between adjacent bright fringes on a screen at distance $L$.

Thin-Film Constructive (one phase shift)

2nt = m\lambda\quad(m = 1,2,3,\ldots)

For films with one reflection phase shift (e.g., soap film in air): constructive when thickness $t$ satisfies this.

Thin-Film Destructive (one phase shift)

2nt = \left(m + \tfrac{1}{2}\right)\lambda\quad(m = 0,1,2,\ldots)

Destructive interference condition with one reflection phase shift.

Worked Example

Young's Double-Slit Fringe Spacing

Problem

In a double-slit experiment, the slit separation is $d = 0.50\text{ mm}$ , the screen is $L = 2.0\text{ m}$ away, and the light has wavelength $\lambda = 589\text{ nm}$ . Find the fringe spacing $\Delta y$ .

Solution

Use the fringe spacing formula.

\Delta y = \frac{\lambda L}{d} = \frac{(589\times10^{-9})(2.0)}{0.50\times10^{-3}}

\Delta y = \frac{1.178\times10^{-6}}{5.0\times10^{-4}} = 2.356\times10^{-3}\text{ m} \approx 2.36\text{ mm}

Answer

\Delta y \approx 2.36\text{ mm}

Practice

Exercises

7 problems

1 of 7

In a double-slit experiment, $\lambda = 500\text{ nm}$ , $d = 0.30\text{ mm}$ , and $L = 2.25\text{ m}$ . What is the fringe spacing $\Delta y$ in mm?

Your answer

2 of 7

In a double-slit setup with $\lambda = 600\text{ nm}$ and $L = 1.5\text{ m}$ , the fringe spacing is measured to be $3.0\text{ mm}$ . What is the slit separation $d$ in mm?

Your answer

3 of 7

What is the angle (in degrees) to the $m=3$ bright fringe in a double-slit experiment with $d = 0.10\text{ mm}$ and $\lambda = 700\text{ nm}$ ?

Your answer

4 of 7

In a double-slit experiment, $\lambda = 550\text{ nm}$ , $d = 0.20\text{ mm}$ , $L = 2.0\text{ m}$ . What is the distance from the central bright fringe to the $m=2$ bright fringe (in mm)?

Your answer

5 of 7

A soap film ( $n = 1.35$ ) in air is illuminated with light of wavelength $\lambda = 540\text{ nm}$ . What is the minimum non-zero thickness (in nm) that gives constructive reflection? (One phase shift occurs at the outer surface.)

Your answer

6 of 7

For the same soap film ( $n=1.35$ , $\lambda=540\text{ nm}$ ), what minimum non-zero thickness gives destructive reflection? (One phase shift at outer surface.)

Your answer

7 of 7

Two waves of wavelength $\lambda = 500\text{ nm}$ arrive at a point with a path length difference of $\Delta L = 750\text{ nm}$ . What is the phase difference $\Delta\phi$ in degrees?

Your answer

Key Takeaways

In Young's double-slit, bright fringes occur when $d\sin\theta = m\lambda$ and fringe spacing on a screen is $\Delta y = \lambda L/d$ .
Thin-film interference depends on the optical path difference $2nt$ and the number of phase-reversing reflections; one phase shift flips the constructive/destructive conditions relative to zero shifts.
Interference requires coherent sources; real lasers and narrow spectral sources are needed to observe stable fringes.