36-4 Diffraction by a Double Slit

36-4 Diffraction by a Double Slit#

Prompts

  • How does the double-slit pattern change when each slit has finite width \(a\) (not \(a \ll \lambda\))? What is the diffraction envelope?

  • Write the intensity formula for double-slit diffraction. Which factor comes from interference between the two slits? Which from diffraction by each slit?

  • The first diffraction minimum occurs at \(a\sin\theta = \lambda\). The double-slit maxima occur at \(d\sin\theta = m\lambda\). How many bright interference fringes fit within the central peak of the diffraction envelope?

  • If \(d/a = 5\), how many bright fringes are in the central peak? In the first side peak?

Lecture Notes#

Overview#

  • Double-slit with finite slit width: The pattern combines double-slit interference (Ch. 35) and single-slit diffraction (Ch. 36). The single-slit pattern acts as an envelope that modulates the intensity of the interference fringes.

  • Intensity: \(I = I_m (\sin\alpha/\alpha)^2 \cos^2\beta\), where the diffraction factor \((\sin\alpha/\alpha)^2\) and the interference factor \(\cos^2\beta\) multiply.

  • Diffraction minima eliminate some double-slit fringes. The ratio \(d/a\) determines how many interference maxima fit within each diffraction peak.

Combining interference and diffraction#

In Chapter 35 we assumed slits so narrow (\(a \ll \lambda\)) that the single-slit diffraction pattern spread over the entire screen—all interference fringes had roughly equal intensity.

For actual slits with width \(a\) comparable to \(\lambda\):

  • Double-slit interference still determines the positions of bright and dark fringes: \(d\sin\theta = m\lambda\) (bright), \(d\sin\theta = (m+\frac{1}{2})\lambda\) (dark).

  • Single-slit diffraction modulates the intensity: the broad central maximum and weaker side maxima of each slit’s diffraction pattern act as an envelope.

  • Where the diffraction pattern has a minimum, the intensity is zero—even at angles that would be interference maxima for infinitely narrow slits. Some double-slit fringes are missing.

Intensity formula#

For two identical slits of width \(a\) and center-to-center separation \(d\):

(305)#\[ I = I_m \left(\frac{\sin\alpha}{\alpha}\right)^2 \cos^2\beta \]

where

(306)#\[ \alpha = \frac{\pi a}{\lambda}\,\sin\theta, \qquad \beta = \frac{\pi d}{\lambda}\,\sin\theta \]

  • Diffraction factor \((\sin\alpha/\alpha)^2\): from single-slit diffraction (section 36-2). Zero when \(a\sin\theta = m\lambda\) (\(m = 1, 2, 3, \ldots\)).

  • Interference factor \(\cos^2\beta\): from two-slit interference (section 35-3). Maxima when \(d\sin\theta = m\lambda\); minima when \(d\sin\theta = (m+\frac{1}{2})\lambda\).

The intensity is the product of both factors. Both must be nonzero for a bright fringe.

Limiting cases

  • \(a \to 0\): \((\sin\alpha/\alpha) \to 1\); reduces to pure two-slit interference (Ch. 35).

  • \(d \to 0\): \(\cos^2\beta \to 1\); reduces to pure single-slit diffraction.

Counting fringes in the central peak#

The central peak of the diffraction envelope extends from the first diffraction minimum on one side to the first on the other: \(a\sin\theta = \pm\lambda\).

Double-slit bright fringes occur at \(d\sin\theta = m\lambda\) (\(m = 0, \pm 1, \pm 2, \ldots\)).

At the first diffraction minimum, \(a\sin\theta = \lambda\). The corresponding double-slit order is \(m = d\sin\theta/\lambda = d/a\). So fringes with \(|m| < d/a\) lie inside the central peak. The number of bright fringes in the central peak is

(307)#\[ 2\lfloor d/a \rfloor + 1 \quad \text{(or } 2d/a - 1 \text{ if } d/a \text{ is an integer)} \]

When \(d/a\) is an integer, the diffraction minimum falls exactly on a would-be interference maximum, so that fringe has zero intensity (“missing order”).

First side peak (between first and second diffraction minima): extends from \(a\sin\theta = \lambda\) to \(a\sin\theta = 2\lambda\). Double-slit orders from \(m = \lceil d/a \rceil\) to \(m = \lfloor 2d/a \rfloor - 1\) (on one side). Typically about \(d/a\) fringes per side peak.

Poll: Slit width from interference pattern

The figure shows the interference pattern from two slits separated by 0.26 mm. What is the width of each slit?

[FIGURE: Double-slit intensity pattern showing diffraction envelope and interference fringes; indicate which orders are missing—e.g., if \(m=10\) is the first missing order, then \(d/a = 10\), so \(a = d/10 = 0.026\) mm]

  • (A) 0.010 mm

  • (B) 0.013 mm

  • (C) 0.020 mm

  • (D) 0.026 mm

Example: \(d/a = 4.8\)

\(d = 19.44\) mm, \(a = 4.05\) mm \(\Rightarrow\) \(d/a \approx 4.8\). Central peak contains \(m = 0, \pm 1, \pm 2, \pm 3, \pm 4\): 9 bright fringes. The \(m = 5\) fringe falls near the first diffraction minimum and is nearly eliminated. First side peak: \(m = 5\) to \(9\) on one side (5 fringes).

Example: Bright fringes between diffraction minima

A double-slit has slit width \(a = 46\,\mu\text{m}\) and slit separation \(d = 0.30\) mm. How many complete bright fringes appear between the two first-order minima of the diffraction pattern?

Solution: First diffraction minima at \(a\sin\theta = \pm\lambda\). At that angle, \(d\sin\theta = d\lambda/a = 0.30\times10^{-3}/(46\times10^{-6}) \approx 6.5\). So interference orders \(m = 0, \pm1, \ldots, \pm6\) lie inside the central diffraction peak. Total: 13 bright fringes (or 11 if \(m=\pm6\) falls exactly on the minimum).

Summary#

  • Double-slit diffraction: interference (\(d\sin\theta\)) modulated by diffraction envelope (\(a\sin\theta\)).

  • Intensity: \(I = I_m (\sin\alpha/\alpha)^2 \cos^2\beta\); \(\alpha = (\pi a/\lambda)\sin\theta\), \(\beta = (\pi d/\lambda)\sin\theta\).

  • Central peak: contains \(\sim 2d/a\) bright fringes (or \(2\lfloor d/a \rfloor + 1\)). Diffraction minima eliminate fringes when \(d/a\) is integer.