33.2.4 Positions of the Fringes

It’s usually easier to measure distances rather than angles, so we can also specify point P by its position on a $y$ -axis with the origin directly across from the midpoint between the slits. You can see from Figure 33.5 that

y = L tan θ

(33.5)

Using the small-angle approximation once again, this time in the form $tan θ \approx θ$ , we can substitute $θ_{m}$ from Equation 33.4 for $tan θ_{m}$ in Equation 33.5 to find that the $m th$ bright fringe occurs at position

\begin{matrix} \begin{matrix} y_{m} = \frac{m λ L}{d} & m = 0, 1, 2, 3, \dots \end{matrix} & (positions of bright fringes) \end{matrix}

(33.6)

The interference pattern is symmetrical, so there is an $m th$ bright fringe at the same distance on both sides of the center. You can see this in Figure 33.4b. As we’ve noted, the $m = 1$ fringes occur at points on the screen where the light from one slit travels exactly one wavelength farther than the light from the other slit.

NOTE Equations 33.4 and 33.6 do not apply to the interference of sound waves from two loudspeakers. The approximations we’ve used (small angles, $L ≫ d$ ) are usually not valid for the much longer wavelengths of sound waves.

Equation 33.6 predicts that the interference pattern is a series of equally spaced bright lines on the screen, exactly as shown in Figure 33.4b. How do we know the fringes are equally spaced? The fringe spacing between the $m$ fringe and the $m + 1$ fringe is

Δ y = y_{m + 1} - y_{m} = \frac{(m + 1) λ L}{d} - \frac{m λ L}{d} = \frac{λ L}{d}

(33.7)

Because $Δ y$ is independent of $m$ , any two adjacent bright fringes have the same spacing.

The dark fringes in the image are bands of destructive interference. They occur at positions where the path-length difference of the waves is a half-integer number of wavelengths:

\begin{matrix} \begin{matrix} Δ r = (​ ​ m + \frac{1}{2} ​ ​) λ & m = 0, 1, 2, \dots \end{matrix} & (destructive interference) \end{matrix}

(33.8)

We can use Equation 33.2 for $Δ r$ and the small-angle approximation to find that the dark fringes are located at positions

\begin{matrix} y_{m}^{'} = (m + \frac{1}{2}) \frac{λ L}{d} & m = 0, 1, 2, \dots & (positions of dark fringes) \end{matrix}

(33.9)

We have used $y_{m}^{'}$ , with a prime, to distinguish the location of the $m th$ minimum from the $m th$ maximum at $y_{m}$ . You can see from Equation 33.9 that the dark fringes are located exactly halfway between the bright fringes.

As a quick example, suppose that light from a helium-neon laser $(λ = 633 nm)$ illuminates two slits spaced 0.40 mm apart and that a viewing screen is 2.0 m behind the slits. The $m = 2$ bright fringe is located at position

y_{2} = \frac{2 λ L}{d} = \frac{2 (633 \times 10^{- 9} m) (2.0 m)}{4.0 \times 10^{- 4} m} = 6.3 mm

Similarly, the $m = 2$ dark fringe is found at $y_{2}^{'} = (2 + \frac{1}{2}) λ L / d = 7.9 mm$ . Because the fringes are counted outward from the center, the $m = 2$ bright fringe occurs before the $m = 2$ dark fringe.

EXAMPLE 33.1 Measuring the wavelength of light

A double-slit interference pattern is observed on a screen 1.0 m behind two slits spaced 0.30 mm apart. Ten bright fringes span a distance of 1.7 cm. What is the wavelength of the light?

MODEL It is not always obvious which fringe is the central maximum. Slight imperfections in the slits can make the interference fringe pattern less than ideal. However, you do not need to identify the $m = 0$ fringe because you can make use of the fact that the fringe spacing $Δ y$ is uniform. Ten bright fringes have nine spaces between them (not ten—be careful!).

VISUALIZE The interference pattern looks like the image of Figure 33.4b.

SOLVE The fringe spacing is

Δ y = \frac{1.7 cm}{9} = 1.89 \times 10^{- 3} m

Using this fringe spacing in Equation 33.7, we find that the wavelength is

λ = \frac{d}{L} Δ y = 5.7 \times 10^{- 7} m = 570 nm

It is customary to express the wavelengths of visible light in nanometers. Be sure to do this as you solve problems.

REVIEW Young’s double-slit experiment not only demonstrated that light is a wave, it provided a means for measuring the wavelength. You learned in Chapter 16 that the wavelengths of visible light span the range 400–700 nm. These lengths are smaller than we can easily comprehend. A wavelength of 570 nm, which is in the middle of the visible spectrum, is only about 1% of the diameter of a human hair.