How Do You Calculate Fringe Spacing in a Double-Slit Experiment?

For IB Physics SL and HL, calculate the fringe spacing using $s = \frac{\lambda D}{d}$, where $\lambda$ is wavelength, $D$ is the slit-to-screen distance, and $d$ is slit separation.

The Reasoning

In a Young's double-slit experiment, light from two coherent sources overlaps and undergoes interference. Constructive interference produces bright fringes, while destructive interference produces dark fringes.

The fringe-spacing equation is:

$$s = \frac{\lambda D}{d}$$

Here, $s$ is the distance between the centres of two adjacent bright fringes or two adjacent dark fringes. The equation assumes that $D$ is much greater than $d$, allowing the small-angle approximation used in its derivation.

To see why the relationship works, consider a bright fringe of order $n$. Constructive interference requires $d\sin\theta=n\lambda$. For small angles, $\sin\theta\approx\tan\theta\approx y/D$, so the position of that fringe is . Subtracting the positions of consecutive fringes gives . This also shows why the spacing is uniform near the centre of the pattern.

For example, suppose $\lambda = 600\text{ nm}$, $D = 2.0\text{ m}$, and $d = 0.50\text{ mm}$. First convert all quantities to SI units:

$$\lambda = 600 \times 10^{-9}\text{ m}, \qquad d = 0.50 \times 10^{-3}\text{ m}$$

Then substitute:

$$s = \frac{(600 \times 10^{-9})(2.0)}{0.50 \times 10^{-3}} = 2.4 \times 10^{-3}\text{ m}$$

Therefore, the fringe spacing is $2.4\text{ mm}$.

In practical data collection, measuring one interval can produce a large percentage uncertainty. Measure from the centre of the first bright fringe to the centre of the eleventh bright fringe; this spans ten intervals. Divide the measured distance by ten to obtain $s$. If the question instead asks for slit separation, rearrange to $d=\lambda D/s$. Keep every distance in metres during calculation, then convert the final answer if appropriate.

A common misconception is to use the distance across several fringes directly as $s$. If ten fringe intervals cover $24\text{ mm}$, then $s = \frac{24}{10} = 2.4\text{ mm}$. Count the , not merely the visible bright lines.

Exam Technique

This is syllabus topic C.3 Wave phenomena. For a Calculate question, show the equation, SI-unit conversions, substitution, and final value. Remember that increasing slit separation makes the fringes closer together because $s \propto \frac{1}{d}$.