In Young’s double-slit experiment, increasing the distance to the screen increases fringe spacing. Increasing slit separation decreases it. If both distances change, divide the screen-distance multiplier by the slit-separation multiplier when the wavelength stays fixed.
The 2027 H2 Physics 9478 syllabus includes solving double-slit interference problems using the fringe-separation equation. This example compares two setups and checks the result by calculating both spacings.
- Keep track of wavelength.
- Identify which distance is in the numerator.
- Calculate both distance multipliers.
- Check the result using SI units.
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Name each quantity in the equation
For the usual small-angle setup, fringe spacing x = λD/a. Here λ is wavelength, D is the distance from the slits to the screen and a is the distance between the slits. Fringe spacing is the distance between neighbouring bright fringes, or between neighbouring dark fringes.
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Worked example | Move the screen and widen the slit separation
Use 600 nm light, a screen distance of 1.50 m and slit separation of 0.300 mm. The first spacing is (600 × 10⁻⁹)(1.50)/(0.300 × 10⁻³) = 3.00 mm.
Now move the screen to 2.00 m and use a slit separation of 0.600 mm. The screen distance is multiplied by 4/3 and the slit separation by 2. The new spacing is 3.00 × (4/3)/2 = 2.00 mm. Direct substitution gives the same result.
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Predict when the spacing would stay the same
If the screen distance and the slit separation were both doubled, their ratio would stay constant. With the same wavelength and the small-angle approximation still valid, the spacing would stay the same.
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Frequently asked questions
What if the wavelength also changes?
Multiply the result by the wavelength factor as well. A shorter wavelength reduces the spacing when both distances are fixed.
How can I read spacing over several fringes?
Measure between two bright fringes and divide by the number of gaps between them. From the first to the fifth bright fringe there are four gaps.