recall and use λ = ax / D for double-slit interference using light

Interference – Double‑slit Experiment

In the double‑slit experiment a monochromatic light source of wavelength λ illuminates two narrow, parallel slits separated by a distance \$a\$. The light from the slits interferes on a screen placed a distance \$D\$ away, producing a series of bright and dark fringes. The position of the \$m^{\text{th}}\$ bright fringe is given by

\$a \sin\theta = m\lambda\$

For small angles (\$\theta\$ is small, which is true when \$D \gg a\$) we can use the approximations \$\sin\theta \approx \tan\theta \approx \theta\$ and \$\tan\theta = x/D\$, where \$x\$ is the distance on the screen measured from the central maximum. Substituting gives the useful relation

\$\boxed{\lambda = \frac{a\,x}{D}}\$

Key \cdot ariables

Derivation of the Formula

1. Path difference between the two waves arriving at a point \$P\$ on the screen is \$a\sin\theta\$.
2. Constructive interference (bright fringe) occurs when the path difference equals an integer multiple of the wavelength: \$a\sin\theta = m\lambda\$.
3. For small angles, \$\sin\theta \approx \tan\theta = x/D\$.
4. Substituting \$\tan\theta\$ for \$\sin\theta\$ gives \$a(x/D) = m\lambda\$.
5. Re‑arranging for the first‑order bright fringe (\$m = 1\$) yields \$\lambda = a x / D\$.

Using the Formula in A‑Level Questions

When a question asks you to find the wavelength, the slit separation, the fringe spacing, or the screen distance, follow these steps:

• Identify which quantities are given and which is required.
• Check that the small‑angle approximation is valid (usually \$D \ge 5a\$).
• Write the appropriate form of the equation:

  • To find \$\lambda\$: \$\lambda = a x / D\$
  • To find \$a\$: \$a = \lambda D / x\$
  • To find \$x\$: \$x = \lambda D / a\$
  • To find \$D\$: \$D = a x / \lambda\$

• Substitute the numerical values, keeping consistent units.
• Perform the calculation and state the answer with the correct number of significant figures.

Example Problem

Question: A double‑slit apparatus uses slits separated by \$a = 0.30\ \text{mm}\$. Monochromatic light of unknown wavelength produces a first‑order bright fringe \$x = 2.5\ \text{cm}\$ from the central maximum on a screen \$D = 2.0\ \text{m}\$ away. Calculate the wavelength of the light.

Solution:

1. Convert all quantities to metres:

   • \$a = 0.30\ \text{mm} = 3.0\times10^{-4}\ \text{m}\$
   • \$x = 2.5\ \text{cm} = 2.5\times10^{-2}\ \text{m}\$
   • \$D = 2.0\ \text{m}\$ (already in metres)

2. Use the formula for the first‑order fringe (\$m=1\$):

   \$\lambda = \frac{a\,x}{D}\$

3. Insert the numbers:

   \$\lambda = \frac{(3.0\times10^{-4}\ \text{m})(2.5\times10^{-2}\ \text{m})}{2.0\ \text{m}}\$

4. Calculate:

   \$\lambda = \frac{7.5\times10^{-6}\ \text{m}^2}{2.0\ \text{m}} = 3.75\times10^{-6}\ \text{m} = 375\ \text{nm}\$

5. Answer: \$\lambda = 3.8\times10^{-7}\ \text{m}\$ (to two significant figures), i.e. \$380\ \text{nm}\$, which lies in the ultraviolet region.

Common Pitfalls

• Forgetting the small‑angle approximation: If \$D\$ is not much larger than \$a\$, use the exact relation \$a\sin\theta = m\lambda\$ and calculate \$\theta\$ from \$\tan\theta = x/D\$.
• Mixing units: Always convert mm and cm to metres before substituting.
• Using the wrong order \$m\$: The formula \$\lambda = a x / D\$ is derived for the first‑order bright fringe (\$m=1\$). For higher orders, replace \$x\$ with the measured distance of the \$m^{\text{th}}\$ fringe and divide by \$m\$: \$\lambda = a x/(m D)\$.
• Significant figures: Propagate the precision of the given data to the final answer.

Summary

The relationship \$\lambda = a x / D\$ provides a quick way to connect the wavelength of light with measurable quantities in a double‑slit experiment, provided the small‑angle approximation holds. Mastery of this formula enables you to solve a wide range of A‑Level physics questions involving interference patterns.