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

Symbol	Quantity	Units
$\lambda$	Wavelength of the light	metre (m)
$a$	Separation between the two slits	metre (m)
$x$	Fringe displacement from the central maximum	metre (m)
$D$	Distance from the slits to the screen	metre (m)
$m$	Order of the bright fringe (integer)	—

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.