Problem with solution: Wavelength of The Monochromatic Light in a Double-Slit Experiment

Distance $ a $ between the double slit and the screen. Interference fringe distance $ x $ and the angle $ \theta $ enclosed by the right-angled triangle are important here.

Monochromatic light hits the double slit in parallel, whose two slits have the distance $ g ~=~ 0.8 \, \mathrm{mm} $ to each other. You observe interference fringes on the screen which are $ a ~=~ 5 \, \mathrm{m} $ away from the double slit. Thereby two bright fringes have the distance $ x ~=~ 3.5 \, \mathrm{mm} $ to each other.

What is the wavelength $ \lambda $ of light?

Solution tips

Use a sketch of the double-slit. It's immensely helpful! But also use your knowledge from the article on the double-slit experiment.

Exercise solution

Since the monochromatic light hits the double slit parallel, the light waves arrive in phase at both slits. Since the waves are in phase, they satisfy the condition for constructive interference, as follows:

$$ \begin{align} \Delta s ~=~ m \, \lambda \end{align} $$

Here, $ \Delta s $ is the path difference. We used the condition for constructive and not destructive interference because the problem statement provided the distance between two bright and not dark fringes.

Also, the distance from the slit to the screen, $ a ~=~ 3 \, \text{m} $, is much larger than the slit separation $ g ~=~ 0.3 \, \text{mm} $. This means that you can use the following approximation:

$$ \begin{align} \tan(\phi) ~\approx~ \sin(\phi) ~=~ \frac{x}{a} \end{align} $$

From the right-angled triangle where the opposite cathetus is the path difference $ \Delta s $, you obtain the information:

$$ \begin{align} \sin(\phi) ~=~ \frac{\Delta s}{g} \end{align} $$

Set 2 and 3 equal and you get:

$$ \begin{align} \frac{x}{a} ~=~ \frac{\Delta s}{g} \end{align} $$

Substitute condition for constructive interference 1 into 4:

$$ \begin{align} \frac{x}{a} ~=~ \frac{m \, \lambda}{g} \end{align} $$

Now you have derived a relationship that only contains quantities given in the problem statement. Now just rearrange 5 for the desired wavelength.

$$ \begin{align} \lambda ~=~ \frac{x \, g}{m \, a} \end{align} $$

Assuming that the bright fringes always have the same spacing and nothing further was mentioned in the problem, you can consider the distance from the 0th to the 1st maximum. Then, $ m ~=~ 1 $. Substituting the given values yields:

$$ \begin{align} \lambda ~&=~ \frac{ 3.5 \cdot 10^{-3}\text{m} ~\cdot~ 0.8 \cdot 10^{-3}\text{m} }{ 1 ~\cdot~ 5\text{m} } \\
~&=~ 5.6 \cdot 10^{-7} \, \text{m} \end{align} $$

This path difference is equivalent to a wavelength of $ 560 \text{nm} $, which is the typical wavelength of green light.