## 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:

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:

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The tangent of the angle is approximated by the sine
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Formula anchor
$$ \begin{align} \tan(\phi) ~\approx~ \sin(\phi) ~=~ \frac{x}{a} \end{align} $$
From the right-angled triangle where the opposite cathetus is the path difference \( \Delta s \), you obtain the information:

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The sine of the angle is equal to the path difference divided by the slit spacing
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Formula anchor
$$ \begin{align} \sin(\phi) ~=~ \frac{\Delta s}{g} \end{align} $$
Set 2

and 3

equal and you get:

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Fringe spacing divided by the screen distance is equal to the path difference divided by the slit spacing
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Formula anchor
$$ \begin{align} \frac{x}{a} ~=~ \frac{\Delta s}{g} \end{align} $$
Substitute condition for constructive interference 1

into 4

:

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

for the desired wavelength.

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:

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Solution: Calculate the wavelength of light for double-slit experiment.
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Formula anchor
$$ \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.