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

Double-slit Experiment - distance, angle, right triangle
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:

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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:

The tangent of the angle is approximated by the sine
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From the right-angled triangle where the opposite cathetus is the path difference \( \Delta s \), you obtain the information:

The sine of the angle is equal to the path difference divided by the slit spacing
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Set 2 and 3 equal and you get:

Fringe spacing divided by the screen distance is equal to the path difference divided by the slit spacing
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Substitute condition for constructive interference 1 into 4:

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

Rearranged double-slit formula for wavelength.
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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:

Solution: Calculate the wavelength of light for double-slit experiment.
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This path difference is equivalent to a wavelength of \( 560 \text{nm} \), which is the typical wavelength of green light.

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