Young's Double Slit Experiment - Find WaveLength

Question:

A Young's interference experiment is performed with monochromatic light. The separation between the slits is 0.5 mm and the interference pattern on a screen 3.3 m away shows the first maximum 3.4 mm from the center of the pattern. What is the wavelength?

Answer:

The maxima will appear on the screen where the fields from the two slits arrive in phase with each other. This happens when the path length from the two slits differ by a multiple of whole wavelengths. Because the screen is much farther away, L, than the slits are apart, d, we can use some simplifications to show that the first maximum distance, y, is related to the wavelength, l, by y=l*L/d. So l*3.3/.0005 = .0034, or l=.0034*.0005/3.3 = 5.15e-7 meters.