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.