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Example from Physics, 17.1 Understanding Diffraction and Interference
Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm, and you find that the third bright line on a screen is formed at an angle of 10.95º relative to the incident beam. What is the wavelength of the light?
Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.
The third bright line is due to third-order constructive interference, which means that m = 3. You are given d = 0.0100 mm and = 10.95º. The wavelength can thus be found using the equation for constructive interference.
The equation is . Solving for the wavelength, , gives
Substituting known values yields
How did it go?