This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Homework 1 Solutions Physics 213 Fall 2003 Note: These solutions are presented symbolically. Use your numbers given to determine your answer. 1. In a Young’s double-slit experiment, the angle that locates the second dark fringe on either side of the central bright fringe is θ degrees. Find the ratio of the slit separation d to the wavelength λ of the light. For the double slit experiment, the bright spots are found at d sin θ = mλ ( m = 0 , ± 1 , ± 2 , . . . ) , but we want the dark spots, which are at d sin θ = m + 1 2 λ. The first dark fringe is at m = 0, and the second dark fringe is at m = 1. Thus, we have d sin θ = 3 2 λ. We want the ratio d/λ , and we find that d λ = 3 2 sin θ . 2. In a Young’s double-slit experiment the separation y between the first-order bright fringe and the central bright fringe on a flat screen is y o m, when the light has a wavelength of λ o nm. Assume that the angles that locate the fringes on the screen are small enough so that sin θ ≈ tan θ . Find the separation y when the light has a wavelength of λ nm. Again, the bright fringe for a double slit experiment is d sin θ = mλ, where m = 1 for the first bright fringe. We are not, however, given the angle θ for either wavelength, but rather the separation y o . That depends on the distance to the screen, which we aren’t given. For now, let’s just call it L . If we knew L we could find θ by tan θ = y L . One thing we do know is that the angle is small, so we can say sin θ ≈ tan θ = y L . This means, for the first fringe, d sin θ = d y o L = λ o ....
View Full Document
- Fall '09
- Work, Wavelength, Thomas Young, bright fringe, first-order bright fringe