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GR9277 #35 |
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Problem
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\prob{35}
Light of wavelength 5200 A is incident normally on a transmission diffraction grating with 2000 lines per cm. The first-order diffraction maximum is at an angle, with respect to the incident beam, that is most nearly
- 3 degrees
- 6 degrees
- 9 degrees
- 12 degrees
- 15 degrees
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Optics }Diffraction Grating
Diffraction gratings have the same formula as 2-slit interference, except each slit is (obviously) much smaller. The condition for maximum is given by  , relating the width of the slit to the wavelength and angle and order m.
The width of each slit is given by the grating ^{-1}=0.5E-5 m) . Thus, plugging in the wavelength one has  .
Now, the approximations to get rid of the trig function. Since  , one can approximate  , where the angle is in radians. Now, convert the angle from radians to degrees.  , as in choice (B).
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Alternate Solutions |
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Comments |
chrisfizzix
2008-10-06 12:24:21 | In the double/N-slit diffraction equation, d is the slit spacing, not the slit width. This problem implicitly assumes that the slits are extremely narrow, so that even though there are 2000 of them in a cm they are much more narrow than they are far apart. Given this interpretation, d = slit spacing = (number of slits in unit length)) = (1 / 2000) cm. The solution is correct besides that. |  | hungrychemist
2007-09-22 18:58:38 | the diffraction formula, w *sin(angle) = m*wavelength gives minima not the maxima. (See pg 893 Halliday)
to approximate the maxima, one can set m = 1.5 (since the maximum occurs in between the two minima).(See pag 894 Halliday)
following yosun's calculation but with m = 1.5, I found the answer to be nearly 9 degrees.
What's going on?
hamood
2007-10-03 17:40:09 |
The minima is for a single slit diffraction pattern. In this question we have a diffraction grating, and the same formula would give the maxima.
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