HW 41 - benavides (jjb2356) homework 41 Turner (59130) 1...

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Unformatted text preview: benavides (jjb2356) homework 41 Turner (59130) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A helium-neon laser (wavelength 393 nm) is used to calibrate a diffraction grating. If the first-order maximum occurs at 30 . 6 , what is the line spacing? Correct answer: 7 . 72039 10 7 m. Explanation: We use the formula d sin = m for m = 1, and obtain d = sin = 393 nm sin(30 . 6 ) = 393 nm . 509042 = 7 . 72039 10 7 m . 002 (part 1 of 2) 10.0 points A diffraction grating is 6 . 66 cm long and con- tains 7150 lines per 2 . 39 cm interval. What is the resolving power of this grating in the third order? Correct answer: 59772 . 8. Explanation: Let : L = 6 . 66 cm , N = 7150 lines , and a = 2 . 39 cm . The diffraction gratings line density is n = N a = (7150 lines) (2 . 39 cm) = 2991 . 63 lines / cm . Applying the formula for the resolving power R of the grating, R = N m = n L m, where m is the order of the diffraction, N = n L is the number of the illuminated lines of the diffraction grating, and L is the length of the diffraction grating, we obtain that for the third order ( m = 3) the resolving power R 3 of the grating is R 3 = n L m = (2991 . 63 lines / cm) (6 . 66 cm) (3) = 59772 . 8 ....
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This note was uploaded on 03/26/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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HW 41 - benavides (jjb2356) homework 41 Turner (59130) 1...

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