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# solution_pdf41 - tovar (jdt436) – homework 41 – Turner...

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Unformatted text preview: tovar (jdt436) – homework 41 – Turner – (59070) 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 307 nm) is used to calibrate a diffraction grating. If the first-order maximum occurs at 19 . 5 ◦ , what is the line spacing? Correct answer: 9 . 19693 × 10 − 7 m. Explanation: We use the formula d sin θ = mλ for m = 1, and obtain d = λ sin θ = 307 nm sin(19 . 5 ◦ ) = 307 nm . 333807 = 9 . 19693 × 10 − 7 m . 002 (part 1 of 2) 10.0 points A diffraction grating is 3 . 74 cm long and con- tains 8270 lines per 3 . 96 cm interval. What is the resolving power of this grating in the third order? Correct answer: 23431 . 7. Explanation: Let : L = 3 . 74 cm , N = 8270 lines , and a = 3 . 96 cm . The diffraction grating’s line density is n = N a = (8270 lines) (3 . 96 cm) = 2088 . 38 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 = (2088 . 38 lines / cm) (3 . 74 cm) (3) = 23431 . 7 ....
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## This note was uploaded on 12/11/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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solution_pdf41 - tovar (jdt436) – homework 41 – Turner...

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