# HW 41 - terry(ect328 – homework 41 – Turner –(59130 1...

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Unformatted text preview: terry (ect328) – 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 548 nm) is used to calibrate a diffraction grating. If the first-order maximum occurs at 21 . 8 ◦ , what is the line spacing? Correct answer: 1 . 47563 × 10 − 6 m. Explanation: We use the formula d sin θ = mλ for m = 1, and obtain d = λ sin θ = 548 nm sin(21 . 8 ◦ ) = 548 nm . 371368 = 1 . 47563 × 10 − 6 m . 002 (part 1 of 2) 10.0 points A diffraction grating is 5 . 88 cm long and con- tains 6270 lines per 0 . 881 cm interval. What is the resolving power of this grating in the third order? Correct answer: 1 . 25542 × 10 5 . Explanation: Let : L = 5 . 88 cm , N = 6270 lines , and a = 0 . 881 cm . The diffraction grating’s line density is n = N a = (6270 lines) (0 . 881 cm) = 7116 . 91 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 = (7116 . 91 lines / cm) (5 . 88 cm) (3) = 1 . 25542 × 10 5 ....
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## This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas.

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HW 41 - terry(ect328 – homework 41 – Turner –(59130 1...

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