HW #13-solutions - Brannon (blb675) – HW #13 – Shih –...

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Unformatted text preview: Brannon (blb675) – HW #13 – Shih – (58250) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A thin film of cryolite ( n c = 1 . 34 ) is applied to a camera lens ( n g = 1 . 53 ). The coating is designed to reflect wavelengths at the blue end of the spectrum and transmit wavelengths in the near infrared. What minimum thickness gives high reflec- tivity at 462 nm? Correct answer: 172 . 388 nm. Explanation: For camera lens coating of cryolite ( n c = 1 . 34) over glass ( n g = 1 . 53), high reflectivity is achieved for 2 n c t 1 = mλ 1 . Here we have taken into account that high reflectivity is achieved for constructive inter- ference. The phase changes at both the “air- cryolite” and the “cryolite-glass” surfaces is φ = 180 ◦ ( n air = 1 < n c < n g ). Note: Two phase changes of 180 ◦ . For minimum thickness m = 1 t 1 = λ 1 2 n c = 462 nm 2 × 1 . 34 = 172 . 388 nm . 002 (part 2 of 2) 10.0 points What minimum thickness gives high trans- mission at 924 nm? Correct answer: 172 . 388 nm. Explanation: Under the same conditions low reflectivity is achieved for 2 n c t = m + 1 2 λ 2 . For minimum thickness m = 0 t 2 = λ 2 4 n c = 924 nm 4 × 1 . 34 = 172 . 388 nm . 003 (part 1 of 2) 10.0 points A pair of narrow, parallel slits separated by . 177 mm are illuminated by green light of wavelength 543 nm. An interference pattern is observed on a screen 1 . 62 m away from the plane of the slits. Calculate the distance from the central maximum to the first bright region on either side of the central maximum. Correct answer: 4 . 96983 mm. Explanation: For the bright band y bright = λ L d , where m = 1 for the first bright region. Thus y = λ L d = (5 . 43 × 10- 7 m) (1 . 62 m) (0 . 000177 m) = 0 . 00496983 m = 4 . 96983 mm . 004 (part 2 of 2) 10.0 points Calculate the distance between the first and second dark bands. Correct answer: 4 . 96983 mm. Explanation: For the dark band y dark = λ L m + 1 2 d . For the first and second dark bands m = 0 and m = 1, respectively, so Δ y = λ L d 1 + 1 2- 0 + 1 2 = λ L d Brannon (blb675) – HW #13 – Shih – (58250) 2 = (5 . 43 × 10- 7 m) (1 . 62 m) (0 . 000177 m) = 0 . 00496983 m = 4 . 96983 mm . This is the same result as for Part 1. 005 10.0 points Find the radius of a star image formed on the retina of the eye if the aperture diameter (the pupil) at night is 0 . 567 cm, and the length of the eye is 3 . 32 cm. Assume the wavelength of starlight in the eye is 451 nm....
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This note was uploaded on 05/03/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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HW #13-solutions - Brannon (blb675) – HW #13 – Shih –...

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