Homework 24 Solutions

Homework 24 Solutions - lim(kl9356 – Homework 24 –...

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Unformatted text preview: lim (kl9356) – Homework 24 – Weathers – (22202) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. THIS ASSIGNMENT IS FOR EXTRA CREDIT. 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 . 68 ). 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 488 nm? Correct answer: 182 . 09 nm. Explanation: For camera lens coating of cryolite ( n c = 1 . 34) over glass ( n g = 1 . 68), 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 = 488 nm 2 × 1 . 34 = 182 . 09 nm . 002 (part 2 of 2) 10.0 points What minimum thickness gives high trans- mission at 976 nm? Correct answer: 182 . 09 nm. Explanation: Under the same conditions low reflectivity is achieved for 2 n c t = parenleftbigg m + 1 2 parenrightbigg λ 2 . For minimum thickness m = 0 t 2 = λ 2 4 n c = 976 nm 4 × 1 . 34 = 182 . 09 nm . 003 10.0 points An oil film with index of refraction 1 . 33 is trapped between two pieces of glass with in- dex of refraction 1 . 43. No light is reflected by such a film when 577 nm light falls on it at normal incidence. What is the minimum thickness of the oil film that will satisfy these conditions? Correct answer: 216 . 917 nm. Explanation: The index of refraction of the glass does not matter. There will be a phase flip at one interface and not at the other interface. The path length is 2 t . The wave length in the oil is λ n . For destructive interference in thin film, 2 t = m λ n For minimum thickness, m = 1, so t min = λ 2 n = 577 nm 2 (1 . 33) = 216 . 917 nm 004 10.0 points A beam of light is diffracted by a single slit. The distance between the positions of zero intensity ( m = ± 1) is 3 . 93 mm. 1 . 965mm 2 . 116 m . 589mm S 1 S 2 θ viewing screen lim (kl9356) – Homework 24 – Weathers – (22202) 2 Estimate the wavelength of the laser light....
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This note was uploaded on 09/07/2010 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.

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Homework 24 Solutions - lim(kl9356 – Homework 24 –...

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