I51zsejc94Ku_1210016877_jwk572

# I51zsejc94Ku_1210016877_jwk572 - homework 13 – KIM, JI...

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Unformatted text preview: homework 13 – KIM, JI – Due: Apr 20 2008, 4:00 am 1 Question 1, chap 35, sect 1. part 1 of 2 10 points A thin film of cryolite ( n c = 1 . 38 ) is ap- plied 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 545 nm? Correct answer: 197 . 464 nm (tolerance ± 1 %). Explanation: For camera lens coating of cryolite ( n c = 1 . 38) 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 = 545 nm 2 × 1 . 38 = 197 . 464 nm . Question 2, chap 35, sect 1. part 2 of 2 10 points What minimum thickness gives high trans- mission at 1090 nm? Correct answer: 197 . 464 nm (tolerance ± 1 %). 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 = 1090 nm 4 × 1 . 38 = 197 . 464 nm . Question 3, chap 35, sect 1. part 1 of 1 10 points An air wedge is formed between two glass plates separated at one edge by a very fine wire as shown. When the wedge is illuminated from above by 528 nm light, 25 dark fringes are observed. The count here includes the dark fringe at the contact point. Calculate the radius of the wire. Correct answer: 3 . 168 μ m (tolerance ± 1 %). Explanation: Given : λ = 528 nm = 5 . 28 × 10 − 7 m and m = 25 . There is a phase reversal upon reflection from the lower surface of the air layer but none at the upper surface, so the condition for destructive interference is 2 t = mλ n = m parenleftbigg λ n air parenrightbigg = mλ , m = 0 , 1 , 2 , ··· so for 25 dark fringes, includ- ing the one where the plates meet, the thick- ness of the gap t = 2 r and r = t 2 = ( m- 1) λ 4 = (25- 1) × bracketleftbigg 5 . 28 × 10 − 7 m 4 bracketrightbigg · 10 6 μ m 1 m = 3 . 168 μ m . Question 4, chap 35, sect 2. part 1 of 1 10 points homework 13 – KIM, JI – Due: Apr 20 2008, 4:00 am 2 Light of wavelength 501 nm is used to cali- brate a Michelson interferometer. By use of a micrometer screw, the platform on which one mirror is mounted is moved 0 . 242 mm. How many fringe shifts are counted? A fringe shift represents the change in going from a minimum to the adjacent maximum or from a maximum to the adjacent minimum. Correct answer: 1932 . 14 (tolerance ± 1 %)....
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## This note was uploaded on 05/05/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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I51zsejc94Ku_1210016877_jwk572 - homework 13 – KIM, JI...

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