# 09HW10 - Niu (qn269) Homework #10 antoniewicz (58855) 1...

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Unformatted text preview: Niu (qn269) Homework #10 antoniewicz (58855) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points An unpolarized light beam with intensity of I passes through 2 polarizers shown in the picture. Transmission axis Polarized lihgt Polarizer Analyzer Unpolarized light E E cos If = 30 ,what is the beam intensity after the second polarizer? 1. I 2 = 3 16 I 2. I 2 = 5 8 I 3. I 2 = 1 4 I 4. I 2 = 3 8 I correct 5. I 2 = 7 16 I 6. I 2 = 1 16 I 7. I 2 = 9 16 I 8. I 2 = 1 8 I 9. I 2 = 5 16 I 10. I 2 = 1 2 I Explanation: The beam intensity after the first polarizer is I 1 = I 2 . We use the formula for the intensity of the transmitted (polarized) light. Thus the beam intensity after the second polarizer is I = I 1 cos 2 = I 2 cos 2 (30 ) = 3 I 8 002 (part 1 of 3) 6.0 points White light is incident normally on a trans- mission grating and the spectrum is observed on a screen 4 m from the grating. In the second-order spectrum, the separation be- tween light of 520 nm wavelength and 590 nm wavelength is 11 . 9 cm. Determine the number of lines per centime- ter of the grating. Correct answer: 2125 cm 1 . Explanation: Let : y = 11 . 9 cm = 0 . 119 m , L = 4 m , 1 = 520 nm , and 2 = 590 nm . 1 2 L y 1 y 2 The grating equation is d sin = m, where m = 0 , 1 , 2 , ... Assuming 2 0 and m = 2, we have sin 2 tan 2 = y L . Substituting, d y L = m y = mL d , Niu (qn269) Homework #10 antoniewicz (58855) 2 so the separation between light of 1 wave- length and 2 wavelength is y = mL 2 d- mL 1 d = mL d ( 2- 1 ) So d = mL ( 2- 1 ) y The number of lines per centimeter of the grating is N = 1 d = y mL ( 2- 1 ) = . 119 m 2 (4 m) (590 nm- 520 nm) 10 9 nm 10 2 cm = 2125 cm 1 . 003 (part 2 of 3) 2.0 points What is the separation between these two wavelengths in the first-order spectrum? Correct answer: 5 . 95 cm. Explanation: Let : N = 2125 cm 1 = 2 . 125 10 5 m 1 . The separation of the wavelengths is y = mL ( 2- 1 ) d = mLn ( 2- 1 ) , so for m = 1 , y = Ln ( 2- 1 ) = (4 m) (2 . 125 10 5 m 1 ) (590 nm- 520 nm) 10 2 cm 10 9 nm = 5 . 95 cm . 004 (part 3 of 3) 2.0 points What is the separation between these two wavelengths in the third-order spectrum?...
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## This note was uploaded on 11/18/2009 for the course PHY 58855 taught by Professor Antoniewicz during the Spring '09 term at University of Texas at Austin.

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09HW10 - Niu (qn269) Homework #10 antoniewicz (58855) 1...

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