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Unformatted text preview: Niu (qn269) Homework #10 antoniewicz (58855) 1 This printout should have 17 questions. Multiplechoice 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 secondorder 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 firstorder 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 thirdorder 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.
 Spring '09
 Antoniewicz
 Work, Light

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