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Unformatted text preview: homework 18 KIM, JI Due: May 5 2008, 4:00 am 1 Question 1, chap 33, sect 3. part 1 of 1 10 points An unpolarized light beam with intensity of I passes through 2 polarizers shown in the picture. I 33 63 I If the first polarizer is rotated at an angle of 33 with respect to vertical and the sec ond at an angle 63 with respect to vertical, what is the beam intensity I after the second polarizer? 1. I = 7 16 I 2. I = 1 8 I 3. I = 1 4 I 4. I = 3 8 I correct 5. I = 5 8 I 6. I = 5 16 I 7. I = 1 2 I 8. I = 9 16 I 9. I = 3 16 I 10. I = 1 16 I Explanation: Let : 1 = 33 and 2 = 63 . The beam intensity after the first polarizer is I 1 = I 2 , since the initial beam is unpolarized. 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 1 cos 2 ( 2 1 ) = I 1 cos 2 (63  33 ) = I 2 cos 2 (30 ) = I 2 parenleftBigg 3 2 parenrightBigg 2 = 3 I 8 . Question 2, chap 33, sect 5. part 1 of 1 10 points At what distance from a 80 W electromag netic wave point source (like a light bulb) is the amplitude of the electric field 13 V / m? All quantities given here are in the SI units. Correct answer: 5 . 32939 m (tolerance 1 %). Explanation: Let : P = 80 W , E max = 13 V / m , and c = 376 . 991 . The wave intensity (power per unit area) is given by I = S av = E 2 max 2 ( o c ) = (13 V / m) 2 2 (376 . 991 ) = 0 . 224143 W / m 2 . The constant o c (impedance of free space) turns out to be 120 . homework 18 KIM, JI Due: May 5 2008, 4:00 am 2 You should verify that (V / m) 2 = W / m 2 . With this we can find the intensity, which multiplied by the area of a circle of radius R must gives us the sources power P , therefore 4 R 2 I = P = 80 W . Solving for the radius we get, R = radicalbigg P 4 I = radicalBigg 80 W 4 (0 . 224143 W / m 2 ) = 5 . 32939 m . Question 3, chap 33, sect 5. part 1 of 1 10 points A 30 kW beam of radiation is incident nor mally on a surface that reflects half of the radiation. The speed of light is 3 10 8 m / s. What is the force on this surface? Correct answer: 0 . 15 mN (tolerance 1 %). Explanation: Let : P = 30 kW = 30000 W and c = 3 10 8 m / s . The total force on the surface is the sum of the force due to the reflected radiation and the force due to the absorbed radiation, so F tot = F r + F a = 2 P 2 c + P 2 c = 3 P 2 c = 3 (30000 W) 2 (3 10 8 m / s) 10 6 mN N = . 15 mN . Question 4, chap 33, sect 5. part 1 of 1 10 points Consider a monochromatic electromagnetic plane wave propagating in the z direction. At a particular point in space, the magnitude of the electric field is pointing in the positive xdirection....
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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.
 Spring '08
 Turner
 Physics, Work, Light

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