j5bYo78eG00w_1210016969_jwk572

j5bYo78eG00w_1210016969_jwk572 - homework 18 KIM, JI Due:...

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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 x-direction....
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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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j5bYo78eG00w_1210016969_jwk572 - homework 18 KIM, JI Due:...

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