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Plane Light Wave of Frequency - Physics for Scientist and Engineers - Solved Past Paper, Exams of Engineering Physics

This is the Solved Past Paper of Physics for Scientist and Engineers which includes Physical Constants, Circular Loop, Magnetic Field, Energy Dissipated by Friction etc. Key important points are: Plane Light Wave of Frequency, Electric Field, Direction of Magnetic Field, Average Value, Energy Flux, Critical Angle of Reflection, Angular Range, Ray Diagram

Typology: Exams

2012/2013

Uploaded on 02/12/2013

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Download Plane Light Wave of Frequency - Physics for Scientist and Engineers - Solved Past Paper and more Exams Engineering Physics in PDF only on Docsity! 1 Solutions , , Last First Initial Name (please print) Student Number UNIVERSITY OF CALIFORNIA AT BERKELEY DEPARTMENT OF PHYSICS PHYSICS 7C FALL SEMESTER 2008 LEROY T. KERTH First Midterm Examination October 2, 2008 This examination is closed book. You may refer to a single 8 1 2 !11 sheet of paper (both sides) that you have prepared. Work all problems. 1 ______________ /20 2 ______________ /20 3 ______________ /20 4 ______________ /20 5 ______________ /20 Total ______________ /100 2 1. A plane light wave of frequency ∑ and wave number k (k = 2π/¬), propagating in the positive z direction. a. If the electric field is in the X̂ direction with amplitude E 0 what is the direction of the magnetic field? b. If the amplitude of the electric field is 30 volts/m what is the amplitude of the magnetic field? c. What is the average value of the energy flux striking a disk of radius 2 m? d. If the wave is reflected, what is the average force on the disk? a. B is in the ŷ direction. b. B = E c , B = 30 3!10 8 = 10 "7 Tesla . c. S = E ! B µ 0 = 1 2 " 0 µ 0 E 2 = 1 2 1 377 30 2 = 1.19W / m 2 Average power flux striking the disk = 4! " S = 15 Watts d. The average pressure is 2 S c . The total force is Pressure X area = 2 !15 3!10 8 = 10 "7 N 5 4. Two Physics grad students were talking and as usual they were trying to out do each other. Joe said that yesterday he had run across the Golden Gate Bridge. Well Jim wanted to brag about his new binoculars. He said “I was on the roof of LeConte Hall and saw you with my brand new 7X50 binoculars. I know it was you because I could see the old Cal shirt you always wear when you go jogging.” The 50 in 7 X 50 means the binoculars have a 50 mm objective lens. The Golden Gate Bridge is about 18km from LeConte hall. The “Cal” on the shirt is about 10 cm high. You may assume 500 nm for the wavelength of the light. Neglect the distortion from temperature variations in the atmosphere. a. In a few words, what is the physics that might limit the ability to have a clear enough image to recognize a person or the script? b. Could Jim have seen an adult person with his binoculars? c. Could he have read the Cal on the shirt? a. The circular aperture of the objective lens diffracts the incoming light through an angle !! = 1.22" D where: D is the diameter of the aperture. This produces a smearing of the image thus rendering details smaller than this impossible to see. b. For these binoculars !" = 1.22 # 500 #10 $9 50 #10 $3 = 1.22 #10 $5 rad Thus, the size of the smearing at the bridge is !! = !" #18km = 1.22 #10 $5 #18 #10 3 = 0.22m A person is of size of the order of 1 to 2 m. Thus, they should be detectable. c. The “Cal ” is smaller than the 22 cm above so would not be resolved. 6 5.A pinhole 0.5 mm diameter is used as a source for a double-slit interference experiment. A sodium lamp (¬ = 590 nm) is used. If the distance from the source to the double-slit is 0.5 m, what is the maximum slit spacing such that interference fringes are just observable? The diffraction spreads the light by an angle !" = 1.22# D as in problem 4. In this case, it must be spread by over a large enough area to send coherent light to each slit. !" = 1.22 # 590 #10 $9 .5 #10 $3 = 1.44 #10 $3 rad The spread at the double-slit screen is = 1.44 !10"3 ! 0.5m = 7.2 !10"4m , or 0.7mm The slits will need to be closer together that 1.4 mm
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