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Midterm 4

# Midterm 4 - Version 003/AAAAD midterm 04 Turner(59130 This...

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Version 003/AAAAD – midterm 04 – Turner – (59130) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider an electromagnetic wave pattern as shown in the figure below. E B The wave is 1. a standing wave and is stationary. 2. traveling left to right. correct 3. traveling right to left. Explanation: The vector E vector and vector B vector are not at the same point on the velocity axis. Pick an instant in time, where the E and B fields are at the same point on the velocity axis. z v x y E B For instance, let us choose the point where the vector E vector is along the x axis, as shown in the above figures. At this same instant, the vector B vector is along the negative y axis (at a point with a phase difference of 360 from the place on the veloc- ity ( z ) axis where the vector E vector is drawn). Then vector E × vector B is along the negative z axis. Therefore, the electromagnetic wave is traveling left to right. 002 10.0 points A thin tungsten filament of length 1 . 26 m radiates 95 . 2 W of power in the form of elec- tromagnetic waves. A perfectly absorbing surface in the form of a hollow cylinder of radius 5 . 09 cm and length 1 . 26 m is placed concentrically with the filament. Assume: The radiation is emitted in the radial direction, and neglect end effects. The speed of light is 2 . 99792 × 10 8 m / s. Calculate the radiation pressure acting on the cylinder. 1. 2.27698e-06 2. 6.22351e-07 3. 7.8804e-07 4. 7.42096e-07 5. 8.64836e-07 6. 3.20558e-07 7. 9.01525e-07 8. 1.77578e-07 9. 9.28679e-07 10. 1.45489e-06 Correct answer: 7 . 8804 × 10 7 N / m 2 . Explanation: Let : r = 5 . 09 cm = 0 . 0509 m , = 1 . 26 m , c = 2 . 99792 × 10 8 m / s , and P = 7 . 8804 × 10 7 N / m 2 . The intensity of the radiation reaching the walls of the cylinder is I = ( S ) = P 2 π r ℓ , so the radiation pressure on the walls is p = I c = P 2 π r ℓ c = 7 . 8804 × 10 7 N / m 2 2 π (0 . 0509 m) (1 . 26 m) (2 . 99792 × 10 8 m / s) = 7 . 8804 × 10 7 N / m 2 . 003 10.0 points A 101 mW laser beam is reflected back upon itself by a mirror.

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Version 003/AAAAD – midterm 04 – Turner – (59130) 2 Calculate the force on the mirror. The speed of light is 2 . 99792 × 10 8 m / s. 1. 5.00346e-07 2. 3.00208e-07 3. 1.26754e-07 4. 4.73661e-07 5. 6.00415e-08 6. 6.47114e-07 7. 4.20291e-07 8. 2.00138e-08 9. 6.73799e-07 10. 3.80263e-07 Correct answer: 6 . 73799 × 10 7 N. Explanation: Let : P = 101 mW = 0 . 101 W and c = 2 . 99792 × 10 8 m / s . The intensity of the laser is P A with P the power and A the area of the beam. The average momentum per unit area per unit time (force per unit area) transferred to the wall is Δ p Δ t A = F A = 2 I c = 2 P A c , so F = 2 P c = 2 (0 . 101 W) 2 . 99792 × 10 8 m / s = 6 . 73799 × 10 7 N . 004 10.0 points The smallness of the critical angle θ c for di- amond means that light is easily “trapped” within a diamond and eventually emerges from the many cut faces. This makes a dia- mond more brilliant than stones with smaller n and larger θ c . Traveling inside a diamond, a light ray is incident on the interface between diamond and air.
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