Lemuel physics 2 1 - Wong Lemuel Homework 1 Due 11:00 pm...

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Wong, Lemuel – Homework 1 – Due: Jan 26 2007, 11:00 pm – Inst: Matt Mackie 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Note: HW is due 11pm central time, which is midnight out time. 001 (part 1 of 2) 10 points A certain blue-green light has a wavelength of 580 nm in air. What is its wavelength in water, where light travels at 75 % of its speed in air? Correct answer: 435 nm. Explanation: Frequency doesn’t change when light trav- els from air to water, but the speed and the wavelength do. wavelength = 0 . 75 (580 nm) = 435 nm . 002 (part 2 of 2) 10 points What is its wavelength in Plexiglas, where light travels at 67 % of its speed in air? Correct answer: 388 . 6 nm. Explanation: wavelength = 0 . 67 (580 nm) = 388 . 6 nm . keywords: 003 (part 1 of 1) 10 points The reflecting surfaces of two intersecting flat mirrors are at an angle of 59 , as shown in the figure. A light ray strikes the horizontal mirror at an angle of 51 with respect to the mirror’s surface. 59 51 φ Figure is not drawn to scale. Calculate the angle φ . Correct answer: 62 . Explanation: Basic Concept: θ incident = θ reflected Solution: θ 1 θ 2 φ θ Figure is to scale. The sum of the angles in a triangle is 180 . In the triangle on the left we have angles θ , 180 - θ 1 2 , and 180 - θ 2 2 , so 180 = θ + 180 - θ 1 2 + 180 - θ 2 2 , or θ 1 + θ 2 = 2 θ . (1) In the triangle on the right we have angles θ 1 , θ 2 , and φ . 180 = θ 1 + θ 2 + φ , so θ 1 + θ 2 = 180 - φ . (2) Combining Eq. 1 and 2, we have φ = 180 - 2 θ = 180 - 2 (59 ) = 62 .
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Wong, Lemuel – Homework 1 – Due: Jan 26 2007, 11:00 pm – Inst: Matt Mackie 2 As a matter of interest, in the upper-half of the figure the angles (clockwise) in the triangles from left to right are 39 , 39 , and 102 ; 78 , 31 , and 71 ; 109 , 20 , and 51 ; 129 , 20 , and 31 ; and in the lower-half of the figure the angles (counter-clockwise) in the triangles from left to right are 20 , 20 , and 140 ; 40 , 31 , and 109 ; 71 , 39 , and 70 ; 110 , 39 , and 31 . keywords: 004 (part 1 of 2) 10 points Consider the case in which light ray A is in- cident on mirror 1 , as shown in the figure. The reflected ray is incident on mirror 2 and subsequently reflected as ray B. Let the an- gle of incidence (with respect to the normal) on mirror 1 be θ A = 35 and the point of incidence be located 20 cm from the edge of contact between the two mirrors. 90 x A B mirror 1 mirror 2 θ A θ B What is the angle of the reflection of ray B (with respect to the normal) on mirror 2 ? Correct answer: 55 . Explanation: If the angle of the incident ray A is θ A , the angle of reflection must also be θ A . Since the mirrors are perpendicular to each other, angle θ B is equal to 90 - θ A θ B = 90 - θ A = 90 - 35 = 55 .
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