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Unformatted text preview: Wong, Lemuel – Homework 1 – Due: Jan 26 2007, 11:00 pm – Inst: Matt Mackie 1 This printout should have 24 questions. Multiplechoice 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 bluegreen 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 ◦ . Wong, Lemuel – Homework 1 – Due: Jan 26 2007, 11:00 pm – Inst: Matt Mackie 2 As a matter of interest, in the upperhalf 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 lowerhalf of the figure the angles (counterclockwise) 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....
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This note was uploaded on 02/17/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Work

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