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# hw 12 - wilson(kw22588 HW12 Mackie(20224 This print-out...

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wilson (kw22588) – HW12 – Mackie – (20224) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The reflecting surfaces of two intersecting flat mirrors are at an angle of 56 , as shown in the figure. A light ray strikes the horizontal mirror at an angle of 50 with respect to the mirror’s surface. 56 50 φ Figure is not drawn to scale. Calculate the angle φ . Correct answer: 68 . 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 (56 ) = 68 . As a matter of interest, in the upper-half of the figure the angles (clockwise) in the triangles from left to right are 40 , 40 , and 100 ; 80 , 34 , and 66 ; 114 , 16 , and 50 ; 130 , 16 , and 34 ; and in the lower-half of the figure the angles (counter-clockwise) in the triangles from left to right are 16 , 16 , and 148 ; 32 , 34 , and 114 ; 66 , 40 , and 74 ; 106 , 40 , and 34 . 002 10.0 points An cylindrical opaque drinking glass has a diameter 5 . 4 cm and height h , as shown in the figure. An observer’s eye is placed as shown (the observer is just barely looking over the rim of the glass). When empty, the observer can just barely see the edge of the bottom of the glass. When filled (with a transparent liquid with an index of refraction of 1 . 26) to the brim, the observer can just barely see the center of the bottom of the glass.

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wilson (kw22588) – HW12 – Mackie – (20224) 2 θ i h 5 . 4 cm θ r e y e Calculate the angle θ r . Correct answer: 63 . 7321 degrees. Explanation: Looking at the figure below, R r R i θ i h r d θ r e y e After filling the glass with liquid, we know from Snell’s law that n liquid sin θ i = n air sin θ r . The radius r is one-half the diameter d , there- fore sin θ i r R i = r r 2 + h 2 .
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hw 12 - wilson(kw22588 HW12 Mackie(20224 This print-out...

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