HW25_solutions - Version 001 – HW 25/26 Reflection&...

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Unformatted text preview: Version 001 – HW 25/26 Reflection & Refraction C&J – sizemore – (21301jtsizemore) 1 This print-out should have 61 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Serway CP 22 33 001 (part 1 of 2) 10.0 points Light with a wavelength of λ = 589 nm travels from a certain material to air. Calculate the critical angle for zircon (index of refraction 1 . 923) when it is surrounded by air (index of refraction 1). Correct answer: 31 . 3336 ◦ . Explanation: Given : n i = 1 . 923 and n r = 1 . 00 . Applying Snell’s law, n i sin θ c = n r sin 90 ◦ = n r sin θ c = n r n i θ c = sin − 1 parenleftbigg n r n i parenrightbigg = sin − 1 parenleftbigg 1 1 . 923 parenrightbigg = 31 . 3336 ◦ . 002 (part 2 of 2) 10.0 points Calculate the critical angle for fluorite (index of refraction 1 . 434) when it is surrounded by air (index of refraction 1). Correct answer: 44 . 2148 ◦ . Explanation: Given : n i = 1 . 434 . sin θ c = n r n i θ c = sin − 1 parenleftbigg n r n i parenrightbigg = sin − 1 parenleftbigg 1 1 . 434 parenrightbigg = 44 . 2148 ◦ . Serway CP 22 21 003 (part 1 of 2) 10.0 points The light ray shown in the figure makes an angle of 13 . 9 ◦ with the normal line at the boundary of linseed oil and water. θ 1 13 . 9 ◦ θ 2 Air Linseed oil Water Find the angle θ 1 . Note that n = 1 . 48 for the linseed oil. Correct answer: 20 . 8264 ◦ . Explanation: Given : n i = 1 . 00 , n r = 1 . 48 , and θ r = 13 . 9 ◦ . Applying Snell’s Law, n i sin θ i = n r sin θ r sin θ i = n r sin θ r n i θ i = sin − 1 parenleftbigg n r sin θ r n i parenrightbigg = sin − 1 parenleftbigg 1 . 48 sin13 . 9 ◦ 1 parenrightbigg = 20 . 8264 ◦ . 004 (part 2 of 2) 10.0 points Find the angle θ 2 . Correct answer: 15 . 4692 ◦ . Explanation: Version 001 – HW 25/26 Reflection & Refraction C&J – sizemore – (21301jtsizemore) 2 Given : n i = 1 . 48 , n r = 1 . 333 , and θ i = 13 . 9 ◦ . Applying Snell’s Law again, sin θ r = n i sin θ i n r θ r = sin − 1 parenleftbigg n i sin θ i n r parenrightbigg = sin − 1 parenleftbigg 1 . 48 sin13 . 9 ◦ 1 . 333 parenrightbigg = 15 . 4692 ◦ . Serway CP 22 52 005 10.0 points A 4 . 62 m long pole stands vertically in a lake having a depth of 2 . 82 m. When the Sun is 68 . 4 ◦ above the horizon, find the length of the pole’s shadow on the bottom of the lake. The index of refraction for water is 1 . 33. Correct answer: 1 . 52494 m. Explanation: θ φ 2 s 2 pole s 1 l φ 1 d Given : ℓ = 4 . 62 m , d = 2 . 82 m , n = 1 . 33 , and θ = 68 . 4 ◦ . Light just passing the top of the pole has an angle of incidence of φ 1 = 90 ◦ − θ = 90 ◦ − 68 . 4 ◦ = 21 . 6 ◦ ....
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HW25_solutions - Version 001 – HW 25/26 Reflection&...

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