GPII - Homework 9, Refraction - solutions

GPII - Homework 9, Refraction - solutions - honea(cth632...

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honea (cth632) – H9Refraction – avram – (1956) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices before answering. 001 10.0 points A shallow pool of a liquid 7 . 4 m deep is not the depth one expects when viewed from over- head. The index of refraction of the liquid is 1 . 35. How deep does it appear to be? Correct answer: 548 . 148 cm. Explanation: Basic Concepts n 1 s 1 + n 2 s 2 = n 2 - n 1 R Solution: Consider a point on the bottom of the pool as our object. In the refraction surface formula n 1 s 1 + n 2 s 2 = n 2 - n 1 R . Because the water surface of a pool is a plane, R = , thus n 2 - n 1 R = 0. The object s 1 is under the water, n 1 = 1 . 35 is the index of refraction of water, and n 2 = 1 is the index of refraction of air. Hence, s 2 = - s 1 n 2 n 1 = - (7 . 4 m)(1) (1 . 35) = - 5 . 48148 m . The image is virtual and on the same side of the interface as the object at a distance d apparent = |- 5 . 48148 m | = 548 . 148 cm. Note: One may also determine the apparent depth in an alternative approach. ±irst sketch a ray diagram for a Fnite incident angle where the refacted ray hits the bottom of the pool at P. Draw a vertical line OP, where O is at the surface of the water, OP is the real depth. The apparent depth is deFned by the depth OP’, where P’ is the intersection between OP and the extrapolated line of the incident ray. We leave it an exercise for the reader to show that OP /OP = n 1 /n 2 . 002 10.0 points Batman and Robin are attempting to escape that dastardly villain, the Joker, by hiding in a large pool of water (refractive index n water = 1 . 333). The Joker stands gloating at the edge of the pool. (His makeup is water- soluble.) He holds a powerful laser weapon y 1 = 1 . 33 m above the surface of the water and Fres at an angle of θ 1 = 29 . 9 to the hor- izontal. He hits the Boy Wonder squarely on the letter “R”, which is located y 2 = 3 . 92 m below the surface of the water. θ x y y 1 1 2 R J Batplastic surface Mirrored Surface water B How far (horizontal distance) is Robin from the edge of the pool? (±ear not, Batfans. The “R” is made of laser-re²ective material.) Correct answer: 5 . 66886 m. Explanation: Basic Concepts: Snell’s law, Total inter- nal re²ection. Solution: Let x 1 be the horizontal distance from the laser to where the laser beam strikes the water and x 2 the horizontal distance from that point to Robin (see the following Fgure). 1 x y 1 x y 2 2 air water J R θ 90-θ ο 1 1 θ 2
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honea (cth632) – H9Refraction – avram – (1956) 2 Then we have x 1 = y 1 tan θ 1 = (1 . 33 m) tan(29 . 9 ) = 2 . 31294 m . From Snell’s law, n air sin(90 - θ 1 ) = n water sin θ 2 . So sin θ 2 = sin(90 - 29 . 9 ) (1 . 333) = 0 . 650335 θ 2 = arcsin(0 . 650335) = 40 . 5669 . Hence, x 2 = y 2 tan θ 2 = (3 . 92 m) tan 40 . 5669 = 3 . 35592 m .
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GPII - Homework 9, Refraction - solutions - honea(cth632...

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