p34_057 - not be seen (provided internally reected light...

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57. (a) The diagram below shows a cross section, through the center of the cube and parallel to a face. L is the length of a cube edge and S labels the spot. A portion of a ray from the source to a cube face is also shown. Light leaving the source at a small angle θ is refracted at the face and leaves the cube; light leaving at a sufficiently large angle is totally reflected. The light that passes through thecubefaceformsac irc le ,therad ius r being associated with the critical angle for total internal reflection. If θ c is that angle, then sin θ c = 1 n where n is the index of refraction for the glass. As the diagram shows, the radius of the circle is given by r =( L/ 2) tan θ c .Now , tan θ c = sin θ c cos θ c = sin θ c p 1 sin 2 θ c = 1 /n p 1 (1 /n ) 2 = 1 n 2 1 and the radius of the circle is r = L 2 n 2 1 = 10 mm 2 p (1 . 5) 2 1
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Unformatted text preview: not be seen (provided internally reected light can be ignored). S . . . . . . . . . . . . . . . . . . ...... ............. . . . . . . . . . . . r L L 2 c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ............................... . . . . . . . . . . . ................... (b) There must be six opaque disks, one for each face. The total area covered by disks is 6 r 2 and the total surface area of the cube is 6 L 2 . The fraction of the surface area that must be covered by disks is f = 6 r 2 6 L 2 = r 2 L 2 = (4 . 47 mm) 2 (10 mm) 2 = 0 . 63 ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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