capa13 - Solution Derivations for Capa #13 1) A super nova...

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Solution Derivations for Capa #13 1) A super nova releases 1 . 3 × 10 45 J of energy. It is 1540 ly from earth. If you were facing the star in question, and your face was a circle 7 cm in radius, how much energy would reach your face? U = Given r 1 = Given r 2 = Given This problem can be done in the same way as on the previous CAPA involving the Poynting vector. You are given the total energy. Calculate the intensity (energy per unit area) and find the proportion of that occupied by the detector. In other words, the amount of energy you feel on your face is the percentage of the total surface area which the energy is spread about. S = energy area = U 4 πr 2 1 The area calculated is the surface area of a sphere since the energy was released in all directions. Also note that the radius was given in light years and you need it to be in meters. To convert, use 1 m = 9 . 46 × 10 15 ly . U detected = S * A detector Since the assumption is being made that your face is a circle of radius r 2 , the area is then A detector = πr 2 2 . Thus, the amount of energy that reaches your face is U detected = U 4 πr 2 1 πr 2 2 = U 4 r 2 1 r 2 2 2) A point source of light is located 4 . 52 m below the surface of a large lake of clear toxic fluid (Lake Ontario, where n = 1 . 50). Find the area of the largest circle on the pool’s surface through which light coming directly from the source can emerge. d = Given n 1 = Given n 2 = Given (air) First, we must find the critical angle as this will be the point at which no addi- tional light will emerge. From Snell’s Law, n 1 sin θ c = n 2 sin θ 2 = n 2 sin 90 = n 2 θ c = sin - 1 ± n 2 n 1 ² = sin - 1 ± 1 n 1 ² 1
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Where the last step follows since the ray is emerging into air. Now, we can find the length of the triangle which will be the radius of the largest circle of which light can still emerge. Thus, φ = 90 - θ c tan φ = d r r = d tan φ = d tan (90 - θ c ) The area of a circle is A = πr 2 = π ± d tan (90 - θ c ) ! 2 3) A laser beam enters a 14 . 0 cm thick glass window at an angle of 46 (from the normal). The index of refraction of the glass is 1 . 50. At what angle from the normal does the beam travel through the glass? Use units of “deg.” w = Given (width) θ 1 = Given n 2 = Given n 1 = Given (air) Thus, using Snell’s Law, n 1 sin θ 1 = n 2 sin θ 2 θ 2 = sin - 1 ² n 1 n 2 sin θ 1 ³ = sin - 1 ² 1 n 2 sin θ 1 ³ 2
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4) How long does it take the beam to pass through the plate? This one is a bit trickier since the ray does not travel straight through the glass.
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This homework help was uploaded on 04/03/2008 for the course PHYS 1120 taught by Professor Rogers during the Fall '08 term at Colorado.

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capa13 - Solution Derivations for Capa #13 1) A super nova...

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