This picture shows a light source at point A, at the top left of the picture. Th

Question

This picture shows a light source at point A, at the top left of the picture. The light which refracts into the lower medium eventually reaches point B, at the bottom right of the picture. Point A is located at x = 0 m, y = +5 m, and point B is located at x = +10 m, y = ?5 m. The upper medium is air, with an index of refraction of n = 1.00. The interface separating the two media lies along the line y = 0. Take the speed of light in vacuum to be 3.00 108 m/s. An interesting fact about light is that the path it takes from A to B always minimizes the total light travel time. In other words, any other path from A to B, including the shortest-distance path, would take the light more time. One can show that this is equivalent to Snell's law. In the case shown in the picture, the time it takes the light to travel from point A to the green point, on the interface, is 30.0 ns. By using Snell's law and considering the geometry, answer the following.

Explanation / Answer

the distance from A to green point is d =speed of light *time taken = 3*10^8*30*10^-9 = 9 m

then sin(90-i) = 5/9

90-i = sin^(-1)(5/9) = 33.74 degrees

tan(90-i) = 5/x

x = 5/tan(90-i) = 5/tan(33.74) = 7.5 m

B) tan(r) = (10-7.5)/5

r = tan^(-1)(2.5/5) = 26.5 degrees

sin(r) = sin(26.5) = 0.446

n1*sin(i) = n2*sin(r)

1*sin(90-33.74) = n2*0.446

n2 = 1.86

C) time taken from green point to B is t2 = sqrt[(2.5^2+5^2)/(c/n2)]

c/n2 = 3*10^8/1.8 = 1.66*10^8 m/s

t2 = sqrt((2.5^2+5^2)/(1.66*10^8)) = 4.33*10^-4 sec

total time taken from A to B is 30*10^-9 + 4.33*10^-4 = 4.33*10^-4 sec

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