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 x 10^8 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 29.0 ns. In other words, the part of the trip that is trhough air takes 29.0 ns.

a. If the light traveled in a striaght line from A to B (half of the distance through air and half through the lower medium), how long would it take?

Explanation / Answer

first of all we need to find the refraction index of medium 2

distance till green point is 29*3*10^8*10^-9=8.7

so horizontal distance of green point is 7.11

tan(theta1)=7.11/5

theta1=54.88

tan(theta2)=0.578

theta2=30

using snells law

n1*sin(theta1)=n2*sin(theta2)

so, n2=1.635

velocity in medium 2=1.834*10^8

time in medium 1=2.35*10^-8 sec

time in medium 2=3.85*10^-8

total time=62nsec

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