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Direct Energy, 2018a

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266 11.8 Problems<br />

a famous result known as Snell's law.<br />

Consider the gure below. Assume that a ray of light travels from<br />

(x 0 ,y 0 ) to (x 1 ,y 1 ) along the path which takes the shortest time. Material<br />

1 has relative permittivity ɛ r1 , so the light travels in that material<br />

at a constant speed √ c<br />

ɛr1<br />

. Material 2 has relative permittivity ɛ r2 ,so<br />

c<br />

the light travels in that material at a constant speed √<br />

ɛr2<br />

. As we<br />

derived in the Prob. 11.6, the light travels along a straight line in<br />

material 1, and it travels along a straight line in material 2. However,<br />

the lines have dierent slopes as shown in the gure. Assume that<br />

the junction of the two materials occurs at x =0.<br />

(a) Find an equation for the total time it takes the light to travel as<br />

a function of h, the vertical distance at which the path crosses<br />

the y axis. Note that you are nding a function here, F (h),<br />

not a functional. You can use the fact that you know the light<br />

follows a straight line inside each material to nd this function.<br />

(b) The path followed by the light takes the minimum time, so the<br />

derivative dF =0. Use this idea to nd an equation for the<br />

dh<br />

unknown vertical height h. Your answer can be written as a<br />

function of the known constants ɛ r1 , ɛ r2 , x 0 , y 0 , x 1 , y 1 ,and c.<br />

You do not need to solve for h here, but instead just evaluate<br />

the derivative and set it to zero.<br />

(c) Use your result in part b above to derive Snell's law :<br />

√<br />

ɛr1 sin θ 1 = √ ɛ r2 sin θ 2<br />

.<br />

Material 1<br />

ɛ r1<br />

Material2<br />

ɛ r2<br />

(x 1 ,y 1 )<br />

(x 0 ,y 0 )<br />

θ 1<br />

θ 2<br />

(0,h)<br />

x=0

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