Consider a beam of light, of frequency omega, traveling through the air and shin

Question

Consider a beam of light, of frequency omega, traveling through the air and shining on a material with index n, where the boundary is set at x = 0. As you know, when the light wave hits the material, some of the wave gets transmitted through, while some gets reflected back. To the left of the material, the total wave is the sum of the incident and reflected wave, while to the right it's only the transmitted wave, y_L(x, t) = A_I cos(k_1x - omega t) + A_R cos (k_1x + omega t) y_R(x, t) = A_T cos (k_2x - omega t), where the incident amplitude A_I is known. Recalling that the velocity of light through a material of index n is v = c/n, and taking n_air = 1, show that k_2/k_1 = n. The boundary conditions say that the waves and their first derivatives with respect to x must match at the boundary (where x = 0). With this, and your results from part (a), show that A_T = (2/1 + n) A_I A_R = (1 - n/1 + n)A_I.

Explanation / Answer

a) yl = Ai cos(k1x - wt) + Ar cos(k1x + wt)
yr = Atcos(k2 - wt)

Here k1 is the angular wavenumber of the wave to the left of the boundary
k2 is the the angular wavenumber of the wave to the right
now, k1 = 2*pi/lambda1
k2 = 2*pi/lambda2
k2/k1 = lambda1/lambda2
but lambda2 =lambda1*n [ where lambda1 is the wavelength of wave in air, annd n is refractive index of material and lambda2 is wavelength of wave in the material]

so
k2/k1 = 1/n

b) waves and first derivatives should match at the boundary ( from boundary conditions)
yl = yr at x = 0
yl' = yr' at x = 0
so,
Ai cos(wt) + Ar cos(wt) = At cos(wt)
Ai + Ar = At

also
yl' = yr' ( wrt x)
-Ai*k1*sin(-wt) - Ar k1 sin(wt) = -At k2 sin(-wt)
Ai*k1 - Ar k1 = At k2
but k1 = n k2
so, (Ai - Ar)*nk2 = At*k2
Ai - Ar = At/n

2Ai = At(n+1/n)

At = 2n*Ai/(n+1)
and Ar = (1-n)Ai/(1+n)

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