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ycow [4]
3 years ago
12

What is the reflectivity of a glass surface (n =1.5) in air (n = 1) at an 45° for (a) S-polarized light and (b) P-polarized ligh

t?
Physics
1 answer:
Goshia [24]3 years ago
3 0

Answer:

a) R_s = 0.092

b) R_p = 0.085

Explanation:

given,

n =1.5 for glass surface

n = 1 for air

incidence angle = 45°

using Fresnel equation of reflectivity of S and P polarized light

R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2

using snell's law to calculate θ t

sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}

cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}

a) R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2

R_s = 0.092

b) R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2

R_p = 0.085

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