Question #120401
A tank containing methanol has walls 2.50 cm thick made of glass of refractive index 1.550. light from the outside air strikes the glass at a 41.3 degree angle with the normal to the glass. Find the angle the light makes with the normal in the methanol.(b) The tang is emptied and refilled with an unknown liquid. if light incident at the same angle as in part (a) enters the liquid in the tank at an angle of 20.2 degree from the normal, what is the refractive index of the unknown liquid?
1
Expert's answer
2020-06-08T19:33:16-0400

Let

ni=n_i= the index of refraction for the material with incident light,

nr=n _r= the index of refraction for the material with refracted light,

θi=\theta_i= the angle of incidence, and

θr=\theta_r= the angle of refraction.

Law of refraction (Shell's Law)


nisinθi=nrsinθrn_i\sin\theta_i=n_r\sin\theta_r

Given na=1,ng=1.550,nm=1.329n_a=1, n_g=1.550, n_m=1.329

a)

From air to glass


nasinθ1=ngsinθ2n_a\sin\theta_1=n_g\sin\theta_21sin41.3°=1.550sinθ21\cdot\sin41.3\degree=1.550\cdot\sin\theta_2θ2=arcsin(sin41.3°1.550)25.2°\theta_2=\arcsin({\sin41.3\degree \over 1.550})\approx25.2\degree

From glass to methanol


ngsinθ2=nmsinθ3n_g\sin\theta_2=n_m\sin\theta_31.550sin25.2°=1.329sinθ31.550\cdot\sin25.2\degree=1.329\cdot\sin\theta_3θ3=arcsin(1.550sin25.2°1.329)29.8°\theta_3=\arcsin({1.550\cdot \sin25.2\degree \over 1.329})\approx29.8\degree

b)

From glass to unknown liquid


ngsinθ2=nusinθu=>nu=sinθ2sinθungn_g\sin\theta_2=n_u\sin\theta_u=>n_u={\sin\theta_2\over\sin\theta_u}\cdot n_gnu=sin25.2°sin20.2°1.5501.911n_u={\sin25.2\degree\over\sin20.2\degree}\cdot1.550\approx1.911

The refractive index of the unknown liquid is 1.9111.911



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