Answer to Question #120401 in Trigonometry for Nimra

Question

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?

Answer 1

Let

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

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

$\theta_i=$ the angle of incidence, and

$\theta_r=$ the angle of refraction.

Law of refraction (Shell's Law)

$$n_i\sin\theta_i=n_r\sin\theta_r$$

Given $n_a=1, n_g=1.550, n_m=1.329$

a)

From air to glass

$$n_a\sin\theta_1=n_g\sin\theta_2$$ $$1\cdot\sin41.3\degree=1.550\cdot\sin\theta_2$$ $$\theta_2=\arcsin({\sin41.3\degree \over 1.550})\approx25.2\degree$$

From glass to methanol

$$n_g\sin\theta_2=n_m\sin\theta_3$$ $$1.550\cdot\sin25.2\degree=1.329\cdot\sin\theta_3$$ $$\theta_3=\arcsin({1.550\cdot \sin25.2\degree \over 1.329})\approx29.8\degree$$

b)

From glass to unknown liquid

$$n_g\sin\theta_2=n_u\sin\theta_u=>n_u={\sin\theta_2\over\sin\theta_u}\cdot n_g$$ $$n_u={\sin25.2\degree\over\sin20.2\degree}\cdot1.550\approx1.911$$

The refractive index of the unknown liquid is $1.911$