Question

4. (a) 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 41.3 degrees angle with the normal to the glass. Find the angle the light makes with the normal in the methanol.
(b) The tank 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 from the normal, what is the refractive index of the unknown liquid?

Answer 1

Answer:

29.775°

1.911

Explanation:

Law of refraction(Snell's law):  

n_1 sin Ф_1 = n_2 sin Ф_2  

n_1 = The index of refraction for material with incident light,

n_2 = The index of refraction for material with refracted light

Ф_1= angle of incidence,

Ф_2 = angle of refraction.  

Note: the angle is measured from the normal: a line drawn perpendicular to the surface at the point where the incident ray strikes the surface.

The Part of the light which continues into the second medium is transmitted rather than reflected, but the transmitted ray changes direction as it crosses the boundary. The transmission of light from one medium to another, but with a change in direction, is called refraction. The angles of the incidence and reflected light is described by (Snells law) which is shown above.  

a)

n_3 = 1.329 —> for metalmol —> from table

1 —> in air

2 —> in the glass

3 —> in methanol  

from air to glass

n_1 sin Ф_1 = n_2 sin Ф_2  

1 x sin 41.3° =  1.550  sin Ф_2  

Ф_2 = arcsin(1*sin41.3°/1.550) = 25.201°

from glass to methanol:

n_2 sin Ф_2 = n_3 sin Ф_3  

1.550 sin 25.201° = 1.329 * sin Ф_3

Ф_3 = arcsin(1.550 sin 25.201°/1.329) = 29.775

from glass to unknown:

n_2 sin Ф_2 = n_3 sin Ф_3  

1.550 sin 25.201° = n_unknown * sin 20.2

n_unknown = 1.550 sin 25.201°/sin 20.2

                     = 1.911