A ray of light impinges from air onto a block of ice (n = 1.309) at a 69.0 degre

Question

A ray of light impinges from air onto a block of ice (n = 1.309) at a 69.0 degree angle of incidence. Assuming that this angle remains the same, find the difference theta_2, ice - theta_2, water in the angles of refraction when the ice turns to water (n = 1.333). theta_2, ice - theta_2, water = The drawing shows a coin resting on the of a beaker filed with an unknown liquid. A ray of light from the coin travels to the surface of the liquid and is refracted as it enters into the air. A person sees the ray as it skims just above the surface of the liquid. How fast is the light traveling in the liquid? (Let x = 5.39 cm and y = 6.56 cm.)

Explanation / Answer

1)
use Snell's law

sin(theta1)/sin(theta2) = n2/n1

sin(69)/sin(theta2_Ice) = 1.309/1

sin(theta2_Ice) = sin(69)/1.309

theta_Ice = sin^-1(0.7132)

= 45.5 degrees

similarly,

sin(theta2_water) = sin(69)/1.333

theta_Ice = sin^-1(0.7132)

= 44.4 degrees

theta2_Ice - theta2_water = 45.5 -44.4

= 1.10 degrees

2)

from the figure

angle of incidence, theta1 = tan^-1(y/x)

= tan^-1(6.56/5.39)

= 50.6 degrees

let n is the refractive index of the liquid.

use Snell's law

sin(theta1)/sin(theta2) = n2/n1

sin(50.6)/sin(90) =1/n

==> n = 1/sin(50.6)

= 1.294

speed of light in the liquid, v = c/n

= 3*10^8/1.294

= 2.32*10^8 m/s

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