# A light ray enters a rectangular block of plastic at an angle of θ1 = 47.0° and emerges at an angle of θ2 = 79.5° as shown in the figure below

A light ray enters a rectangular block of plastic at an angle of θ1 = 47.0° and emerges at an angle of θ2 = 79.5° as shown in the figure below

(a) Determine the index of refraction of the plastic.

(b) If the light ray enters the plastic at a point L = 50.0 cm from the bottom edge, what time interval is required for the light ray to travel through the plastic?
ns

asked Apr 7, 2012 in Physics

based on the information given redraw the diagram as follows:

Given that θ1 = 47° and θ2 = 79.5° .
Snell’s law at the first surface gives
n sinα = 1.00 sin 47.0°                                      Equation 1

Also using the geometry, you can conclude that the angle of incidence of the second surface is:

β = 90.0°− α

Thus using snell's law we get:

n sinβ = n sina90.0°−α f = 1.00 sin79.5°

and so simplying we get:

n cosα = sin79.5°                                              Equation 2

We can find α, by dividing equation 1 by equation 2:

tanα = sin θ1 /θ2

α = inverse tan (sin θ1 /θ2)

and so we can plug α back in to equation 1 to solve for n.

Part B:

From the sketch, observe that the distance the light travels in the plastic is

d = L / sinα

and teh speed of light in the plastic is:

v = c / n

So the time required to travel through the plastic is:

Δt = d / v

answered Apr 8, 2012 by ~Expert~ (1,216 points)
selected Sep 13, 2012 by kirby