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

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asked Apr 7, 2012 by jason1111

 

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

 

1 Answer

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answered Apr 8, 2012 by Joey33

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


 

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