- 128

- 1

**1. Homework Statement**

The figure below shows the path of a beam of light through several layers (n1 = 1.58, n2 = 1.42, n3 = 1.20 and n4 = 1.00) of different indices of refraction.

https://loncapa2.physics.sc.edu/res/brookscole/serway/College_Physics_7ed/Chap22/graphics/serw2244.gif

a) If θ1 = 30.3 deg, what is the angle, θ2, of the emerging beam?

**(5.29×10^1 deg)**

b) What must the incident angle, θ1, be in order to have total internal reflection at the surface between the n3 = 1.20 medium and the n4 = 1.00 medium?

**2. Homework Equations**

For part a)

n1sin(theta1)=n2sin(alpha)

n2sin(alpha)=n3sin(beta)

n3sin(beta)=n4sin(theta2)

For part b) I'm not too sure.

**3. The Attempt at a Solution**

__For part a)__

n1sin(theta1)=n2sin(alpha)

(1.58)sin(30.3)=(1.42)sin(alpha)

alpha=34.15099

n2sin(alpha)=n3sin(beta)

(1.42)sin(34.15099)=(1.20)sin(beta)

beta=41.628239

n3sin(beta)=n4sin(theta2)

(1.20)sin(41.628)=(1)sin(theta2)

theta2=

**52.90 deg**---> this is the correct answer

(the formulas can also be simplified to... theta2=arcsin(n1sin(theta1)/n4)

__For part b)__

This is the part I"m having difficulties with.

(1.20)sin(theta)=(1.00)sin(90)

sin(theta)=(1.00)/(1.20)

theta=56.44269 deg

I know that this isn't the answer. Am I right in connecting the total internal reflection with sin(90)? Is the critical angle equivalent to the incident angle...are they the same thing?