Help with some optics questions?

Question

1. Light in air from a laser is incident on a slab of glass with angle of incidence 30 degrees, as shown in the diagram. The index of refraction of the glass is 1.50. Sketch the reflected ray in the diagram, calculate the angle of reflection, and mark both the angle of incidence and the angle of reflection on the diagram.

http://i14.photobucket.com/albums/a304/organizedkonfusion/optics.jpg

2. Sketch the refracted ray in (1) and calculate the angle of refraction, mark it on the diagram.

3. Is it possible for total internal reflection to take place? Explain.

4. What is the critical angle for the glass in (1)

Don't expect a 'sketch' but an explanation would be best. Thanks.

Answer 1

1. Angle of incidence = Angle of reflection



2. Ray tracing and Snell's Law

n1 * sin(theta 1) = n2 * sin(theta 2)

1.00 * sin(30°) = 1.50 * sin(theta 2 )



3. If the internal ray hits the surface at an angle greater than the critical angle it will be internally reflected. (Fiber optics is dependent on total internal reflection!)



4. The critical angle can be calculated from Snell's law by setting the refraction angle (theta 2) equal to 90°.

Answer 2

The angle of incidence is the angle between the line perpendicular to the surface, and the incoming ray.



So, you should be able to figure out angle of incidence from the diagram...



Angle of reflection is always equal to angle of incidence (just on the other side of the perpendicular line). So that should take care of item 1.



Now, surely you've learned about Snell's law? n1*sin(theta1) = n2*sin(theta2) (this is the easiest form to remember for me -- sometimes textbooks list as as sin(theta1)/n2 = sin(theta2)/n1)



You can replace the "1"s above with incident, and the "2"s with refracted ... so n1 = 1 (air), n2 = 1.5 (glass), and theta1 = 30 degrees...Now just solve for theta2.



For total internal reflection to occur, when you try to solve the above equation, you would end up with sin(theta) > 1, which obviously can never happen. In this case, there is no refraction, only reflection.



If you look at the equation, it should be obvious that this can only happen when going from a higher index to a lower index -- such as going from glass to air, but never from air to glass.

Also, given the two indecies are fixed, the incident angle must be above a certain critical angle.



Assuming the slab of glass has parallel sides, then the original refracted ray (you calculated the angle in part 2) has the same angle of incidence onto the backside of the slab. So snell's law works in reverse, and you end up with a beam coming out the back that was parallel to the original beam, and no problem with TIR. If the slab was instead angled (more of a wedge than a slab), then TIR would be possible.



The critical angle for glass is found by solving snell's law where the refracted angle is just 90 degrees (sin theta=1).

Since we're going from glass to air, this would be solving 1.5*sin(theta_crit) = 1*sin(90) = 1

so the arcsine of (1/1.5).

Answer 3

Other sources cannot be used in fiber optics, because optics means 'light', and if there is any other source it would not be included in fiber optics. Anyhow, radio waves need a medium to travel in and the rarer the medium the better it can travel without attenuation. We can use sound signals somehow in fiber optics because sound travel faster in solids than in air because it needs a denser medium. Still the draw back is that it cannot be sent very far.