Total Internal Reflection

runfast220

1. The problem statement, all variables and given/known data
The drawing shows a crystalline quartz slab with a rectangular cross-section. A ray
of light strikes the slab at an incident angle of 1=34o, enters the quartz and travels to
point Po (Figure2). This slab is surrounded by a fluid with a refractive index n. What
is the maximum value of n such that total internal reflection occurs at point Po?

@1= 34deg n2= 1.544




2. Relevant equations

n1sin@1 = n2sin@2
@c = sin-1 (n2/n1)

3. The attempt at a solution
Well I know in order to find n1 I need to know the critical angle, but I don't know how to find the critical angle without n1. My only idea was to use Snell's law of refraction to find n1 by setting @1 at 90deg?

Attached Thumbnails

 

cepheid

Hi runfast220,

The ray of light crosses an interface twice: once on the way into the quartz, and once while exiting it (unless, of course, it is totally internally reflected!).

This means that you need to apply Snell's law twice: once for each crossing. The first crossing, on the way into the quartz, tells you what [itex] \theta_2 [/itex] is. Granted, you can't solve for a numerical value for it just yet, but you can assume that the parameters you need (the refractive index of quartz, the refractive index of the fluid, n, and the incident angle [itex] \theta_1 [/itex]) are known, and can therefore include them algebraically (as symbols) in Snell's law. You will therefore end up with an algebraic expression for [itex] \theta_2 [/itex] in terms of these three known quantities.

Now, because the light ray enters the quartz at this angle ([itex] \theta_2 [/itex]), the angle of incidence upon the second (exiting) surface will be closely related to [itex] \theta_2 [/itex] (can you see how)? At that point, applying Snell's law for the exiting light ray, you can figure out what condition (on the known parameters) must be satisfied in order for this incident angle to be greater than or equal to the critical angle.