Total Internal Reflection of quartz

[runfast220]

Homework Statement

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

Homework Equations

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

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?

 

[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 $\theta_2$ 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 $\theta_1$) are known, and can therefore include them algebraically (as symbols) in Snell's law. You will therefore end up with an algebraic expression for $\theta_2$ in terms of these three known quantities.

Now, because the light ray enters the quartz at this angle ($\theta_2$), the angle of incidence upon the second (exiting) surface will be closely related to $\theta_2$ (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.

 

[kerimek]

I would suggest approaching this problem by first understanding the concept of total internal reflection. Total internal reflection occurs when a ray of light traveling from a medium with a higher refractive index to a medium with a lower refractive index strikes the interface at an angle greater than the critical angle. In this scenario, the ray of light will be reflected back into the higher refractive index medium rather than being refracted into the lower refractive index medium.

In this case, the critical angle can be calculated using Snell's law as @c = sin-1(n2/n1), where n1 and n2 are the refractive indices of the two media. Since we are given the incident angle and the refractive index of the surrounding fluid (n2), we can rearrange the equation to solve for n1.

n1 = n2/sin(@1)

Substituting the given values, we get n1 = 1.544/sin(34deg) = 2.930.

Therefore, for total internal reflection to occur at point Po, the refractive index of the surrounding fluid should be less than or equal to 2.930. Any value of n greater than 2.930 will result in refraction rather than total internal reflection.