Angle of Incidence for Internal Reflection

1. Jul 30, 2008

cgaleb

1. The problem statement, all variables and given/known data
In the figure shown below, a light ray enters a glass slab (with index of refraction nglass = 1.65 ) at point A
and then undergoes total internal reflection at point B. The medium that surrounds the glass slab is a liquid whose
index of refraction, nliquid = 1.30. Find the value of the initial angle of incidence i for which total internal reflection
can take place.

2. Relevant equations
Snell's Law At point A: n (liquid) sin $$\Theta$$ i = n (glass) sin $$\Theta$$ 2

At Point B: n (liquid) sin $$\Theta$$ 3 = n (liquid) sin 90

$$\Theta$$3=90-$$\Theta2$$

3. The attempt at a solution

Using the above equations I got the $$\Theta$$i=15.5, $$\Theta$$2=12.12, and $$\Theta$$3=77.88

Now what has thrown me off is the HINT the professor gave , which was tan$$\Theta$$=sin $$\Theta$$/cos$$\Theta$$.

What is this equation for? I can't seem to find the relevance in my text or notes. AND what value would I use for $$\Theta$$, the $$\Theta$$ for i ????

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2. Jul 30, 2008

alphysicist

Hi cgaleb,

It's probably just a typo, but the first n(liquid) should be n(glass).

I don't believe the 77.88 degrees is correct for the critical angle $\theta_3$ (using the corrected version of your point B equation). If you are still getting that number, can you show the details of how you got it?

3. Jul 30, 2008

cgaleb

Yeah-you are right about the typo. The professor said that $$\Theta$$3=90-$$\Theta$$2, so I took 90o-12.12 (my $$\Theta$$2=77.88.

All of my angles add up to 180o. (o=degrees) 90o+12.12o+77.88o=180.

We covered all of this stuff 3 chapters today, and he really only touched on the topics, giving us "hints" to select problems. So, I'm really trying to learn this myself, as well as relearn basic algebra/trig. He didn't even mention a critical angle, and I've tried to look it up, but need the definition dumbed down, I guess...

How does my work look?

4. Jul 30, 2008

cgaleb

Oh, guess I'll add this as it may help

N(glass)sin$$\Theta$$3=n(liquid)sin90o
So, sin $$\Theta$$3=sin(90o-$$\Theta$$2)
Therefore
1.65 (1-sin$$\Theta$$2)=1.3 (1)
and sin$$\Theta$$2=.21
So, $$\Theta$$2=12.12

Is my logic sound?

5. Jul 30, 2008

alphysicist

Yes, this last post is exactly what you needed to show. Without your work I was not able to see how you got your answers.

This is where the numerical error is, because sin(90-theta) is not equal to (1- sin(theta)).

However, you don't need to go that far. Look at your first equation:

N(glass)sin$$\Theta$$3=n(liquid)sin90o

In that equation, you know everything except $\theta_3$ so you can solve for that angle first. Then finding $\theta_2$ and $\theta_1$ is straightforward from the other two equations.

6. Jul 30, 2008

cgaleb

Alright, I plugged in the work just like you said, and came up with $$\Theta$$ i =51.1,
$$\Theta$$ 2 =37.8, and $$\Theta$$ 3 =52.2. Hoping that this is right....

If so, what do I do next, I'm still confused about my professor's tan$$\Theta$$ = sin $$\Theta$$ / cos $$\Theta$$

Which # would I use for $$\Theta$$ in this scenario, and what does this equation solve for?

BTW, Thanks for the help!!!

Crossing fingers...

7. Jul 30, 2008

alphysicist

Those numbers are close to what I got (I got 51.9877 degrees for $\theta_3$).

I think that would give an alternative way of finding $\theta_1$. For example, you can use the fact that $\theta_2+\theta_3=90^{\circ}$, and $\sin(90^{\circ}-\theta)=\cos(\theta)$, and then divide the Snell's law equations to relate $\theta_3$ and $\theta_1$.

The wording of the problem question seems a bit strange to me. Just to check: did it ask for the value of the initial angle of incidence, or did it ask for the values of the initial angle of incidence? In other words, was it asking for a single answer or a range of answers?

Either way, you already have the angle of incidence (at point A) that is the crossover between total reflection (at point B) happening or not. So if they do want a range of values, all you have to decide is if angles of incidence (at point A) that are greater than or less than $\theta_1$ also give total reflection at point B.

8. Jul 30, 2008

cgaleb

I believe its just a single answer problem. The wording "find the value of the initial angle of incidence for which total internal reflection can take place" is a bit tricky. But I feel great having finally sovled the intial angle of incidence, regardless. THANK YOU so much for all of your help!!!