# Snell's Law x2: Double-checking trig?

1. Aug 1, 2007

### exi

1. The problem statement, all variables and given/known data

In this drawing:

http://img245.imageshack.us/img245/6559/physgv8.png [Broken]

Index of refraction for glass: 1.52
Index of refraction for surrounding carbon disulfide: 1.63
Incident angle at point A: 43.0°
At what angle does the ray leave the glass at point B?

2. Relevant equations

$$n_{1}sin\theta_{1} = n_{2}sin\theta_{2}$$ (twice)

3. The attempt at a solution

Please double-check my conceptual understanding of this.

Part 1, as light travels through the carbon disulfide into the glass:

$$1.63sin43 = 1.52sin\theta_{2}$$

$$\theta_{2} = sin^{-1}\frac{1.63sin43}{1.52}$$

So Θ within the glass is 47°.

Drawn out and making a triangle with the upper left hand corner of the glass, a triangle forms with point B as one of its own points, meaning that the angle with a normal line at that point would be 180 - (90 + 47) = 43°.

Part 2, as light travels through the glass back into the carbon disulfide:

$$1.52sin43 = 1.63sin\theta_{2}$$

$$\theta_{2} = sin^{-1}\frac{1.52sin43}{1.63}$$

Which yields 39.4923°.

Last edited by a moderator: May 3, 2017
2. Aug 1, 2007

### hage567

Looks OK to me.

3. Aug 1, 2007

### exi

Thanks for taking a peek - turned out to be correct.

4. Aug 1, 2007

### nrqed

Looks correct to me too.