# Thread: more angle of refraction.

1. ## more angle of refraction.

The drawing shows a rectangular block of glass (n = 1.52) surrounded by liquid carbon disulfide (n = 1.63). A ray of light is incident on the glass at point A with a = 36.0° angle of incidence. At what angle of refraction does the ray leave the glass at point B?

Please explain a way to reach the solution for this problem. I tried using the equation: n2*sin(36) / n1

pic: http://img128.imageshack.us/img128/900/13113445ga1.png

it does not seem to work with the problems i'm working with. I may not be using the equation correctly.

2. Originally Posted by rcmango
The drawing shows a rectangular block of glass (n = 1.52) surrounded by liquid carbon disulfide (n = 1.63). A ray of light is incident on the glass at point A with a = 36.0° angle of incidence. At what angle of refraction does the ray leave the glass at point B?

Please explain a way to reach the solution for this problem. I tried using the equation: n2*sin(36) / n1

pic: http://img128.imageshack.us/img128/900/13113445ga1.png

it does not seem to work with the problems i'm working with. I may not be using the equation correctly.
Probably you are only a little bit confused because you use 2 different types of media.

The $CS_2$ is the first (surrounding) medium, the glass is the second one. Then the law of Snellius states:

$\frac{\sin(\alpha)}{\sin(\beta)} = \frac{n_2}{n_1}$

You want to get $\beta$ :

$\sin(\beta) = \frac{n_1}{n_2} \cdot \sin(\alpha)~\implies~ \sin(\beta) = \frac{1.63}{1.52} \cdot \sin(36^\circ)~\implies~\beta \approx 39.1^\circ$

That means the angle of refraction is greater than the angle of incidence.

3. Okay, i'm following you,

Is, 39.1 the angle of refraction that exits the glass for part A?

so now need to find for part B??