Refraction Problem: Find Angles with Triangles

[khoais]

Homework Statement

http://img98.imageshack.us/img98/371/physicsproblemrh8.jpg

Homework Equations

The Attempt at a Solution

Basically, I tried to use triangles to find the angles. But I don't know how as it seems I'm only given one side of the triangles.

 

[LowlyPion]

Homework Statement

Homework Equations

The Attempt at a Solution

Basically, I tried to use triangles to find the angles. But I don't know how as it seems I'm only given one side of the triangles.

I think they want you to consider the effect of Snell's Law here.

$$\frac{n_1}{n_2} = \frac {Sin \theta_1}{Sin \theta_2}$$

 

[khoais]

I think they want you to consider the effect of Snell's Law here.

$$\frac{n_1}{n_2} = \frac {Sin \theta_1}{Sin \theta_2}$$

You mean
$$\frac{n_1}{n_2} = \frac {Sin \theta_2}{Sin \theta_1}$$

So let's say n1 = n2 when the glass is empty because the angle of incident is equal to the angle of refraction. When the glass is full, n2 = 1.22 and everything else is unknown. So how would I go about finding just theta r, which is theta_1 in this case?

 

[LowlyPion]

You mean
$$\frac{n_1}{n_2} = \frac {Sin \theta_2}{Sin \theta_1}$$

So let's say n1 = n2 when the glass is empty because the angle of incident is equal to the angle of refraction. When the glass is full, n2 = 1.22 and everything else is unknown. So how would I go about finding just theta r, which is theta_1 in this case?

Yes. Sorry for the hurried typo.

With respect to the problem you know the angle of sight over the top is h/(h² + d²)^1/2

And the angle it makes in the liquid is h/(h² + d²/4)^1/2

And you know they are related by the 1.22 then don't you?

You know D so solve for h. Having h in hand then you can figure the angle directly.

 

[khoais]

Yes. Sorry for the hurried typo.

With respect to the problem you know the angle of sight over the top is h/(h² + d²)^1/2

And the angle it makes in the liquid is h/(h² + d²/4)^1/2

And you know they are related by the 1.22 then don't you?

You know D so solve for h. Having h in hand then you can figure the angle directly.

Thanks a lot lowlypion.

I had a little trouble figuring out why your values didn't work, but in the end I found that the values were actually:

d/(d² + h²)^1/2 for the top angle

and

(d/2)/(d²/4 + h²)^1/2 for the liquid angle.

All in all though, you greatly helped me. Thank you so much!

 

[LowlyPion]

Thanks a lot lowlypion.

I had a little trouble figuring out why your values didn't work, but in the end I found that the values were actually:

d/(d² + h²)^1/2 for the top angle

and

(d/2)/(d²/4 + h²)^1/2 for the liquid angle.

All in all though, you greatly helped me. Thank you so much!

My apologies. I relied on the Hyperphysics link that shows the angle with the surface rather than with the normal.

http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/refr.html#c3