What is the actual elevation angle of the Sun above the horizontal?

[roam]

Homework Statement

An underwater scuba diver sees the Sun at an apparent angle of 42.0° above the horizontal. What is the actual elevation angle of the Sun above the horizontal? (Use 1.333 for the index of refraction of water.)

The correct answer must be 7.86° above the horizon.

Homework Equations

Snell's law

The Attempt at a Solution

[tex]sin \theta = 1.333 \ sin (42)[/tex]

[tex]sin^{-1} (0.89) = 62.87[/tex]

Now to find the actual elevation angle [tex]90- 62.87= 27.13[/tex]

But this is wrong since the correct answer must be 7.86°. What's wrong?

 

[Doc Al]

An underwater scuba diver sees the Sun at an apparent angle of 42.0° above the horizontal.

So what's the angle of incidence? (You're using the wrong angles in Snell's law.)

 

[roam]

So what's the angle of incidence? (You're using the wrong angles in Snell's law.)

Okay, I see. Here's what I did now:

[tex]sin(42) = 1.333 \ sin (\theta)[/tex]

[tex]\theta = sin^{-1} (\frac{sin(42)}{1.333}) = 30.13[/tex]

90 - 30.13=59.86

Why is it still not right? (it should be 7.86°)

 

[Doc Al]

Okay, I see. Here's what I did now:

[tex]sin(42) = 1.333 \ sin (\theta)[/tex]

No. You were closer before.

Please draw a diagram for yourself. The angle of incidence is with respect to the normal, not the horizontal.

 

[roam]

No. You were closer before.

Please draw a diagram for yourself. The angle of incidence is with respect to the normal, not the horizontal.

Yes, here's a diagram:

[PLAIN]http://img534.imageshack.us/img534/909/diverh.jpg

What I've marked in red is the elevation angle of the Sun above the horizontal I'm required to find.

[tex]n_1 sin \theta_1 = n_2 sin \theta_2[/tex]

[tex]sin \theta_1 = 1.333 sin 42[/tex]

[tex]\theta_1 = 63.11[/tex]

Then to get the marked angle 90-63.11=26.88. But why is this wrong?

 

[Doc Al]

Yes, here's a diagram:

[PLAIN]http://img534.imageshack.us/img534/909/diverh.jpg

What I've marked in red is the elevation angle of the Sun above the horizontal I'm required to find.

Good.

[tex]n_1 sin \theta_1 = n_2 sin \theta_2[/tex]

Good.

[tex]sin \theta_1 = 1.333 sin 42[/tex]

Here's your mistake. 42 degrees is the apparent angle with the horizon, not the angle of refraction (θ₂). (The angles of incidence and refraction are with respect to the normal, not the horizontal.)

 

[roam]

Good.


Good.


Here's your mistake. 42 degrees is the apparent angle with the horizon, not the angle of refraction (θ₂). (The angles of incidence and refraction are with respect to the normal, not the horizontal.)

Ohhh... So [tex]\theta_1 = sin^{-1} (1.33 \ sin (90-42))= 81.3[/tex], and 90-81.3=8.7. Thank you very much for the explanation. :)

 

[Doc Al]

Now you've got it.

 

[PeterO]

Ohhh... So [tex]\theta_1 = sin^{-1} (1.33 \ sin (90-42))= 81.3[/tex], and 90-81.3=8.7. Thank you very much for the explanation. :)

A general suggestion/guide:
Application of Snell's Law is a fairly straight forward activity. Once you have done the early standard questions, the only way spice up the application is to not tell you the angle of incidence/refraction - at least not directly. Instead you are given some other angle(s) which you can use to calculate the angle you need for the formula.
Always read a question carefully to check how to interpret which angle is which.