1. ## angle of refraction.

A ray of light impinges from air onto a block of ice (n = 1.309) at a 63.0° angle of incidence. Assuming that this angle remains the same, find the difference 2, ice - 2, water in the angles of refraction when the ice turns to water (n = 1.333).

theta2 ice - theta2, water

I have tried using this formula n1* sin(63) / n2

and then using n2*sin(63) / n1

then i subtract the theta and used inverse sin to get an answer.

I have 1.85 degrees which isn't correct.

anyone who can correct me on this one please.

2. Could you word that better... It's hard to understand what you're asking.

I think it's just asking to find this difference:

$(1.333)sin(63) - (1.309)sin(63) = 0.021384157$

And then:

$arcsin(0.021384157) = 1.23 \ degrees$

Maybe that's correct?

3. no, unfortunately, I also tried using that formula without dividing by n1 or n2 and that wasn't correct either.

I need to try and find the angle of refraction from the difference of ice and water. ice - water.

still haven't reached the answer for this one.

4. Originally Posted by rcmango
no, unfortunately, I also tried using that formula without dividing by n1 or n2 and that wasn't correct either.

I need to try and find the angle of refraction from the difference of ice and water. ice - water.

still haven't reached the answer for this one.
I'm still not certain of what you are expected to find.

Here's my guess: You need to find the angle of refraction using the ice, then subtract from that the angle of refraction using the water?

If that's true then Aryth's answer is correct. (Assuming he did the math correctly, I didn't check the numbers.)

-Dan

5. I tried the same math and did get a different answer, however, it was still incorrect. I will post the correct answer for this one, when I figure one. Thanks.

6. not sure why, it is: 0.951 degrees