Finding Sin r from 1.52 and 20 Degrees

  • Thread starter Thread starter Novus Dakota
  • Start date Start date
  • Tags Tags
    Degrees Sin
Click For Summary

Homework Help Overview

The discussion revolves around applying Snell's Law to find the sine of an angle (sin r) given certain refractive indices and an angle of incidence. The participants are attempting to clarify the relationships between these values and the correct approach to solve for sin r.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are discussing the application of Snell's Law, questioning the notation used, and exploring different interpretations of the problem. There are attempts to clarify the process of finding sin r and the relationship between the given values.

Discussion Status

The discussion includes various interpretations of the problem and attempts to clarify the steps involved in solving for sin r. Some participants provide guidance on using a calculator and applying inverse sine, while others express confusion about the notation and the process.

Contextual Notes

There appears to be some uncertainty regarding the notation and the values being used, as well as differing opinions on the expected outcome for sin r. Participants are navigating these ambiguities while adhering to the constraints of the homework context.

Novus Dakota
Messages
9
Reaction score
0
so my given is 1.52 20degrees = 1.10 sin r

im trying to find sin r which comes out to 18.9 degrees but I am not sure how find that, can somebody help me out please?

so you do 1.10/1.52

do you divide that into sin 20?
 
Last edited:
Physics news on Phys.org
I don't quite understand your notation...

Is that first part : 1.52 * 20 deg?
 
snells law is n1 sin i = n2 sin r

n1= 1.52
sin i = 20 deg

n2 = 1.10

sin r = ?

sin r actually equals 28.2 deg but i don't understand the process
 
Isn't this just simple algebra? What do you get when you solve for the unknown variable?
 
I got 27.6 as my answer...
 
...

[tex]\sin{x}[/tex] does not give you an answer in degrees.

[tex]n_1 \sin{\theta_1} = n_2 \sin{\theta_2} \Longrightarrow \theta_1 = \sin^{-1}\left( \frac{n_2}{n_1} \sin{\theta_2}\right)[/tex]
 
Isn't the question what is sin r? Not what is the angle...
 
Well, in that case he's in for a surprise if he wants it to "come out to [tex]18.9^\circ[/tex]"
 
but how do you do it

1.10/1.52 =.723

sin 20=.723

then I am stuck...
 
  • #10
Look for the [tex]\sin^{-1}[/tex] button on your calculator. Apply it as in the equation I put above.
 
  • #11
first calculate sin(20)
then multiply that by 1.52
then divide by 1.10
then take the arcsin/inverse sin of that number
make sure its in degrees and yorue done.

28.2
 

Similar threads

Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 18 ·
Replies
18
Views
6K
Replies
28
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K