Snell's Law HELP TEST TOMMOROW

[bflatt]

Snell's Law HELP! TEST TOMMOROW!

Homework Statement

Hi, I have a problem with Snell's Law...i understand that it is:
n1(sin)theta1 = n2(sin)theta2

but for the test, we need to know how to manipulate the problem to find the angle of refraction...i will have a calculator to plug thing in but i just don't understand how to! Apparently after you find one thing you need to use an inverse sin (sin-1) or something...PLEASE HELP!

An example problem for this is:
A Beam of light passes from glass (n=1.51) to an unknown medium X (nX=?).
If the angle of incidence is 40 and the angle of refraction is 29 then the index of refraction of the unknown medium is closest to:
a)1.17, b)1.38, c)1.63, d)1.83 or e)2.00

Followed by: The beam of light exists medium X and enters air. If it has an incident angle of 29 at the X-Air boundary, find the angle of refranction in the air?

The Attempt at a Solution

for example my notes say: Nw(sin)theta W = Na(sin)theata A
w=water=1.333
a=air=1
1.333(sin)theta W=1 x sin(37)(angle of incidence)
sin thetaW=sin(37)/1.333
Theta W= (sin -1)/(inverse sin) [sin(37)/1.333]

WHY IS IT GO FROM SIN(theta W) TO INVERSE SIN!?

I Know there is a lot fo questions and info here...these are questoins from a quiz i did and i did very poorly...my test is tommorow and help would be greatly appretiated THANKS SOOO MUCH!

 

[hage567]

To explain the inverse trig function part:

Starting with this line of your example:

sin(theta_w)=(sin37/1.333)

The sin^-1 (or inverse sin, also called arcsin) is just a function that you use to "undo" the sin(theta_w) to get just theta_w. You apply it to both sides of your equation like you would anything else when trying to isolate a variable. So you would have

sin^-1{sin(theta_w)} = sin^-1{sin37/1.333}

the left side then becomes just theta_w, and by calculating the right side you get the angle of refraction.

 

[bflatt]

ooooo thanks so much...the test was still REALLY hard but what can you do...thanks a ton i appreatiate it