The ratio involves taking the sine of the angle of incidence, and the sine of the angle of refraction. Remember doing sine, cosine and tangent on the calculator in module 1? Well, this is one of the reasons that we did it. Get out you calculator, and tell me what the sin 45 degrees is? How about sin 12? What angle has a sin that is equal to 0.3420? (That one was usually written sin a = 0.3420, a = ?)

Did you get 0.7071 for sin 45? And did you get sin 12 = 0.2079? The angle whose sin is 0.3420 is 20 degrees. Did you get these answers? If not, RIGHT NOW go to  module 1 trigonometry  and review how to do this. You are going to be doing this a lot this semester, so review it careful now. If you have trouble with it, call your Instructor. When you are through there press the BACK button to get back here -- there is no direct link between that course and this one.

Now, back to that ratio -- the one shown in the diagram above.  Remember cross multiplication? That is all we do to get the second equation in that diagram from the first one.

First, we have the index of refraction of the material that the ray of light is travelling from or exiting.  In this case, the ray is coming from air.  That will not always be the case.  What did you find out about the index of refraction of air?  That it is equal to 1?

Next we have the sine of the angle of incidence. That is the angle that the ray of light is making with the normal to the surface of the lens (or whatever material the ray is entering.)

Third we have the index of refraction of the material that the ray is travelling  to or entering.  In this case it will be the index of the lens material; when we discuss the ray EXITING the lens the second index will be for air.

Last we have the sine of the angle of refraction. This is going to show us what direction the ray is travelling in with respect to the normal to the surface once the ray has entered the material.

We are going to use  to represent the index of refraction of the incident material, or the index of the material that the light ray is coming from or exiting. We are going to use  for the index of refraction of the refracting material, or the index of the material that the ray is travelling to or entering.

Ready? 

Here is a ray of light that is entering CR39 from air. Its angle of incidence is 10 degrees. What will happen to it?

Well, the formula says that  times sin i is equal to  times sin r.
What is ?
What is i? What is sin i?
What is ?
What is sin r? What is r?

We will do all of these problems the way we did the angle problems last week: first, read the problem, then write down the equation and a list of the variables, then read the question phrase by phrase, substituting numbers for letters in the equation, then solve the equation.

So, the formula is () (sin i) = () (sin r)
and the variables are
=
<i=          sin i =
=
<r =        sin r=

Reading the problem phrase by phrase, we see that the ray is "entering CR39" -- entering means that is where it is going TO, so it is the refracting or r side, so  is the index of CR39. Lets call this 1.50.
= 1.5
Well, it also says "from air" so air is the incident or from side, and so  is the index of air. What is the index of refraction of air?
= 1
We are next told that "the angle of incidence is 10 degrees." So, i = 10.
          sin i = ;  sin 10 = ?
And r is what we do not know.
          sin r = ?

Ready? Using the formula
       () (sin i) = ( ) (sin r)
we now have
       (1) (sin 10) = (1.50) (sin r)
Since you found out that sin 10 = 0.1736, that gives us
       (1) (0.1736) = (1.50) (sin r)
Solving this equation for sin r we have to divide both sides by 1.50. Now we have
       (0.1736)(1.50) = (sin r)
So in your calculator, first you did sin 10, which gave you 0.1736 . . ., now you punch in  1.5 = and you get 0.1157 . . . That is the sine of r. What is r? press shift or 2nd or inv or whatever your calculator has, then sin, and you should have . . . 6.6 degrees, which we will round to 7 degrees.

This type of problem is easiest if you draw yourself a little diagram showing which side is the dense material and which is the rare material, and then decide which angle will be smaller and which will be larger. You do not need anything fancy -- just enough to picture what is happening.

OK, Lets do another one, and get a good look at what we are doing.

Here we have a ray leaving the CR39, index 1.50, and entering air. The angle of incidence this time is 12 degrees. What is the angle of refraction?

First list the formula, then list the variables. Then read the problem again, phrase by phrase, and fill in what you know.
       () (sin i) = () (sin r)
=
<i =           sin i =
=
<r =          sin r =
" . . . a ray leaving the CR39" means the CR39 is where the ray is coming from, so CR39 is the incident side, and the n = 1.50 is which n?
= 1.50
" . . . and entering air." OK, we know what that means.
= 1
"The angle of incidence this time is 12 degrees." So,
          sin i = sin 12 = ?
and then the questions asks what r is.
          sin r = ?

       ( ) (sin i) = () (sin r)
= 1.50
<i = 12;  sin i = sin 12 = 0.2079
= 1
sin r = ?

        (1.50) (sin 12) = (1) (sin r)
       (1.50) (0.2079) = sin r
       sin r = 0.3118 . . .
       r = 18 degrees.

Right.

What did you do with your calculator when you did this one? You punched in "12" "sin" "x" "1.5" "=" "2nd" "sin", and magically you got 18...... Do you see WHY you are entering the information in that order?

(PS -- for those of you with the OTHER type of calculator, you needed to punch in "sin" "12" "=" "x" "1.5" "=" "2nd" "sin" "=". Right?)

There is more practice for you on Snell's law here.

Read pages (23-24 / 29-31) in the Optical Formulas book, do the exercises on page (24 / 31), and check those answers in the back of the book.  Also read pages 375-377 in Systems for Ophthalmic Dispensing and do exercises 1-3 on page 384.