Monochromatic Light and a 30-60-90 Prism Beam

1. Apr 26, 2013

craighenn

1. The problem statement, all variables and given/known data
Red light is incident in air on a 30o - 60o - 90 o prism as shown. The incident beam is directed at an angle of φ1 = 44o with respect to the horizontal and enters the prism at a height h = 16 cm above the base. The beam leaves the prism to the air at a distance d = 41.5 along the base as shown.

What is φ3, the angle the transmitted beam makes with the horizontal axis?

http://imgur.com/poYBTbE
φ2 = 49.22 degrees
n2, index of refraction inside the prism = 1.47

2. Relevant equations
Snell's Law: n1sinφ1 = n2sinφ2
angle geometry

3. The attempt at a solution
I dropped a normal line between φ2 and φ3, so that I could find the angle of incident as α = 90-φ2 = 40.78. Then I used Snell's Law to calculate the angle of refraction β = 73.76 degrees. This angle is complementary to φ3, so 90 - β =16.24. For some reason this isn't the answer, though everything I know tells me it should be.

Last edited: Apr 26, 2013
2. Apr 26, 2013

Simon Bridge

How did you find $\phi_2$?

You are given $\phi_1$, $n$, $h$, and $d$ ...
If you put a coordinate axis origin on the bottom-left point of the prism, then you should be able to find the coordinates where the ray enters and leaves the prism.

3. Apr 26, 2013

craighenn

In order to find ϕ2 I first had to put a normal down from where the ray first enters the prism, then find the values of the parts of D using the given 30 degree corner and h, the height of the incident.

How would the coordinate axis help in finding ϕ3? I don't think I'm supposed to use h and d beyond what I already have (finding ϕ2)

4. Apr 26, 2013

Simon Bridge

The coordinates allow you to find the angle of incidence to the base of the prism without having to use $\phi_1$. The ray inside the prism forms the hypotenuse of a handy right-angled triangle.

What is this D if which you speak? I see a d - this what you mean?

You wanted to use snells law at the first interface and then again at the second - basically tracing the ray through the prism? - then I get you. But the devil is in the details: this is where you probably went wrong. If you don't share the details you'll just have to triple and quadruple check it yourself.

i.e. how did you get the angle of refraction at the first surface?
How did you turn that into the angle of incidence at the second interface?

Another approach to check your working is to go through the prism backwards. Optics is supposed to work in both directions but the math is often easier one way than the other.

5. Apr 26, 2013

craighenn

I just noticed a typo in my initial post. ϕ2 =49.22, not ϕ1 (that equals 44)
Sorry, here are the steps I took:

1. Dropped a vertical line from where the ray first intersects the prism (let's call that point B). The point C will be where ϕ2 is, where the ray exits the prism. The point C will be at the given 30 degree corner. The point P will be where that vertical line from point B intersects with the base of the prism.

2. Using h and the given angle 30, I calculated the line segment AP, which is a portion of the length d.

3. Now that I have AP, I can calculate the segment PC from "d - AP"

4. With h and PC I can calculate ϕ2 using tanϕ2 = (h/PC) = 49.24

Now I calculate n2, the index of refraction inside the prism

5. I first find α1, the angle of incidence of the ray entering the prism. This is the thin sliver of angle left between ϕ1 and the normal. The normal is 90 degrees, ϕ1 is 44, and the other remaining angle is also 30 due to angle rules, so α1 = 90- 30 - 44 = 16.

6. To find β1, the angle of refraction that coincides with α1, I simply look at the triangle ABC and calculate the remaining angle of the triangle, the B corner. That is 100.12 degrees. The small sliver next to the normal is β1 and is found by β1 = 100.12 - 90 = 10.12

7. I can now use Snell's Law to calculate n2 in the prism. n2 = 1.47

This is where I get stuck.

8. After dropping a normal line at the second angle of incident, I found α2 to be complementary to ϕ2. Therefore, α2 = 90-49.24 = 40.76

9. Knowing n1, n2, and α2, I calculate β2, the angle of refraction outside of the prism. It ends up being 74.26 degrees. ϕ3 is complementary to β2, so ϕ3 = 90 - 74.26 = 15.74

That should be the answer: ϕ3 = 15.74 degrees. I thought that maybe since the angle is below the axis it would be -15.74, but that still isn't correct (according to smart physics)

Hope this wasn't too wordy.

6. Apr 26, 2013

haruspex

I agree with your value for ϕ2, but I get about half a degree more than 100.12 for the angle ABC. That makes a small difference to the refractive index. There seems to be another numerical error in calculating β2, again by about half a degree.

7. Apr 26, 2013

Simon Bridge

I'mhaving trouble with your geometry there...

A is where the ray enters the prism.
C is where the ray exits the prism.
B is directly below A so ABC is a right-angled triangle - B is the right angle.
The angle at C is $\phi_2$?

Then you say:
... so... $\phi_2$=30deg?
Or do you intend that the 30deg corner on the prism should have the label C - as well as the place the ray exits the prism? Perhaps it should have a different label - D?

This makes the rest of your statement difficult to follow - every time you mention point C I don't know what you mean. But I can see how you may have got turned around someplace.

OK - the whole problem seems to be about Snells law when the angle is to the surface rather than to the normal. i.e. applying Snell's law at the second surface you get:
$n\cos\phi_2=\cos\phi_3$ (check)
... when you realize that, you can find a similar equation relating $\phi_2$ to $\phi_1$ ... then you use a trig relation to cancel out the $\phi_2$'s.

I still think the coordinate geometry approach is easier.
Knowing d and h and that 30deg slope is the same as knowing $\phi_2$.

[edit - since the refractive index is not given - a third equation is needed.]

Last edited: Apr 26, 2013
8. Apr 26, 2013

rl.bhat

With reference to your post #5, the calculation of β1 is not correct.
Let ABC is the right angled triangle with A the point of incidence, B is the right angle and C the angle ∅2. Let the normal at A meets BC at D. Then the angle ADB = 60° = β1 + 49.24°.
You can directly find the final required angle with out finding the refractive index by applying Snell's law at A and C and equating them.

Last edited: Apr 26, 2013