# Light + prisma

1. Mar 18, 2005

### Kahsi

Hi!

I'm trying to figure out what angel the light-wave will leave the glasprisma (it enters the glasprisma from the air).

http://home.tiscali.se/21355861/bilder/prisma1.JPG [Broken]

So I need to use Snell's law

$$n_{1}sin\theta_1 = n_{2}sin\theta_2$$

n for the glasprisma equals 1.6

So when it enters the prisma me should get

$$\theta_2 = sin^{-1}{\left(\frac{1*sin0}{1.6}\right)} = 0$$

So when it enters the glasprisma the angle still is 0. When it's leaving the prisma we get this.

http://home.tiscali.se/21355861/bilder/prisma2.JPG [Broken]

Where v = 90 - 35 = 55degres.

So

$$\theta_2 = sin^{-1}{\left(\frac{1.6*sin-55}{1}\right)} = Error!$$

What am I doing wrong?

Thank you.

Last edited by a moderator: May 1, 2017
2. Mar 18, 2005

### Staff: Mentor

total internal reflection

It just means that the angle of incidence on that surface exceeds the critical angle for total internal reflection. The light will reflect from that surface and exit from the bottom surface.

3. Mar 18, 2005

### Kahsi

So in this case it's total reflection? Then I've misscalculated.

4. Mar 18, 2005

### Staff: Mentor

What makes you think you've miscalculated? You still can calculate the angle that the light leaves the prism.

5. Mar 18, 2005

### Kahsi

The change of the direction should be 57degrees. If it leaves the prism from the bottom surface the change of direction is probably more then 100degrees. Or maybe I've missunderstod you.

Maybe it should be something like this.

http://home.tiscali.se/21355861/bilder/prisma3.JPG [Broken]

Last edited by a moderator: May 1, 2017
6. Mar 18, 2005

### Staff: Mentor

Don't guess, calculate.

Exactly.

Last edited by a moderator: May 1, 2017
7. Mar 18, 2005

### Kahsi

Thank you Doc Al. I'll calculate until I get the correct answer. :)

8. Mar 18, 2005

### zanazzi78

Kashi,
Have you tried using; n= [sin( (A+D)/2)] / sin(A/2) where A is the angle of the prism and D the maximum angle of deviation. when you calculate the max angle of deviation you can use; 2i=A+D where i is the angle of incidence on the prism.

hope this helps

9. Mar 18, 2005

### Kahsi

Hi zanazzi78.

Sorry but you lost me there. Do you think you cold point out int the picture where A, D and i is? It would be really helpfull if you could attach the picture

10. Mar 18, 2005

### zanazzi78

I`m new to this forum thing !!! how do i post pictures?

11. Mar 18, 2005

### Kahsi

Look at the attachment.

http://home.tiscali.se/21355861/bilder/picture.PNG [Broken]

Last edited by a moderator: May 1, 2017
12. Mar 18, 2005

### zanazzi78

Oops here you go

13. Mar 18, 2005

### Kahsi

zanazzi78 it looks like it doesn't work. I got the same problem but I thought it only wasn't working from me. You can upload the picture here

Choose "Bilder".

Last edited by a moderator: Apr 21, 2017
14. Mar 18, 2005

### zanazzi78

I can't get the pic small enough to post!

The angle A is the interal angle between the sides of the prism.

The angle D is the change in angle, between, where the ray would have gone(if not refracted), and the actulla path that the ray leaves the prism!

The angle i is the angle the ray makes with the prim.

15. Mar 18, 2005

### zanazzi78

Sorry for the delay My pc crashed (Damn technology!)

16. Mar 18, 2005

### Kahsi

Thank you zanazzi78. Now I understand :)

17. Mar 18, 2005

### Staff: Mentor

That's the right attitude. But you're almost done. Figure out (from geometry) the angle of incidence that the light makes with the bottom surface, then apply Snell's law one last time.