Figuring Out the Refraction Angle of Light Through a Glasprisma

[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

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

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.

 

[Doc Al]

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.

 

[Kahsi]

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

 

[Doc Al]

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

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

 

[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

 

[Doc Al]

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.

Don't guess, calculate.

Maybe it should be something like this.

http://home.tiscali.se/21355861/bilder/prisma3.JPG

Exactly.

 

[Kahsi]

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

 

[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

 

[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

 

[zanazzi78]

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

 

[Kahsi]

Look at the attachment.

http://home.tiscali.se/21355861/bilder/picture.PNG

 

[zanazzi78]

Oops here you go

 

[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
http://213.214.228.251/upp/upload_css.php

Choose "Bilder".

 

[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.

 

[zanazzi78]

Uploaded!

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

 

[Kahsi]

Thank you zanazzi78. Now I understand :)

 

[Doc Al]

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

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.