Light refracted trough a prism

[xinlan]

Homework Statement

Light is incident along the normal to face AB of a glass prism of refractive index 1.47

Find the alpha max, the largest value of the angle alpha such that no light is refracted out of the prism at face AC if the prism is immersed in air.

Homework Equations

n1*sin thetha1 = n2 * sin thetha2

The Attempt at a Solution

since no light is refracted out of the prism, then total internal reflection is involved.
sin critical alpha = (n2/n1)*sin 90
= 1.33/147
alpha = 64.8

but I got it wrong,
please someone help me..
thanks..

 

[Doc Al]

What you found is the angle that the incident ray makes with the normal to surface A-C. That's not alpha, but you can use it to find alpha.

 

[xinlan]

I just got it..
I just realized that the n for air should be 1.00 not 1.33.
then to get the alpha --> 90 - the angle that the incident ray makes with the normal to surface A-C..

anyway.. thanks.. :)

 

[Doc Al]

I just got it..

Good!

I just realized that the n for air should be 1.00 not 1.33.

D'oh! (I didn't even notice that. I must have been sleeping.)

then to get the alpha --> 90 - the angle that the incident ray makes with the normal to surface A-C..

Right.

 

[blue_shark]

can someone explain a little more abour this problem ... i do no understand why is necessary to use 90 ... i just got one answer but its wrong this is the feedbak ... You have given the angle between the light ray and the normal to the surface AB, not the angle for alpha

 

[Doc Al]

can someone explain a little more abour this problem ... i do no understand why is necessary to use 90 ... i just got one answer but its wrong this is the feedbak ... You have given the angle between the light ray and the normal to the surface AB, not the angle for alpha

What you gave, I presume, is the critical angle, which is not the angle alpha. You've got to use the critical angle to figure out alpha. Draw a clear diagram and look for the triangles.