# Having trouble finding the angle of deflection of light

1. Dec 6, 2009

### kliker

we have this picture

http://img190.imageshack.us/img190/1128/60283381.jpg [Broken]

how can i find the angle of deflection?

i must show that it is: sin(A+δ)/2 = n*sinA/2

Last edited by a moderator: May 4, 2017
2. Dec 6, 2009

### kuruman

You need to apply Snell's Law at the the two interfaces. Note that since the incident angle with respect to the normal is the same as the exit angle with respect to the normal, symmetry demands that the ray travel parallel to the base inside the prism. Assume that the prism angle (at top) is known because the angle of deflection depends on it.

3. Dec 6, 2009

### kliker

thanks but how can we know if the incident angle with respect to the normal is the same as the exit angle?

it doesnt say anything like that, how can we assume this

also

i applied snell but i have sina/sinb = n and sinc/sind = 1/n

also i have proven that A = b + c

http://img130.imageshack.us/img130/6987/50944329.png [Broken]

im having trouble finding that sin(A+δ)/2 = n*sinA/2 :(

thank you again

Last edited by a moderator: May 4, 2017
4. Dec 6, 2009

### kuruman

As best as I can say from your posted diagram at

http://img190.imageshack.us/img190/1128/60283381.jpg [Broken]

both incident and exit angles are labeled θa. Is this not true?

Last edited by a moderator: May 4, 2017
5. Dec 6, 2009

### kliker

no I think the second one is θb where b = β(the greek letter)

6. Dec 6, 2009

### kliker

if we supposed that it is the same how could we reach the desired solution?

7. Dec 6, 2009

### ehild

Referring to your drawing, it is easy to find that the deflection is a+d-A.
The formula you have to prove is valid only for the minimal deflection. The deflection is minimal when the incident and the exit rays are symmetrical: a=d.

ehild

8. Dec 6, 2009

### kliker

ok guys i figured everything out

the teacher made a mistake it was θα the second one too, it was easy to solve i thought they were different

thanks everyone