# Snell's Law variation

• Apr 19th 2007, 12:49 PM
perrien
Snell's Law variation
I'm a hobby gem faceter and am trying to figure out some of the math involved. What I want to know is assuming I have a piece of glass, refractive index of 1.54 (or anything for that matter) at what angle would the light need to enter it to be deviated by a certain amount?

For instance, using Snell's law, I know that if a beam of light goes into glass at a 45 degree angle (from the norm), it will travel through the glass at about 27 degrees. The light has deviated by 18 degrees. What angle would I need it to enter if I wanted it to deviate by 25 degrees?

Thanks for any help.
• Apr 19th 2007, 01:11 PM
Jhevon
Quote:

Originally Posted by perrien
I'm a hobby gem faceter and am trying to figure out some of the math involved. What I want to know is assuming I have a piece of glass, refractive index of 1.54 (or anything for that matter) at what angle would the light need to enter it to be deviated by a certain amount?

For instance, using Snell's law, I know that if a beam of light goes into glass at a 45 degree angle (from the norm), it will travel through the glass at about 27 degrees. The light has deviated by 18 degrees. What angle would I need it to enter if I wanted it to deviate by 25 degrees?

Thanks for any help.

Remember what Snell's law says:

sin(x1)/sin(x2) = v1/v2 = L1/L2 = n2/n1

where: - x1 is the angle between the normal and the ray of light when entering the new material
- x2 is the angle between the normal and the ray of light when after entering the new material
- v1 is the speed of light in material 1
- v2 is the speed of light in material 2
- L1 is the wavelength of light in material 1
- L2 is the wavelength of light in material 2
- n1 is the refractive index of material 1
- n2 is the refractive index of material 2

i suppose you're going from air to glass. the refractive index of air is 1.0008, let's just use 1

you only need
sin(x1)/sin(x2) = n2/n1
for your question. so fill in everything except what you want to find.

you want an angle of 25 degrees after entering the new material, so

sin(x1) = (n2/n1) * sin(x2)
= 1.54/1 * sin(25)
= 0.650832123

=> x1 = arcsin(0.650832123) = 40.6 degrees ......this is the angle you must enter the new medium at
• Apr 19th 2007, 01:23 PM
perrien
Quote:

you only need
sin(x1)/sin(x2) = n2/n1
for your question. so fill in everything except what you want to find.

you want an angle of 25 degrees after entering the new material, so

....
=> x1 = arcsin(0.650832123) = 40.6 degrees ......this is the angle you must enter the new medium at
I've got that much but that's not what I'm asking. What I need to know is at what angle must the light enter for it to CHANGE by 25 degrees. In the example above, the change was 15 degrees.
• Apr 19th 2007, 02:18 PM
Jhevon
Quote:

Originally Posted by perrien
I've got that much but that's not what I'm asking. What I need to know is at what angle must the light enter for it to CHANGE by 25 degrees. In the example above, the change was 15 degrees.

well in that case the problem becomes a bit more complicated. i'm in a rush not so i can't do everything.

let's say you want it to increase by 25 degrees, then you'd solve:

sin(x1)/sin(x1 + 25) = n2/n1

if you want to decrease by 25, solve:
sin(x1)/sin(x1 - 25) = n2/n1

let's see how the first works out:

sin(x1)/sin(x1 + 25) = n2/n1
=> sin(x1) = n2sin(x1 + 25)/n1
=> sin(x1) = n2(sin(x1)cos(25) + sin(25)cos(x1))/n1
=> sin(x1) = [(n2/n1)*cos(25)]sin(x1) + [(n2/n1)*sin(25)]cos(x1)
=> [(n2/n1)*cos(25)]sin(x1) + [(n2/n1)*sin(25)]cos(x1) - sin(x1) = 0
=> [(n2/n1)*cos(25) - 1]sin(x1) + [(n2/n1)*sin(25)]cos(x1) = 0
=> [(n2/n1)*cos(25) - 1]sin(x1) + [(n2/n1)*sin(25)]sin(90 - x1) = 0
.
.
.
so this will get messy, i can't see a better way to do it at the moment, neither can i finish this, i REALLY have to leave now.

see where you get with this, or a similar approach, good luck
• Apr 19th 2007, 02:25 PM
topsquark
Quote:

Originally Posted by Jhevon
well in that case the problem becomes a bit more complicated. i'm in a rush not so i can't do everything.

let's say you want it to increase by 25 degrees, then you'd solve:

sin(x1)/sin(x1 + 25) = n2/n1

if you want to decrease by 25, solve:
sin(x1)/sin(x1 - 25) = n2/n1

let's see how the first works out:

sin(x1)/sin(x1 + 25) = n2/n1
=> sin(x1) = n2sin(x1 + 25)/n1
=> sin(x1) = n2(sin(x1)cos(25) + sin(25)cos(x1))/n1
=> sin(x1) = [(n2/n1)*cos(25)]sin(x1) + [(n2/n1)*sin(25)]cos(x1)
=> [(n2/n1)*cos(25)]sin(x1) + [(n2/n1)*sin(25)]cos(x1) - sin(x1) = 0
=> [(n2/n1)*cos(25) - 1]sin(x1) + [(n2/n1)*sin(25)]cos(x1) = 0

I'm going to pick up here, but I'm going to change Jhevon's last line a bit. Divide both sides by cos(x1):
=> [(n2/n1)*cos(25) - 1]sin(x1)/cos(x1) + [(n2/n1)*sin(25)] = 0

Then, of course, sin(x1)/cos(x1) = tan(x1):
=> [(n2/n1)*cos(25) - 1]tan(x1) + [(n2/n1)*sin(25)] = 0

=> [(n2/n1)*cos(25) - 1]tan(x1) = -[(n2/n1)*sin(25)]

=> tan(x1) = -[(n2/n1)*sin(25)]/[(n2/n1)*cos(25) - 1]

=> tan(x1) = [(n2/n1)*sin(25)]/[1 - (n2/n1)*cos(25)]
(Note: This is positive since (n2/n1)*cos(25) < 1 for this problem.)

=> x1 = atn{[(n2/n1)*sin(25)]/[1 - (n2/n1)*cos(25)]}

-Dan
• Apr 20th 2007, 01:21 PM
perrien
Quote:

Originally Posted by topsquark
=> x1 = atn{[(n2/n1)*sin(25)]/[1 - (n2/n1)*cos(25)]}

That's exactly what I was looking for. I've been trying to figure this out for quite a while and the trig was just a bit above what I remember. Many thanks.