Pavilion Rays
Pavilion - Angles Of Reflected Rays
This figure shows a flat-top gem with a typical ray reflected off both sides of the pavilion so that it is returned to the viewer. Equating angles of incidence and reflection (relative to the surface normal) gives:
| P + q1 = q2 - P | .......... | q1 - q2 = - 2P | (1a) | |
| P + q3 = 180o - q2 - P | .......... | q3+q2 = 180o - 2P | (1b) | |
| Adding these eliminates q2, thus | .......... | q1+q3 = 180o - 4P | (1c) | |
| or | .......... | q1+q3 = 4(45o - P) |
Eq.(1c) shows that the angle (q1+ q3) between rays q1 and q3 is constant for each pavilion angle P; thus, as angle q1 decreases, angle q3 increases by the same amount. This means that one ray is at a maximum value when the other is at a minimum:
| q max+q min = 180o-4P | .......... | (1d) |
Pavilion - Limits Of Reflected Rays
A ray is reflected only if its angle to the surface normal is more than the critical angle C; thus, for reflection to occur:
| P+qmin = C | ......or...... | qmin = C-P | (2a) |
Substituting this into Eq.(1d) gives:
| qmax+ C-P = 180o- 4 P | .....thus..... | qmax = 180o-C-3P | (2b) |
Pavilion Angle Limits To Reflect All Incident Rays
Light incident to the table from all directions enters the stone within the critical angle. For all of this light to be reflected back to the table,
qmin <= - C, qmax >= + C;
thus, per Eq.(2a,b):
| -C >= (qmin = C-P) | .....whereby..... | P min = 2C | |
| +C <= (qmax = 180o-C-3P) | .....whereby..... | P max = 60o-2C/3 |
For diamond, C=24.43o; this gives:
P min = 48.86o, P max = 43.71o
The minimum is more than the maximum, which means that there is no solution for diamond which returns all incident light to the table (a solution exists only for C <= 22.5o = synthetic rutile).