Back of the Brilliant
 
  Let us now pass to the consideration of the other alternative, i.e. where the top surface is a horizontal plane AB and where the bottom surface AC is inclined at an angle (to the horizontal (fig. 25)). As before, we have to introduce a third plane BC to have a symmetrical section.  
  First Reflection  
  Let a vertical ray PQ strike AB. As the angle of incidence is zero, it passes into the stone without refraction and meets plane AC at R. Let RN be the normal at that point, then, for total reflection to occur, angle NRQ = 2426'. But, angle NRQ = angle QAR - a as AQ and QR, AR and RN are perpendicular. Therefore, for total reflection of a vertical ray,
 
Fig. 25
a = 2426'. Let us now incline the ray PQ so that it gradually changes from a vertical to a horizontal direction, and let P'Q' be such a ray. Upon passing into the diamond it is refracted, and strikes AC at an angle Q'R'N' where R'N' is the normal to AC. When P'Q' becomes horizontal, the angle of refraction T'Q'R' becomes equal to 2426'. This is the extreme value attainable by that angle; also, for total reflection, angle Q'R'N' must not be less than 2426'. If we draw R'V, vertical angle VR'Q' = R'Q'T' = 2426', and angle VR'N' = VR'Q' + Q'R'N' =2426' + 2426' = 4852' as before, a = angle VR'N', and therefore a = 4852 (9).
 
 

For absolute total reflection to occur at the first facet, the inclined facets must make an angle of not less that 4852' with the horizontal.

 
 
Second Reflection
 
 
 
Fig. 26
When the tray of light is reflected from the first inclined facet AC (fig. 26), it strikes the opposite one BC. Here too the light must be totally reflected, for otherwise there would be a leakage of light through the back of gem stone. Let us consider, in the first instance, a ray of light vertically incident upon the stone. The path of the ray will be PQRST. If RN and SN' are the normals at R and S respectively, then for total reflection, angle N'SR = 2426'. Let us find the value of a to fulfil that condition: angle QRN = angle QAR = a as having perpendicular sides. angle SRN = angle QRN as angles of incidence and reflection. Therefore angle NRS = a. Now let angle N'SR = c. Then, in triangle RSC,
 
 
angle SRC = 90 - a
 
 
angle RSC = 90 - c
 
 
angle RCS = 2 x angle RCM = 2 x ARQ = 2 (90 - a).
 
  The sum of these three angles equals two right angles,  
 
90 - a + 90 - c + 180 - 2a = 180, or
 
 
3a+ c = 180 3a = 180 - c.
 
 
 
Fig. 27


Now, c is not less than 2426', therefore a is not greater than a = (180 - 2426')/3 = 5151'. Let us again incline PQ from the vertical until it becomes horizontal, but in this case in the other direction, to obtain the inferior limit. Then (fig.27) the path will be PQRS. Let QT, RN, SN' be the normals at Q, R, and S respectively. At the extreme case, TQR will be 2426'. Draw RV vertical at R.

 
 
Then angle QRV = angle TQR = 2426' angle VRN = a.
 
  As before, in triangle RSC,  
 
angle SRC = 90 - NRS = 90 - a - 2426'
 
 
angle RCS = 2 a (90 - a)
 
 
angle RSC = 90 - c.
 
  Then  
 
90 - a - 2426' + 180 - 2a + 90 - c = 180
 
 
3a + x = 180 - 2426' = 15534'
 
 
3a = 15534' - c.
 
  In the case now considered,  
 
c = 2426'.
 
  Then  
 
3a = 15534' - 2426' = 131 8'
 
 
a = 4343' (10)
 
 

For absolute total reflection at the second facet, the inclined facets must make an angle of not more than 4343' with the horizontal.

 
 
We will note here that this condition and the one arrived at on "First Reflection" are in opposition. We will discuss this later, and will pass now to considerations of refraction.
 
  Author: Marcel TOLKOWSKY.
This material is a part of "Diamond Design" book,
was published E & F.N. Spon, Ltd., London, 1919.
 
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