

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 = 24°26'. But, angle NRQ = angle QAR  a
as AQ and QR, AR and RN are perpendicular. Therefore,
for total reflection
of a vertical ray,




Second Reflection




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.




Then
angle QRV = angle TQR = 24°26' angle VRN = a.


As before, in triangle RSC,  
angle SRC = 90°
 NRS = 90°  a  24°26'


angle RCS = 2 a (90°  a)


angle RSC = 90°  c.


Then  
90  a  24°26' + 180°  2a
+ 90°  c = 180°


3a + x = 180°  24°26' = 155°34'


3a = 155°34'  c.


In the case now considered,  
c = 24°26'.


Then  
3a = 155°34'  24°26' = 131°
8'


a = 43°43'
(10)




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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