OPTICAL PROPERTIES IN CRYSTALS 187 



the Z axis, the last of equations (79) shows that the birefringence is equal 

 to 



3 



5. = I VCft + Ai - ^2 - ^,y + 4(A6)2 (96) 



Since, for an isotropic substance /3i = /^2 , we have, after substituting the 

 \-alue of Ai and A2 , with the appropriate photoelastic constants from equa- 

 tion (95), (last tensor): 



3 



5. = I (tTu - TTis) V(7^] - 7^2)2 + 4r62 (97) 



If we transform to axes rotated by an angle d about Z, the values of Tn 

 and r22 are given by 



Tn = cos 207^1 + 2 sin cos OT^ + sin^ dT^ 



Til = sin ""QTx — 2 sin cos QT^ -f cos ^T<i, 

 If, now, we choose the angle Q so that Tn is a maximum, we find 



(98) 



tan IQ = -±^ (99) 



-i 1 — -t 2 



Inserting this value of tan 20 in (98) we find 



t', = ^^"^ ^' - W{Ti - T,f + ^T,^ 



(100) 



and, hence, 



t[ - rj = ^y{T^ - T^y + 47^62 (101) 



Hence the birefringence obtained in stressing a material is proportional to 

 the difference in the principal stresses. By observing the isoclinic lines of a 

 photoelastic picture, methods^ are available for determining the stresses 

 in a model. A photograph^ of a stressed disk is shown by Fig. 3. The high 

 concentration of lines near the surface shows that the shearing stress is 

 \ery high at these points. By counting the number of fines from the edge 

 and knowing the stress optical constant, the stress can be calculated at any 

 point. 



If we apply a single stress Ti , the birefringence is given by the equation 



3 

 Bz= ^ (tth - 7ri2)ri (102) 



^ See Photoelasticity, Coker and Filon, Cambridge University Press, 1931. 

 ^ This photograph was taken by T. F. Osmer. 



