How to measure water and organic enamel volumes from enamel birefringence?

How to measure water and organic enamel volumes from enamel birefringence? An example follows below.

First, enamel ground sections have to be microradiographed to obtain mineral volume at selected histological sites (~ 15x15 micrometers). Given a mineral volume fraction (V1) of 0.80 (80 %) – pore volume fraction (V2) of 0.20 (20%) - and observed birefringence in water of – 0.0020 (measured under polarizing microcopy), the theoretical observed birefringence is given by (Sousa et al., J Microsc 221: 79-83, 2006):

-0.0065*0.85+[0.80*0,2*(1.62²-n2²)²]/{2*(0.8*1.62+0.20*n₂)*[(1+0.80)*n₂² + 0.20*1.62²]}= -0.002    (1)

Where -0.0065 is the corrected intrinsic birefringence of the mineral phase and 0.85 is the mean crystallite alignment. 1.62 is the refractive index (n1) of hydroxyapatite. The variable n2 is given by (Sousa et al., J Microsc 221: 79-83, 2006; Macena et al., Arch Oral Biol, 2014):

n₂ = [1.33*(a₁/V₂)]+[n_i*(a₂/V₂)]+[1.56*(b/V₂)]  (2)

Where alpha1 and alpha2 are the firmly and loosely bound water volumes (alpha1 + alpha2 = total water volume, alpha), respectively, and ni is the refractive index of the immersion medium (water in this case; 1.33). Beta is the organic volume fraction. Thus

n₂ = [1.33*(a₁/0.20)]+[n_i*(a₂/0.20)]+[1.56*(b/0.20)]   (3)

Which is inserted into Eq.(1). And

V₂ = a + b   (4)

The next step is to consider that there are two unknowns (beta and alpha) and two equations: (1) and (4). Thus, the values can be found. The results are alpha = 7.7535% and beta = 12.2465%. So, V1 + alpha +beta = 100%.