Question

The index of refraction of quartz is anisotropic. Suppose that visible light is passing from one grain to another of different crystallographic orientation and at normal incidence to the grain boundary. Calculate the reflectivity at the boundary if the indices of refraction for the two grains are 1.544 and 1.553.

Answer 1

Answer:

The reflectivity at the boundary of the two grains is $2.9$×${10}^{-3}$

Step-by-step explanation:

Given :

Refractive index of one grains ($n_{1}$) = 1.553

Refractive index of second grains ($n_{2}$) = 1.544

According to the fresnel equation,

∴ general equation of refraction ( reflectivity ) $r = \frac{ n_{1} cos\alpha _{i} - n_{2}cos_{t} }{ n_{1} cos\alpha _{i} + n_{2}cos_{t} }$

where $cos\alpha _{i}$ = incident angle, $cos\alpha _{t}$ = transmitted angle

but for normal incidence $\alpha = 0$°

so our equation become,

⇒   $r = \frac{n_{1} - n_{2} }{n_{1} + n_{2}}$

     $r = \frac{1.553 - 1.544}{1.553 + 1.544}$

     $r = \frac{0.009}{3.097}$

     $r = 2.9$×$10^{-3}$

Therefor reflectivity at the boundary is given by $r = 2.9$×$10^{-3}$.