Light cannot normally be transmitted through two coaxial polarising filters if their polarisation ax

Question

Light cannot normally be transmitted through two coaxial polarising filters if their polarisation axes are orthogonally oriented. However, light can be made to pass by placing a uniform plate of uniaxial birefringent material between and parallel to the filters. The plate has a refractive index $n_1$ for light polarized parallel to its optic axis $e$, and $n_2$ for light polarized perpendicular to $e$. 
The system is illuminated perpendicularly by unpolarized monochromatic light of wavelength $\lambda = 600$ nm. The difference in refractive indices is $|n_1 - n_2| = 0.005$. To maximize the fraction of incident light transmitted through the entire system, the optic axis $e$ is optimally oriented at 45° to the first polarizer's axis. 
Calculate the minimum required thickness $d$ of the birefringent plate (in $\mu$m) to achieve this maximum possible transmission.

Professor PC Thomas Classes · Answer

Let the initial unpolarized intensity be $I_0$. The first polarizer transmits plane-polarized light of intensity $I_1 = I_0 / 2$.
For the light to perfectly pass through the second (crossed) polarizer, the birefringent plate must rotate the plane of polarization by exactly 90°. A birefringent plate achieves this if its optic axis is at 45° to the incident polarization plane, and if it acts as a half-wave plate.
A half-wave plate introduces a phase difference of $\pi$ between the ordinary and extraordinary rays.
Phase difference $\Delta \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} |n_1 - n_2| d = \pi$
Therefore, the path difference must be $\lambda / 2$.
$|n_1 - n_2| d = \frac{\lambda}{2}$
$d = \frac{\lambda}{2 |n_1 - n_2|}$
Substitute the given values:
$d = \frac{600 	imes 10^{-9} 	ext{ m}}{2 	imes 0.005} = \frac{600 	imes 10^{-9}}{0.01} = 60 	imes 10^{-6} 	ext{ m}$
Thus, the minimum thickness is 60 $\mu$m.