Question

A linearly polarized light of wavelength 590 nm is incident normally on the surface of a 20 μm thick quartz film. The plane of polarization makes an angle 30° with the optic axis. Refractive indices of ordinary and extraordinary waves differ by 0.0091, resulting in a phase difference of fπ between them after transmission. The value of f (rounded off to two decimal places) and the state of polarization of the transmitted light is

• A. 0.62 and linear
• B. 0.62 and elliptical
• C. -0.38 and elliptical
• D. 0.5 and circular

Answer 1

The Correct Option is B

To solve the problem, we need to determine the phase difference \(f\pi\) and the resulting state of polarization when linearly polarized light passes through the quartz film. 

1. First, calculate the phase difference induced due to the difference in refractive indices between the ordinary and extraordinary waves. The refractive index difference given is \(0.0091\).
2. The formula for the phase difference \(\Delta = \left( \frac{2\pi}{\lambda} \right) t (n_e - n_o)\), where \(\lambda\) is the wavelength, \(t\) is the thickness of the film, \(n_e\) and \(n_o\) are the refractive indices of the extraordinary and ordinary waves respectively.
3. Substitute the values \(\lambda = 590 \, \text{nm} = 590 \times 10^{-9} \, \text{m}\), \(t = 20 \, \mu\text{m} = 20 \times 10^{-6} \, \text{m}\), and \(n_e - n_o = 0.0091\) into the formula.
4. Calculate the phase difference: \(\Delta = \left( \frac{2\pi}{590 \times 10^{-9}} \right) \times (20 \times 10^{-6}) \times 0.0091 = \frac{2\pi \times 20 \times 0.0091}{590} = \frac{0.182\pi}{590} = 0.62 \pi\)
5. The phase difference becomes \(f\pi\) where \(f = 0.62\).
6. Now, let's consider the state of polarization:
7. Initially, the light is linearly polarized and makes an angle of \(30^\circ\) with the optic axis.
8. After passing through the quartz film, the components of the light (ordinary and extraordinary) accumulate a phase difference.
9. A non-integral multiple of \(\pi\) (\(0.62\pi\) here) generally results in elliptical polarization, as the phase difference between components leads to an elliptical trace in the field vector.

Based on these calculations, the correct answer is 0.62 and elliptical.