Discussion
Light of wavelength 632 nm is passing through an optically active medium of thickness 20 cm. The optical rotation exhibited by the medium is 18°.
Which of the following options correctly states the magnitude of the difference in refractive indices corresponding to the left and the right circularly polarized light?
Answer: B
The difference in refractive indices is \(3.16 \times 10^{-7}\).
In an optically active medium the two circular components travel with indices \(n_L\) and \(n_R\). Over a thickness \(d\) they build up a phase difference \(\frac{2\pi}{\lambda}(n_L - n_R)d\), and the plane of polarisation turns through half of it:
\[ \begin{aligned} \theta &= \frac{\pi d}{\lambda}\,|n_L - n_R| \\ |n_L - n_R| &= \frac{\theta \lambda}{\pi d} \end{aligned} \]
With \(\theta = 18^\circ = \pi/10\) rad, \(\lambda = 632 \times 10^{-9}\) m and \(d = 0.20\) m:
\[ \begin{aligned} |n_L - n_R| &= \frac{(\pi/10)(632 \times 10^{-9})}{\pi \times 0.20} \\ &= 3.16 \times 10^{-7} \end{aligned} \]
Putting the angle in as 18 degrees instead of radians gives \(1.81 \times 10^{-5}\), and the other two options are exactly double these two values, so they do not satisfy the rotation formula either.
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