Optics Question 567
Question: In the visible region of the spectrum the rotation of the place of polarization is given by $ \theta =a+\frac{b}{{{\lambda }^{2}}} $ . The optical rotation produced by a particular material is found to be 30° per mm at $ \lambda =5000 $ Å and 50° per mm at $ \lambda =4000 {\AA} $ . The value of constant a will be
Options:
A) $ +\frac{50{}^\circ }{9} $ per mm
B) $ -\frac{50{}^\circ }{9} $ per mm
C) $ +\frac{9{}^\circ }{50} $ per mm
D) $ -\frac{9{}^\circ }{50} $ per mm
Show Answer
Answer:
Correct Answer: B
Solution:
$ \theta =a+\frac{b}{{{\lambda }^{2}}} $
$ 30=a+\frac{b}{{{(5000)}^{2}}} $ and $ 50=a+\frac{b}{{{(4000)}^{2}}} $ .
Solving for a, we get $ a=-\frac{50{}^\circ }{9}permm $
