Question

In the visible region of the spectrum the rotation of the plane of polarization is given by
$\theta =a+\frac{b}{{\lambda }^2}$
The optical rotation produced by a particular material is found to be ${30}^o$ per mm at $\lambda =5000\mathring{A}$ and ${50}^o$ per mm at $\lambda =4000\mathring{A}$. The value of constant $a$ will be

• A. $+\frac{{50}^o}{9}$ per mm
• B. $-\frac{{50}^o}{9}$ per mm
• C. $+\frac{9^o}{50}$ per mm
• D. $-\frac{9^o}{50}$ per mm

Answer 1

The correct option is B $-\frac{{50}^o}{9}$ per mm

Given

$\theta =a+\frac{b}{{\lambda }^2}$

Substituting values,

$30=a+\frac{b}{(5000\times {10}^7{)}^2}...........\left(i\right)$

$50=a+\frac{b}{(4000\times {10}^7{)}^2}...........\left(ii\right)$

(i) / $16$ - (ii) / $25$ gives:

$\frac{30}{16}-\frac{50}{25}=\frac{a}{16}-\frac{a}{25}$

$30\times 25-50\times 16=9a$

$a=-\frac{{50}^o}{9}/mm$