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 ${30}^0$ per $mm$ at $\lambda =5000A^o$ and ${50}^0$ per $mm$ at $\lambda =4000A^o$. The value of constant $a$ will be

• A. $+\frac{{50}^0}{9}$ per $mm$
• B. $-\frac{{50}^0}{9}$ per $mm$
• C. $+\frac{9^0}{50}$ per $mm$
• D. $-\frac{9^0}{50}$ per $mm$

Answer 1

Answer: (B)

$-\frac{{50}^0}{9}$ per $mm$

$\text{Solution}$

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

$⟹30=a+\frac{b}{{5000}^2}$ and $50=a+\frac{b}{{4000}^2}$

Solving for a, we get

$a=\frac{{50}^o}{9}permm$

$\text{Hence B is the correct option}$