Explanation

(d) For liquid A
$L_{1} = 20\ cm,\ \theta_{1} = 38{^\circ};concentration = C_{1}$
Specific rotation $a_{1} = \frac{\theta_{1}}{L_{1}C_{1}} = \frac{38{^\circ}}{20\ \times \ C_{1}}$
Similarly, for liquid B
$L_{2} = 30\ cm,\ \theta_{2} = - 24{^\circ},\ concentration = C_{2}$
Specific rotation $a_{2} = \frac{\theta_{2}}{L_{2}C_{2}} = \frac{( - 24{^\circ})}{30\ \times \ C_{2}}$
The mixture has 1 part of liquid A and 2 parts of liquid B,
$\therefore C_{1}^{'}:C_{2}^{'} = 1\ :2$
$\theta = \left\{ \ a_{1}C_{1}^{'} + a_{2}C_{2}^{'} \right\} l$
$= \left\{ \frac{38{^\circ}}{20 \times C_{1}} \times \frac{C_{1}}{3} + \frac{( - 24{^\circ})}{30 \times C_{2}} \times \frac{2C_{2}}{3} \right\} \times 30$
$= 19{^\circ} - 16{^\circ} = 3{^\circ}$
Thus, the optical rotation of mixture is $+ 3{^\circ}$ in right had direction.