A cylindrical tube of length \(0.2 \mathrm{~m}\) and radius \(R\) with sugar solution of concentration \(C\) produce a rotation of \(\theta\) in the plane of vibration of a plane polarized light. The same sugar solution is transferred to
another tube of length \(0.3 \mathrm{~m}\) of same radius. The remaining gap is filled by distilled water. Now the optical rotation produced is

We know the specific rotation,
\(S=\frac{\theta}{l \cdot C}=\frac{\theta \cdot v}{l \cdot m}\)
where \(\theta=\) rotation produced
\(I=\) length of the tube
\(m=\) mass of sugar dissolved
\(V=\) volume of sugar solution
\(\Rightarrow \quad \theta=\frac{S \cdot 1 \cdot m}{V}\)
For first case, \(\quad \theta=\frac{S / m}{V}...(i)\)
For second case, \(\quad \theta_{1}=\frac{S I_{1} m_{1}}{V_{1}}...(ii)\)
Here, \(\quad I=0.2 \mathrm{~m}\) and \(\quad l_{1}=0.3 \mathrm{~m}\) also, \(\quad V_{1}=V+\frac{V}{2}=1.5 V=\frac{3}{2} V\)
On dividing Eq. (ii) by Eq. (i), we get
\(\begin{aligned}
\quad \frac{\theta_{1}}{\theta}=& \frac{\frac{S l_{1} m}{V_{1}}}{\frac{S I m}{V}}=\frac{l_{1}}{l} \times \frac{V}{V_{1}}=\frac{l_{1}}{l} \times \frac{2 V}{3 V}=\frac{0.3}{0.2} \times \frac{2}{3}=1 \\
\Rightarrow \quad \theta_{1} &=\theta
\end{aligned}\)