1. Intensity after the polarizer

Since the incoming light is unpolarized and the polarizer-analyser system is assumed to have no absorption,

• the polarizer transmits half the original intensity.

So if original intensity is $I_0$, then after the polarizer: $I_p = \frac{I_0}{2}$

2. Use Malus' law for the analyser

If the angle between the transmission axes of polarizer and analyser is $\theta$, then the intensity after the analyser is $I = I_p \cos^2\theta = \frac{I_0}{2}\cos^2\theta$

Given that this is $10\%$ of the original intensity: $\frac{I_0}{2}\cos^2\theta = 0.1 I_0$

Cancel $I_0$: $\frac{1}{2}\cos^2\theta = 0.1$ $\cos^2\theta = 0.2$

Thus, $\cos\theta = \sqrt{0.2} = \frac{1}{\sqrt{5}}$

So, $\theta = \cos^{-1}\left(\frac{1}{\sqrt{5}}\right) \approx 63.4^\circ$

3. Condition for zero intensity

For zero intensity after the analyser, the analyser must be perpendicular to the polarizer axis: $\theta_{zero} = 90^\circ$

Therefore, the analyser must be rotated further by $\Delta \theta = 90^\circ - 63.4^\circ = 26.6^\circ$

4. Compare with options

The required angle is $26.6^\circ$

This value is not present in the given options.

Hence none of the listed options is correct. In particular, option C ($18.4^\circ$) is not correct.

5. Possible source of the stored answer

If one incorrectly assumes $I = I_0\cos^2\theta = 0.1 I_0$ then $\cos^2\theta = 0.1 \Rightarrow \theta \approx 71.6^\circ$ so further rotation to $90^\circ$ would be $90^\circ - 71.6^\circ = 18.4^\circ$ which matches option C.

But this ignores the factor $\tfrac{1}{2}$ loss at the polarizer for unpolarized incident light, so it is not physically correct here.