A polarizer - analyser set is adjusted such that the intensity of light coming out of the analyser is just 36% of the original intensity. Assuming that the polarizer - analyser set does not absorb any light, the angle by which the analyser needs to be rotated further, to reduce the output intensity to zero, is ($$\sin^{-1}\left(\frac{3}{5}\right) = 37^\circ$$)

Let the intensity of the unpolarised light incident on the polariser be $$I_0.$$

Because the statement tells us that the polariser-analyser system “does not absorb any light”, we treat the polariser as changing only the state of polarisation, not the intensity. Hence, after the light passes through the polariser its intensity is still $$I_0,$$ but it is now plane-polarised along the transmission axis of the polariser.

If the analyser’s transmission axis makes an angle $$\theta$$ with the polariser’s axis, Malus’ law gives the transmitted intensity

$$I \;=\; I_0 \,\cos^2\theta.$$

According to the question this intensity is 36 % of the original, so

$$I = 0.36\,I_0.$$

Substituting this value into Malus’ law, we have

$$0.36\,I_0 \;=\; I_0 \,\cos^2\theta.$$

Dividing both sides by $$I_0$$ eliminates the common factor:

$$\cos^2\theta = 0.36.$$

Taking the positive square root (the angle is between 0° and 90°),

$$\cos\theta = \sqrt{0.36} = 0.6 = \dfrac{3}{5}.$$

Hence

$$\theta = \cos^{-1}\!\left(\dfrac{3}{5}\right) = 53^\circ.$$

To make the transmitted intensity fall to zero, the analyser’s axis must become perpendicular to the polariser’s axis, i.e. the angle between them must be $$90^\circ.$$ At present the angle is $$\theta = 53^\circ.$$ Therefore the analyser must be rotated through an additional angle

$$\phi = 90^\circ - \theta = 90^\circ - 53^\circ = 37^\circ.$$

The numerical value is confirmed by the given relation $$\sin^{-1}\!\left(\dfrac{3}{5}\right) = 37^\circ.$$ So the analyser needs to be rotated through $$37^\circ$$ to extinguish the light completely.

Hence, the correct answer is Option B.