A beam of unpolarised light of intensity $$I_0$$ is passed through a polaroid A and then through another polaroid B which is oriented so that its principle plane makes an angle of 45° relative to that of A. The intensity of the emergent light is:

We begin with an unpolarised light beam whose initial intensity is denoted by $$I_0$$.

First, the beam passes through polaroid A. For unpolarised light, the standard result is that a single ideal polaroid transmits exactly one-half of the incident intensity. We can state this as:

Formula: “When unpolarised light passes through one perfect polaroid, the transmitted intensity becomes half of the incident intensity.”

Using this, we write $$I_1 = \frac{I_0}{2},$$ where $$I_1$$ is the intensity of the light emerging from polaroid A.

Now this partially polarised light encounters polaroid B. The transmission through a second polaroid is governed by Malus’ Law.

Formula: “If polarised light of intensity $$I$$ is incident on a polaroid whose transmission axis makes an angle $$\theta$$ with the plane of polarisation of that light, the emergent intensity $$I'$$ is given by $$I' = I \cos^2\theta.$$”

Here, the transmission axis of B is at $$\theta = 45^\circ$$ with respect to that of A, and the intensity incident on B is $$I_1 = \dfrac{I_0}{2}.$$ Applying Malus’ Law gives

$$ I_2 = I_1 \cos^2 45^\circ = \frac{I_0}{2} \times \cos^2 45^\circ. $$

We know the numerical value $$\cos 45^\circ = \frac{1}{\sqrt2}.$$ Substituting this value step by step, we obtain

$$ I_2 = \frac{I_0}{2} \times \left(\frac{1}{\sqrt2}\right)^2 = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4}. $$

Thus the intensity of the light emerging from polaroid B is $$\dfrac{I_0}{4}.$$

Hence, the correct answer is Option A.