A source of light is placed in front of a screen. The intensity of light on the screen is $$I$$. Two Polaroids $$P_1$$ and $$P_2$$ are so placed in between the source of light and screen that the intensity of light on the screen is $$\frac{I}{2}$$. Then the $$P_2$$, should be rotated by an angle of _________ (degrees) so that the intensity of light on the screen becomes $$\frac{3I}{8}$$.

We begin with an un-polarised beam of light that produces an intensity $$I$$ on the screen when nothing obstructs it. Un-polarised light contains vibrations in all possible planes perpendicular to the direction of propagation.

When such a beam passes through a single ideal Polaroid, only the vibrations lying in the transmission axis of the Polaroid are allowed to pass. As a result, exactly one-half of the incident intensity survives. Stating this result,

$$I_{\text{after first Polaroid}} \;=\;\dfrac{I}{2}.$$

We now place a second Polaroid $$P_2$$ after the first one $$P_1$$. Let the angle between their transmission axes be $$\theta$$. For a beam already plane-polarised by the first Polaroid, the intensity transmitted by the second Polaroid is governed by Malus’ law.

Malus’ Law (to be stated explicitly): If a plane-polarised light of intensity $$I_0$$ is incident on a Polaroid and the angle between the light’s plane of polarisation and the Polaroid’s transmission axis is $$\theta$$, the emerging intensity $$I$$ is

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

Here the incident intensity on $$P_2$$ is $$I_0 = \dfrac{I}{2}$$ and the emergent intensity is given to be $$\dfrac{I}{2}$$. Substituting these values in Malus’ law, we have

$$\dfrac{I}{2} \;=\; \left(\dfrac{I}{2}\right)\cos^2\theta.$$

Dividing both sides by $$\dfrac{I}{2}$$ gives

$$\cos^2\theta \;=\; 1.$$

So $$\cos\theta = \pm 1$$ and therefore $$\theta = 0^\circ \ (\text{or }180^\circ).$$ Physically this means that the two Polaroids were initially parallel (their axes coincided), which is the simplest arrangement that keeps the intensity unchanged at $$\dfrac{I}{2}$$.

Next, we rotate only the second Polaroid $$P_2$$ through an additional angle $$\phi$$ while keeping $$P_1$$ fixed. The new angle between their axes now is $$\phi$$. After this rotation the first Polaroid still transmits $$\dfrac{I}{2}$$, and the second Polaroid again follows Malus’ law. Therefore the new intensity on the screen becomes

$$I_{\text{new}} \;=\;\left(\dfrac{I}{2}\right)\cos^2\phi.$$

According to the problem this new intensity must equal $$\dfrac{3I}{8}$$. Equating the two expressions,

$$\left(\dfrac{I}{2}\right)\cos^2\phi \;=\; \dfrac{3I}{8}.$$

We now cancel the common factor $$I$$ from both sides:

$$\dfrac{1}{2}\cos^2\phi \;=\; \dfrac{3}{8}.$$

Multiplying every term by 8 to clear the denominators:

$$4\cos^2\phi \;=\; 3.$$

Dividing by 4,

$$\cos^2\phi \;=\; \dfrac{3}{4}.$$

Taking the positive square root (since we are looking for a small physical rotation, $$0^\circ \le \phi \le 90^\circ$$),

$$\cos\phi \;=\; \dfrac{\sqrt{3}}{2}.$$

The angle whose cosine equals $$\dfrac{\sqrt{3}}{2}$$ is

$$\phi \;=\; 30^\circ.$$

Hence, the correct answer is Option 30°.