Two beams, A and B of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through 30° makes the two beams appear equally bright. If the initial intensities of the two beams are $$I_A$$ and $$I_B$$ respectively, then $$\frac{I_A}{I_B}$$ equals:

We recall Malus’s law, which states that when plane-polarized light of incident intensity $$I_0$$ meets a polaroid whose transmission axis makes an angle $$\theta$$ with the light’s plane of polarization, the emerging intensity is

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

Let the transmission axis of the polaroid be initially aligned with the plane of polarization of beam A. Thus, for beam A the angle is $$\theta_A = 0^\circ$$, while for beam B, whose polarization is perpendicular to that of A, the angle is $$\theta_B = 90^\circ.$$

Applying Malus’s law to the initial position, we have

$$I_{A\,(\text{through})} = I_A \cos^2 0^\circ = I_A \times 1 = I_A,$$

$$I_{B\,(\text{through})} = I_B \cos^2 90^\circ = I_B \times 0 = 0.$$

This matches the given condition: beam A is brightest and beam B is extinguished.

Now the polaroid is rotated through $$30^\circ$$. The planes of polarization of the two beams stay fixed in space; only the axis of the polaroid turns. Consequently,

• For beam A, the new angle with the axis is $$\theta_A' = 30^\circ.$$
• For beam B, originally at $$90^\circ$$ from A, the new angle is $$\theta_B' = 90^\circ - 30^\circ = 60^\circ.$$

Using Malus’s law again, the transmitted intensities become

$$I_{A}' = I_A \cos^2 30^\circ,$$

$$I_{B}' = I_B \cos^2 60^\circ.$$

According to the statement, the two beams now appear equally bright, so

$$I_{A}' = I_{B}'.$$

Substituting the expressions,

$$I_A \cos^2 30^\circ = I_B \cos^2 60^\circ.$$

We insert the numerical values $$\cos 30^\circ = \frac{\sqrt3}{2}$$ and $$\cos 60^\circ = \frac12$$:

$$I_A\left(\frac{\sqrt3}{2}\right)^2 = I_B\left(\frac12\right)^2,$$

$$I_A \left(\frac{3}{4}\right) = I_B \left(\frac{1}{4}\right).$$

Dividing both sides by $$\dfrac14$$ gives

$$3I_A = I_B.$$

Rearranging, we find

$$\frac{I_A}{I_B} = \frac13.$$

Hence, the correct answer is Option D.