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 ntensity), a rotation of polaroid through `30^circ` makes the two beams appear equally bright. If the initial intensities of the two beams are `I_A` and `I_B` respectively, then `(I_A)/(I_B)` equals:

Options

A. 3
B. `(3)/(2)`
C. 1
D. `(1)/(3)`

Explanation

To solve the problem, we will use Malus's Law, which states that when polarized light passes through a polarizer, the intensity of the transmitted light is given by:

\[ I = I_0 \cos^2(\theta) \]

where:
- \( I \) is the transmitted intensity,
- \( I_0 \) is the initial intensity of the polarized light,
- \( \theta \) is the angle between the light's polarization direction and the axis of the polarizer.

### Step-by-Step Solution:

1. **Initial Condition**:
- We have two beams of light, A and B, with initial intensities \( I_A \) and \( I_B \) respectively. The planes of polarization of these beams are mutually perpendicular.

2. **Maximum Intensity for Beam A**:
- When the polarizer is aligned with beam A, the intensity of beam A is maximum, which means:
\[ I_A' = I_A \quad \text{(when the polarizer is aligned with A)} \]
- The intensity of beam B at this position is zero since its plane of polarization is perpendicular to the polarizer.

3. **Rotating the Polarizer**:
- When the polarizer is rotated by \( 30^\circ \), the intensity of beam A becomes:
\[ I_A' = I_A \cos^2(30^\circ) \]
- The intensity of beam B becomes:
\[ I_B' = I_B \cos^2(60^\circ) \]

4. **Using Malus's Law**:
- We can calculate the new intensities:
\[ I_A' = I_A \left(\frac{\sqrt{3}}{2}\right)^2 = I_A \cdot \frac{3}{4} \]
\[ I_B' = I_B \left(\frac{1}{2}\right)^2 = I_B \cdot \frac{1}{4} \]

5. **Setting Intensities Equal**:
- According to the problem, after rotating the polarizer, the two beams appear equally bright:
\[ I_A' = I_B' \]
- Therefore, we can set the equations equal to each other:
\[ I_A \cdot \frac{3}{4} = I_B \cdot \frac{1}{4} \]

6. **Solving for the Ratio**:
- Rearranging the equation gives:
\[ \frac{I_A}{I_B} = \frac{1/4}{3/4} = \frac{1}{3} \]

### Final Answer:
Thus, the ratio of the initial intensities of the two beams is:
\[ \frac{I_A}{I_B} = \frac{1}{3} \]