At what angle should the axes of two polaroids be placed so as to reduce the intensity of incident unpolarized light to `1//3`.

Answer

To solve the problem of determining the angle at which the axes of two polaroids should be placed to reduce the intensity of incident unpolarized light to \( \frac{1}{3} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- We have unpolarized light incident on the first polaroid. The intensity of this light is denoted as \( I_0 \).

2. **Intensity After First Polaroid**:
- When unpolarized light passes through a polaroid, the intensity is reduced to half. Therefore, the intensity after the first polaroid (let's call it \( I_1 \)) is given by:
\[ I_1 = \frac{I_0}{2} \]

3. **Intensity After Second Polaroid**:
- The second polaroid is at an angle \( \theta \) with respect to the first. The intensity after passing through the second polaroid (let's call it \( I_2 \)) can be calculated using Malus's Law:
\[ I_2 = I_1 \cos^2(\theta) \]
- Substituting \( I_1 \):
\[ I_2 = \frac{I_0}{2} \cos^2(\theta) \]

4. **Set the Condition for Final Intensity**:
- We want the final intensity \( I_2 \) to be \( \frac{I_0}{3} \):
\[ \frac{I_0}{2} \cos^2(\theta) = \frac{I_0}{3} \]

5. **Cancel \( I_0 \)**:
- Since \( I_0 \) is present on both sides, we can cancel it out:
\[ \frac{1}{2} \cos^2(\theta) = \frac{1}{3} \]

6. **Solve for \( \cos^2(\theta) \)**:
- Multiply both sides by 2:
\[ \cos^2(\theta) = \frac{2}{3} \]

7. **Find \( \theta \)**:
- Taking the square root of both sides gives:
\[ \cos(\theta) = \sqrt{\frac{2}{3}} \]
- To find the angle \( \theta \), take the inverse cosine:
\[ \theta = \cos^{-1}\left(\sqrt{\frac{2}{3}}\right) \]

### Final Answer:
The angle \( \theta \) at which the axes of the two polaroids should be placed is:
\[ \theta = \cos^{-1}\left(\sqrt{\frac{2}{3}}\right) \]