A polarized light of intensity `I_(0)` is passed through another polarizer whose pass axis makes an angle of `60^(@)` with the pass axis of the former, What is the intensity of emergent polarized light from second polarizer?

Options

A. `I=I_(0)`
B. `I=(I_(0))/(6)`
C. `I=(I_(0))/(5)`
D. `(I_(0))/(4)`

Answer

To solve the problem, we will use Malus's Law, which describes how the intensity of polarized light changes as it passes through a polarizer.

### Step-by-Step Solution:

1. **Understand Malus's Law**:
Malus's Law states that when polarized light passes through a polarizer, the intensity of the transmitted light (I) is given by:
\[ I = I_0 \cos^2(\theta) \]
where:
- \(I_0\) is the intensity of the incident polarized light,
- \(\theta\) is the angle between the light's polarization direction and the axis of the polarizer.

2. **Identify Given Values**:
- The intensity of the incident light: \(I_0\)
- The angle between the pass axis of the first polarizer and the second polarizer: \(\theta = 60^\circ\)

3. **Apply Malus's Law**:
Substitute the known values into Malus's Law:
\[ I = I_0 \cos^2(60^\circ) \]

4. **Calculate \(\cos(60^\circ)\)**:
We know that:
\[ \cos(60^\circ) = \frac{1}{2} \]
Therefore:
\[ \cos^2(60^\circ) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

5. **Substitute Back into the Equation**:
Now substitute \(\cos^2(60^\circ)\) back into the equation:
\[ I = I_0 \cdot \frac{1}{4} \]

6. **Final Result**:
Thus, the intensity of the emergent polarized light from the second polarizer is:
\[ I = \frac{I_0}{4} \]

### Conclusion:
The intensity of the emergent polarized light from the second polarizer is \(\frac{I_0}{4}\).

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