When an unpolarized light of intensity `I_0` is incident on a polarizing sheet, the intensity of the light which dows not get transmitted is

Options

A. x
B. `x/2`
C. `x/4`
D. Zero

Answer

To solve the problem, we need to determine the intensity of light that does not get transmitted through a polarizing sheet when unpolarized light of intensity \( I_0 \) is incident on it.

### Step-by-Step Solution:

1. **Understanding Unpolarized Light**:
- Unpolarized light consists of waves vibrating in multiple planes. When it passes through a polarizing sheet, only the component of light aligned with the axis of the polarizer will be transmitted.

2. **Intensity Reduction**:
- According to Malus's Law, when unpolarized light passes through a polarizing filter, the intensity of the transmitted light is reduced to half of the incident intensity. Therefore, the transmitted intensity \( I_t \) can be expressed as:
\[ I_t = \frac{I_0}{2} \]

3. **Calculating Non-Transmitted Intensity**:
- The intensity of light that does not get transmitted through the polarizing sheet can be calculated by subtracting the transmitted intensity from the initial intensity:
\[ I_{not\ transmitted} = I_0 - I_t \]
- Substituting the expression for \( I_t \):
\[ I_{not\ transmitted} = I_0 - \frac{I_0}{2} \]

4. **Simplifying the Expression**:
- Simplifying the equation gives:
\[ I_{not\ transmitted} = \frac{I_0}{2} \]

5. **Conclusion**:
- Therefore, the intensity of light that does not get transmitted through the polarizing sheet is:
\[ I_{not\ transmitted} = \frac{I_0}{2} \]

### Final Answer:
The intensity of light which does not get transmitted is \( \frac{I_0}{2} \).