Light from sodium lamp is made to pass through two polaroids placed one after the other in the path of light. Taking the intensity of the incident light as `100%`, the intensity of the out coming light that can be varied in the range

Options

A. `0% " to " 100%`
B. `0% " to " 50%`
C. `0% " to " 25%`
D. `0% " to " 75%`

Explanation

To solve the problem, we need to determine the intensity of light that emerges after passing through two polaroids, given that the intensity of the incident light is 100%. 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 incident intensity,
- \( \theta \) is the angle between the light's polarization direction and the axis of the polarizer.

### Step-by-step Solution:

1. **Identify the Incident Intensity**:
- The intensity of the incident light from the sodium lamp is given as \( I_0 = 100\% \).

2. **First Polaroid**:
- When the light passes through the first polaroid, the intensity is reduced to half. According to the properties of polaroids, we have:
\[ I_1 = \frac{I_0}{2} = \frac{100\%}{2} = 50\% \]

3. **Second Polaroid**:
- Let \( \theta \) be the angle between the transmission axes of the two polaroids. The intensity after the second polaroid can be calculated using Malus's Law:
\[ I_2 = I_1 \cos^2(\theta) \]
- Substituting \( I_1 \):
\[ I_2 = \left(\frac{I_0}{2}\right) \cos^2(\theta) = \left(\frac{100\%}{2}\right) \cos^2(\theta) = 50\% \cos^2(\theta) \]

4. **Determine the Range of \( I_2 \)**:
- The angle \( \theta \) can vary from \( 0^\circ \) to \( 90^\circ \).
- At \( \theta = 0^\circ \):
\[ I_2 = 50\% \cos^2(0) = 50\% \cdot 1 = 50\% \]
- At \( \theta = 90^\circ \):
\[ I_2 = 50\% \cos^2(90) = 50\% \cdot 0 = 0\% \]
- Therefore, as \( \theta \) varies from \( 0^\circ \) to \( 90^\circ \), \( I_2 \) varies from \( 50\% \) to \( 0\% \).

5. **Conclusion**:
- The intensity of the outgoing light can vary in the range from \( 0\% \) to \( 50\% \).

### Final Answer:
The intensity of the outgoing light can be varied in the range of **0% to 50%**.