if the intensity of light is made `4I_0`, then the saturation current will become

Options

A. `0.40xx1muA`
B. `0.40xx2muA`
C. `0.40xx4muA`
D. `0.40xx8muA`

Explanation

To solve the problem, we need to understand the relationship between the intensity of light and the saturation current in the context of the photoelectric effect.

### Step-by-step Solution:

1. **Identify the Given Information**:
- Initial saturation current \( I_s = 0.4 \, \mu A \).
- Initial intensity \( I_0 \).
- New intensity \( I = 4I_0 \).

2. **Understand the Relationship**:
- The saturation current \( I_s \) is directly proportional to the intensity of light. This means:
\[ I_s \propto I \]
- If the intensity increases, the saturation current will also increase proportionally.

3. **Set Up the Proportionality**:
- Let \( k \) be the proportionality constant. Then we can write:
\[ I_s = k \cdot I \]
- For the initial case:
\[ I_{s1} = k \cdot I_0 \]
- For the new case with intensity \( 4I_0 \):
\[ I_{s2} = k \cdot (4I_0) \]

4. **Relate the Two Saturation Currents**:
- From the proportionality, we can express the new saturation current in terms of the old saturation current:
\[ I_{s2} = 4 \cdot I_{s1} \]

5. **Substitute the Known Values**:
- Substitute \( I_{s1} = 0.4 \, \mu A \):
\[ I_{s2} = 4 \cdot 0.4 \, \mu A \]

6. **Calculate the New Saturation Current**:
- Perform the multiplication:
\[ I_{s2} = 1.6 \, \mu A \]

### Final Answer:
The saturation current when the intensity of light is made \( 4I_0 \) will be \( 1.6 \, \mu A \).

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