In a photo cell 4 unit photo electric current is flowing, the distance between source and cathode is 4 unit. Now distance between source and cathode becomes 1 unit. What will be photo electric current now ?

Answer

To solve the problem, we need to understand the relationship between the photoelectric current and the distance from the light source to the cathode. The photoelectric current (I) is directly proportional to the intensity (I) of light, which is inversely proportional to the square of the distance (R) from the light source.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Initial photoelectric current (I1) = 4 units
- Initial distance (R1) = 4 units

2. **Identify the New Conditions**:
- New distance (R2) = 1 unit
- We need to find the new photoelectric current (I2).

3. **Understand the Relationship**:
- The intensity of light is inversely proportional to the square of the distance from the source:
\[ I \propto \frac{1}{R^2} \]
- Therefore, we can express the relationship between the two currents and their distances as:
\[ \frac{I1}{I2} = \frac{R2^2}{R1^2} \]

4. **Substitute Known Values**:
- Substitute the known values into the equation:
\[ \frac{4}{I2} = \frac{1^2}{4^2} \]
- This simplifies to:
\[ \frac{4}{I2} = \frac{1}{16} \]

5. **Cross-Multiply to Solve for I2**:
- Cross-multiply to isolate I2:
\[ 4 \cdot 16 = I2 \implies I2 = 64 \text{ units} \]

6. **Conclusion**:
- The new photoelectric current (I2) when the distance is reduced to 1 unit is 64 units.

### Final Answer:
The photoelectric current now will be **64 units**.