When a point source of light is at a distance of 50 cm from a photoelectric cell , the cut-off voltage is found to be ` V_(0)` . If the same source is placed at a distance of 1 m from the cell , then the cut-off voltage will be

Options

A. `V_(0)//4`
B. `V_(0) //2 `
C. `V_(0)`
D. ` 2 V_(0)`

Answer

To solve the problem, we need to understand the relationship between the intensity of light and the distance from the source, as well as how this affects the cut-off voltage in a photoelectric cell.

### Step-by-Step Solution:

1. **Understand the Inverse Square Law**: The intensity of light (I) from a point source is inversely proportional to the square of the distance (d) from the source. This can be expressed as:
\[ I \propto \frac{1}{d^2} \]
This means that if the distance from the source increases, the intensity of light at that distance decreases.

2. **Calculate the Initial Intensity**: When the light source is at a distance of 50 cm (0.5 m), let the intensity be \( I_1 \).
\[ I_1 \propto \frac{1}{(0.5)^2} = \frac{1}{0.25} = 4 \]

3. **Calculate the New Intensity**: When the source is moved to a distance of 1 m, the intensity \( I_2 \) becomes:
\[ I_2 \propto \frac{1}{(1)^2} = 1 \]

4. **Determine the Ratio of Intensities**: The ratio of the intensities at the two distances is:
\[ \frac{I_1}{I_2} = \frac{4}{1} = 4 \]

5. **Relate Intensity to Cut-off Voltage**: The cut-off voltage \( V_0 \) in a photoelectric cell is directly proportional to the intensity of light. Therefore, if we denote the cut-off voltage at 50 cm as \( V_0 \), then at 1 m, the new cut-off voltage \( V' \) can be expressed as:
\[ V' = \frac{V_0}{4} \]

6. **Conclusion**: Thus, when the source is moved from 50 cm to 1 m, the new cut-off voltage will be:
\[ V' = \frac{V_0}{4} \]

### Final Answer:
The cut-off voltage when the source is placed at a distance of 1 m from the cell will be \( \frac{V_0}{4} \).