The electric field intensity produced by the radiations coming from 100W bulbs at a 3m distance is E. The electric field intensity produced by the radiations coming from 50W bulb at the same distance is

Options

A. `E//2`
B. `2E`
C. `E//sqrt2`
D. `sqrt2 E`

Explanation

To solve the problem, we need to understand how the electric field intensity (E) produced by a bulb relates to its power output. The electric field intensity is directly proportional to the square root of the power of the source.

### Step-by-Step Solution:

1. **Understand the relationship between power and electric field intensity**: The electric field intensity (E) produced by a bulb is proportional to the square root of its power (P). This can be expressed mathematically as:
\[ E \propto \sqrt{P} \]

2. **Identify the electric field intensity for the 100W bulb**: Given that the electric field intensity produced by the 100W bulb at a distance of 3m is E, we can express this as:
\[ E_{100} = k \sqrt{100} \]
where \( k \) is a proportionality constant.

3. **Calculate the electric field intensity for the 50W bulb**: For the 50W bulb, we can express the electric field intensity as:
\[ E_{50} = k \sqrt{50} \]

4. **Relate the two electric field intensities**: We can now relate \( E_{50} \) to \( E_{100} \):
\[ E_{50} = k \sqrt{50} = k \sqrt{(50/100) \cdot 100} = \sqrt{0.5} \cdot k \sqrt{100} = \sqrt{0.5} \cdot E \]

5. **Simplify the expression**: Since \( \sqrt{0.5} = \frac{1}{\sqrt{2}} \), we can write:
\[ E_{50} = \frac{E}{\sqrt{2}} \]

### Final Answer:
The electric field intensity produced by the radiations coming from the 50W bulb at a 3m distance is:
\[ E_{50} = \frac{E}{\sqrt{2}} \]