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

Options

A. `(E)/(2)`
B. 2E
C. `(E)/(sqrt(2))`
D. `sqrt(2)E`

Explanation

To solve the problem, we need to understand the relationship between the electric field intensity (E) produced by a bulb and its power output (P). The electric field intensity is proportional to the square root of the power of the bulb.

### Step-by-step Solution:

1. **Understanding the Relationship**:
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. **Setting Up the Ratios**:
For two bulbs with powers \( P_1 \) and \( P_2 \), the electric field intensities \( E_1 \) and \( E_2 \) can be related as follows:
\[ \frac{E_1}{E_2} = \sqrt{\frac{P_1}{P_2}} \]

3. **Assigning Values**:
Let \( P_1 = 100 \, W \) (for the first bulb) and \( P_2 = 50 \, W \) (for the second bulb). We know that the electric field intensity from the 100 W bulb at a distance of 3 m is \( E \). Therefore, we can write:
\[ E_1 = E \quad \text{and} \quad E_2 = ? \]

4. **Calculating the Electric Field Intensity for the 50 W Bulb**:
Using the relationship established in step 2:
\[ \frac{E}{E_2} = \sqrt{\frac{100}{50}} \]
Simplifying the right side:
\[ \sqrt{\frac{100}{50}} = \sqrt{2} \]
Thus,
\[ \frac{E}{E_2} = \sqrt{2} \]

5. **Rearranging to Find \( E_2 \)**:
Rearranging the equation gives:
\[ E_2 = \frac{E}{\sqrt{2}} \]

6. **Final Result**:
The electric field intensity produced by the 50 W bulb at a distance of 3 m is:
\[ E_2 = \frac{E}{\sqrt{2}} \]

### Summary:
The electric field intensity produced by the radiations coming from the 50 W bulb at the same distance is \( \frac{E}{\sqrt{2}} \).