Correct exposure for a photographic print is 10 s at a distance of one metre from a point source of 20 cd. For an equal fogging of the print placed at a distance of 2 m from a 16 cd source, the necessary time for exposure is

Options

A. (a)100 s
B. (b)25 s
C. (c)50 s
D. (d)75 s

Explanation

To solve the problem, we need to use the formula that relates the exposure time, intensity of the light source, and the distance from the light source. The formula is:

\[ L_1 \cdot R_1^2 \cdot T_1 = L_2 \cdot R_2^2 \cdot T_2 \]

Where:
- \( L_1 \) and \( L_2 \) are the luminous intensities of the light sources.
- \( R_1 \) and \( R_2 \) are the distances from the light sources.
- \( T_1 \) and \( T_2 \) are the exposure times.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- For the first scenario:
- \( L_1 = 20 \, \text{cd} \)
- \( R_1 = 1 \, \text{m} \)
- \( T_1 = 10 \, \text{s} \)
- For the second scenario:
- \( L_2 = 16 \, \text{cd} \)
- \( R_2 = 2 \, \text{m} \)
- \( T_2 = ? \) (This is what we need to find)

2. **Plug the Values into the Formula:**
\[ 20 \cdot (1)^2 \cdot 10 = 16 \cdot (2)^2 \cdot T_2 \]

3. **Calculate the Left Side:**
\[ 20 \cdot 1 \cdot 10 = 200 \]

4. **Calculate the Right Side:**
\[ 16 \cdot (2^2) = 16 \cdot 4 = 64 \]
So, we have:
\[ 200 = 64 \cdot T_2 \]

5. **Solve for \( T_2 \):**
\[ T_2 = \frac{200}{64} = \frac{25}{8} = 3.125 \, \text{s} \]

### Final Answer:
The necessary time for exposure \( T_2 \) is approximately **3.125 seconds**.