A photographic plate placed at a distance of 5 cm from a weak point source is exposed for 3 s. If the plate is kept at a distance of 10 cm from the source, the time needed for the same exposure is

Options

A. A. 3s
B. B. 12s
C. C. 24s
D. D. 48s

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between time and distance
The time of exposure (t) is inversely proportional to the illuminance (E) received by the photographic plate. The illuminance from a point source is given by the formula:

\[ E \propto \frac{1}{d^2} \]

where \( d \) is the distance from the source.

### Step 2: Set up the initial conditions
From the problem, we have:
- Distance \( d_1 = 5 \) cm
- Time of exposure \( t_1 = 3 \) seconds

### Step 3: Set up the second condition
Now, we want to find the time of exposure \( t_2 \) when the distance is increased to:
- Distance \( d_2 = 10 \) cm

### Step 4: Use the relationship between time and distance
Since the illuminance is inversely proportional to the square of the distance, we can write:

\[ E_1 = \frac{E_0}{d_1^2} \]
\[ E_2 = \frac{E_0}{d_2^2} \]

The time of exposure is inversely proportional to the illuminance, which gives us:

\[ t_1 \propto \frac{1}{E_1} \]
\[ t_2 \propto \frac{1}{E_2} \]

### Step 5: Relate the two times
From the proportionality, we can write:

\[ \frac{t_1}{t_2} = \frac{E_2}{E_1} \]

Substituting the expressions for \( E_1 \) and \( E_2 \):

\[ \frac{t_1}{t_2} = \frac{d_1^2}{d_2^2} \]

### Step 6: Substitute the known values
Now substituting the values we have:

\[ \frac{3}{t_2} = \frac{5^2}{10^2} \]

Calculating the squares:

\[ \frac{3}{t_2} = \frac{25}{100} = \frac{1}{4} \]

### Step 7: Solve for \( t_2 \)
Cross-multiplying gives:

\[ 3 \cdot 4 = t_2 \]

Thus,

\[ t_2 = 12 \text{ seconds} \]

### Final Answer
The time needed for the same exposure at a distance of 10 cm is **12 seconds**.

---