The exposure time of a camera lens at the `(f)/(2.8)` setting is `(1)/(200) ` second. The correct time of exposure at `(f)/(5.6)` is

Options

A. 0.04 s
B. 0.20 s
C. 0.40 s
D. 0.02 s

Explanation

To solve the problem, we need to determine the correct time of exposure at an f-stop of \( f/5.6 \) given that the exposure time at \( f/2.8 \) is \( \frac{1}{200} \) seconds.

### Step-by-Step Solution:

1. **Understand the relationship between exposure time and f-stop:**
The exposure time \( t \) is directly proportional to the square of the f-number (f-stop). This can be expressed mathematically as:
\[ t \propto f^2 \]
This means that if the f-stop changes, the exposure time will change according to the square of the ratio of the f-stops.

2. **Set up the equation:**
Let \( t_1 \) be the exposure time at \( f/2.8 \) and \( t_2 \) be the exposure time at \( f/5.6 \). We can write:
\[ \frac{t_2}{t_1} = \left(\frac{f_2}{f_1}\right)^2 \]
where \( f_1 = 2.8 \) and \( f_2 = 5.6 \).

3. **Substitute known values:**
We know that \( t_1 = \frac{1}{200} \) seconds. Now substituting the values:
\[ \frac{t_2}{\frac{1}{200}} = \left(\frac{5.6}{2.8}\right)^2 \]

4. **Calculate the ratio of f-stops:**
Calculate \( \frac{5.6}{2.8} \):
\[ \frac{5.6}{2.8} = 2 \]
Therefore,
\[ \left(\frac{5.6}{2.8}\right)^2 = 2^2 = 4 \]

5. **Substitute back to find \( t_2 \):**
Now substitute this back into the equation:
\[ \frac{t_2}{\frac{1}{200}} = 4 \]
This implies:
\[ t_2 = 4 \times \frac{1}{200} = \frac{4}{200} = \frac{1}{50} \text{ seconds} \]

6. **Convert to decimal form:**
Converting \( \frac{1}{50} \) seconds to decimal form gives:
\[ t_2 = 0.02 \text{ seconds} \]

### Final Answer:
The correct time of exposure at \( f/5.6 \) is \( 0.02 \) seconds.

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