The camera lens has an aperture of f and the exposure time is (1/60)s. What will be the new exposure time if the aperture become 1.4f ?

Options

A. `(1)/(42)`
B. `(1)/(56)`
C. `(1)/(72)`
D. `(1)/(31)`

Explanation

To solve the problem, we need to understand the relationship between the exposure time (T) and the aperture size (f). The exposure time is directly proportional to the square of the aperture size.

### Step-by-Step Solution:

1. **Identify the relationship**:
The exposure time (T) is directly proportional to the square of the aperture (f). This can be expressed mathematically as:
\[ T \propto f^2 \]
or
\[ \frac{T_1}{f_1^2} = \frac{T_2}{f_2^2} \]

2. **Assign known values**:
From the problem, we know:
- Initial exposure time, \( T_1 = \frac{1}{60} \) seconds
- Initial aperture, \( f_1 = f \)
- New aperture, \( f_2 = 1.4f \)

3. **Set up the equation**:
Substitute the known values into the equation:
\[ \frac{1/60}{f^2} = \frac{T_2}{(1.4f)^2} \]

4. **Simplify the equation**:
We can simplify \( (1.4f)^2 \):
\[ (1.4f)^2 = 1.96f^2 \]
Now, substituting this back into the equation gives:
\[ \frac{1/60}{f^2} = \frac{T_2}{1.96f^2} \]

5. **Cancel out \( f^2 \)**:
Since \( f^2 \) is common in both the numerator and denominator, we can cancel it:
\[ \frac{1}{60} = \frac{T_2}{1.96} \]

6. **Solve for \( T_2 \)**:
Rearranging the equation to solve for \( T_2 \):
\[ T_2 = \frac{1}{60} \times 1.96 \]
\[ T_2 = \frac{1.96}{60} \]

7. **Convert to a simpler fraction**:
To simplify \( \frac{1.96}{60} \):
\[ T_2 = \frac{196}{6000} = \frac{98}{3000} \approx \frac{1}{30} \]

8. **Final answer**:
Therefore, the new exposure time \( T_2 \) is approximately:
\[ T_2 \approx \frac{1}{30} \text{ seconds} \]