Two thin lenses of focal length `f_(1)` and `f_(2)` are in contact and coaxial. The power of the combination is

Options

A. `sqrt((f_(1))/(f_(2)))`
B. `sqrt((f_(2))/(f_(1)))`
C. `(f_(1)+f_(2))/(f_(1)f_(2))`
D. `(f_(1)+f_(2))/(f_(1)f_(2))`

Explanation

To find the power of a combination of two thin lenses in contact, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Concept of Power and Focal Length**:
- The power \( P \) of a lens is defined as the reciprocal of its focal length \( f \) (in meters):
\[ P = \frac{1}{f} \]
- The focal length of a lens is the distance from the lens at which parallel rays of light converge or appear to diverge.

2. **Identify the Focal Lengths of the Lenses**:
- Let the focal lengths of the two lenses be \( f_1 \) and \( f_2 \).

3. **Use the Formula for Equivalent Focal Length**:
- When two lenses are in contact and coaxial, the equivalent focal length \( F \) of the combination can be calculated using the formula:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
- Rearranging gives:
\[ F = \frac{f_1 f_2}{f_1 + f_2} \]

4. **Calculate the Power of the Combination**:
- The power \( P \) of the combined lens system can be calculated using the equivalent focal length:
\[ P = \frac{1}{F} \]
- Substituting the expression for \( F \):
\[ P = \frac{1}{\left(\frac{f_1 f_2}{f_1 + f_2}\right)} = \frac{f_1 + f_2}{f_1 f_2} \]

5. **Final Expression for Power**:
- Therefore, the power of the combination of the two lenses is given by:
\[ P = \frac{f_1 + f_2}{f_1 f_2} \]

### Summary of the Solution
The power of the combination of two thin lenses in contact is calculated using the formula:
\[ P = \frac{f_1 + f_2}{f_1 f_2} \]