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Step-by-Step Solution
Step 1: Write down the combined lens formula
For two thin lenses of focal lengths $f_1$ and $f_2$ separated by a distance $d$, the resultant focal length $f$ is given by:
$
\frac{1}{f}
= \frac{1}{f_1} + \frac{1}{f_2}
- \frac{d}{f_1 f_2}.
$
Step 2: Substitute the given values
We are given:
Resultant focal length, $f = 10\,\text{cm}$
Separation between the lenses, $d = 2\,\text{cm}$
Hence the formula becomes:
$
\frac{1}{10}
= \frac{1}{f_1} + \frac{1}{f_2}
- \frac{2}{f_1 f_2}.
$
Step 3: Rearrange the equation
Combine the terms on the right-hand side by placing them over a common denominator $f_1 f_2$:
$
\frac{1}{10}
= \frac{f_2 + f_1 - 2}{f_1 f_2}.
$
Multiply both sides by $f_1 f_2$:
$
f_1 f_2 \times \frac{1}{10}
= f_1 + f_2 - 2.
$
$
\frac{f_1 f_2}{10} = f_1 + f_2 - 2.
$
Multiply both sides by 10:
$
f_1 f_2 = 10f_1 + 10f_2 - 20.
$
Step 4: Identify the correct pair of focal lengths
By checking the given options, the pair $f_1 = 18\,\text{cm}$ and $f_2 = 20\,\text{cm}$ satisfies:
$
18 \times 20 = 10 \times 18 + 10 \times 20 - 20,
$
$
360 = 180 + 200 - 20,
$
$
360 = 360.
$
Hence, the correct focal lengths are $18\,\text{cm}$ and $20\,\text{cm}$.
Final Answer
The focal lengths of the component lenses are $18\,\text{cm}$ and $20\,\text{cm}$.
