Answer:
[tex]\huge\boxed{\sf f \approx 6.7 \ cm}[/tex]
Explanation:
Given data:
Object distance = p = 10 cm
Image distance = q = 20 cm
Required:
Focal length = f = ?
Formula:
[tex]\displaystyle \frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
Solution:
Put the given data in the above formula.
[tex]\displaystyle \frac{1}{f} = \frac{1}{10} + \frac{1}{20} \\\\\frac{1}{f} = 0.1 + 0.05\\\\\frac{1}{f} = 0.15\\\\f = 1 / 0.15\\\\f \approx 6.7 \ cm\\\\\rule[225]{225}{2}[/tex]
