Question

A converging lens having magnitude of focal length as $f_{1}$ is kept coaxially in contact with a diverging lens having magnitude of focal length as $f_{2}$ . The focal length of the combination would be:

• A. $\frac{f_{1} f_{2}}{f_{1} - f_{2}}$
• B. $\frac{f_{1} + f_{2}}{f_{1} \, f_{2}}$
• C. $\frac{f_{1} - f_{2}}{f_{1} \, f_{2}}$
• D. $\frac{f_{1} \, f_{2}}{f_{1} + f_{2}}$
  
  
  
  
  
  
  
  
  
  

Answer 1

Answer: (A)

We have to consider sign convention as only magnitude of focal length is given.
If $f_{1}$ is focal length of convex (converging) lens and $f_{2}$ is focal length of concave (diverging) lens. Then their equivalent focal length F would be
$\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{- f_{2}}=\frac{f_{2} - f_{1}}{f_{1} f_{2}}\text{ }\left[\right. \because \frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \left]\right. \\ \therefore \, F=\frac{f_{1} f_{2}}{f_{2} - f_{1}}$