question_answer

An object is placed 30 cm away from a convex lens of focal length 10 cm and a sharp image is formed on a screen. Now a concave lens is placed in contact with the convex lens. The screen now has to be moved by 45 cm to get a sharp image again. The magnitude of focal length of the concave lens is (in cm)

A)  72    

B)  60    

C)  36

D)  20

Correct Answer: D

Solution :

 For the first condition \[\frac{1}{{{f}_{1}}}=\frac{1}{v}-\frac{1}{u}\Rightarrow \frac{1}{10}=\frac{1}{v}+\frac{1}{30}\] \[\frac{1}{v}=\frac{1}{10}-\frac{1}{30}\] \[\Rightarrow \] \[v=\frac{30}{2}=15\,cm\] where \[{{f}_{1}}=10\,cm,\,u=-30\,cm\] For the second condition when concave lens is placed \[v=(15+45)cm=60\,cm\] \[\frac{1}{F}=\frac{1}{v}-\frac{1}{u}\]  (where \[F=\]focal length of combination) \[\therefore \] \[\frac{1}{F}=\frac{1}{60}+\frac{1}{30}\] \[F=\frac{60}{3}cm=20cm\] The magnitude of focal length of concave lens \[\frac{1}{F}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\Rightarrow \frac{1}{20}=\frac{1}{10}+\frac{1}{{{f}_{2}}}\Rightarrow {{f}_{2}}=-20\,cm\](Negative sign for concave lens)