question_answer

A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is:   JEE Main Solved Paper-2017

A)  real and at a distance of 40 cm from the divergent lens

B)  real and at a distance of 6 cm from the convergent lens

C)  real and at a distance of 40 cm from convergent lens

D)  virtual and at a distance of 40 cm from convergent lens.               [JEE Main 2017]

Correct Answer: C

Solution :

 As parallel beam incident on diverging lens if forms virtual image at \[{{v}_{1}}=-25\,cm\]from the diverging lense which works as a object for the converging lense (f = 20 cm)                 So for converging lens\[u=-40cm,f=20\,cm\] \[\therefore \]  Finance image \[\frac{1}{V}-\frac{1}{-40}=\frac{1}{20}\] V = 40 cm from converging lenses.