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  • Need explanation for: 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 :

 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 :

 

  • Option 1)

    real and at a distance of 40 cm from convergent lens.

     

  • Option 2)

    virtual and at a distance of 40 cm from convergent lens.

     

  • Option 3)

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

     

  • Option 4)

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

     

 

Answers (1)

As we learnt in

Thin lens formula -

\frac{1}{v}-\frac{1}{u}= \frac{1}{f}

 

- wherein

u \, and \, v are object and image distance from lens.

 

 Given focal length of concave lens f = - 25 cm

Focal length of convex lens f' = 20 cm

The inmage of diverging lens will form at F. i.e. at focal length of concave lens. 

Now this image will serve as the object for convex lens, it is at twice the focal length of the convex lens (i.e. - 2f) which is at a distance of 40 cm from the convergent lens.

Correct option is 3.

 


Option 1)

real and at a distance of 40 cm from convergent lens.

 

This is an incorrect option.

Option 2)

virtual and at a distance of 40 cm from convergent lens.

 

Reflection 

Option 3)

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

 

Reflection 

Option 4)

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

 

Reflection 

Posted by

Vakul

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