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Question

A beam of light converges to a point P. A lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is :
(a) a convex lens of focal length 20 cm,
(b) a concave lens of focal length 16 cm.


Solution

As a lens is placed in the path of the convergent beam, the point $$ P $$ would lie on the right of the lens and acts as a virtual object.

Object distance, $$ u = 12\ cm $$

(a)  Focal length, $$ f = 20\ cm $$ (Convex Lens)

Using the lens formula, 

$$ \dfrac{1}{f} =   \dfrac{1}{v} -  \dfrac{1}{u} $$

we have, 

$$ \dfrac{1}{20} =   \dfrac{1}{v} -  \dfrac{1}{12} $$

$$ \dfrac{1}{v} = \dfrac{1}{20} +  \dfrac{1}{12} = \dfrac{3 + 5}{60} = \dfrac{8}{60} $$

$$ \Rightarrow v = \dfrac{60}{8} = 7.5\ cm $$ 

It is located at $$ 7.5\  cm $$ from the lens and it is a real image. 

(b) Focal length, $$ f = -16\ cm $$ (Concave Lens)

Applying lens formula, we have

$$ \dfrac{1}{-16} =   \dfrac{1}{v} -  \dfrac{1}{12} $$

$$ \dfrac{1}{v} = \dfrac{-1}{16} +  \dfrac{1}{12} = \dfrac{-3 + 4}{48} = \dfrac{1}{48} $$

$$ \Rightarrow v = 48\ cm $$

It is located at $$ 48\  cm $$ from the lens and it is a real image.

Physics

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