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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