A beam of light converges at a point P. Now a | vedclass

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(C) The incident beam is convergent,so the object is virtual. The distance of the point $P$ from the lens is $u = +12 \, cm$.The focal length of the c

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$48$

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(C) The incident beam is convergent,so the object is virtual. The distance of the point $P$ from the lens is $u = +12 \, cm$.The focal length of the concave lens is $f = -16 \, cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$Substituting the values: $\frac{1}{-16} = \frac{1}{v} - \frac{1}{12}$$\frac{1}{v} = \frac{1}{12} - \frac{1}{16} = \frac{4 - 3}{48} = \frac{1}{48}$Therefore,$v = +48 \, cm$.The beam converges at a point $48 \, cm$ from the lens.

$A$ beam of light converges at a point $P$. Now 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 concave lens of focal length $16 \, cm$ (in $, cm$)?

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