The figure given below shows a beam of light | vedclass

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(D) For a concave lens,the focal length $f = -16 \ cm.$ Since the light rays are converging at point $P$ behind the lens,the object for the lens is a

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$48$

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(D) For a concave lens,the focal length $f = -16 \ cm.$ Since the light rays are converging at point $P$ behind the lens,the object for the lens is a virtual object. Therefore,the object distance $u = +12 \ cm.$ Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ Substituting the values: $\frac{1}{-16} = \frac{1}{v} - \frac{1}{12}$ Rearranging to solve for $v$: $\frac{1}{v} = \frac{1}{12} - \frac{1}{16}$ Finding a common denominator: $\frac{1}{v} = \frac{4 - 3}{48} = \frac{1}{48}$ Thus,$v = 48 \ cm.$ Therefore,the beam converges at a distance $x = 48 \ cm$ from the lens.

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