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

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(B) For a convergent beam,the point $P$ acts as a virtual object for the lens. Since the beam is converging towards $P$,the object distance $u$ is tak

Vedclass · Accepted Answer

$7.5$

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(B) For a convergent beam,the point $P$ acts as a virtual object for the lens. Since the beam is converging towards $P$,the object distance $u$ is taken as positive. Given: $u = +12 \, cm$ and $f = +20 \, cm$ (for a convex lens). Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$. Substituting the values: $\frac{1}{20} = \frac{1}{v} - \frac{1}{12}$. Rearranging for $v$: $\frac{1}{v} = \frac{1}{20} + \frac{1}{12}$. Calculating the sum: $\frac{1}{v} = \frac{3 + 5}{60} = \frac{8}{60} = \frac{2}{15}$. Therefore,$v = \frac{15}{2} = 7.5 \, cm$. The beam converges at a distance of $7.5 \, cm$ 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 convex lens of focal length $20 \, cm$ (in $, cm$)?

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