A converging beam of rays is incident on a di | vedclass

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(D) The rays are converging towards a point $15 \, cm$ behind the lens,so the image distance $v = +15 \, cm$.If the lens is removed,the rays meet $5 \

Vedclass · Accepted Answer

$-30$

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(D) The rays are converging towards a point $15 \, cm$ behind the lens,so the image distance $v = +15 \, cm$.If the lens is removed,the rays meet $5 \, cm$ closer to the lens,meaning the original convergence point was at $15 - 5 = 10 \, cm$ from the lens position. Since the rays were converging towards this point,the object distance $u = +10 \, cm$.Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$Substituting the values: $\frac{1}{15} - \frac{1}{10} = \frac{1}{f}$$\frac{2 - 3}{30} = \frac{1}{f}$$-\frac{1}{30} = \frac{1}{f}$$f = -30 \, cm$

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