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

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(D) Let the position of the converging beam's focus without the lens be at distance $x$ from the lens. When the lens is present,the image is formed at

Vedclass · Accepted Answer

$-30$

Vedclass · Answer

(D) Let the position of the converging beam's focus without the lens be at distance $x$ from the lens. When the lens is present,the image is formed at $v = +15\, cm$.When the lens is removed,the rays meet at a point $5\, cm$ closer,so the object distance for the lens is $u = 15 - 5 = +10\, cm$.Since the beam is converging,the object is virtual,hence $u$ is positive.Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Substituting the values: $\frac{1}{15} - \frac{1}{10} = \frac{1}{f}$.$\frac{2 - 3}{30} = \frac{1}{f}$.$\frac{-1}{30} = \frac{1}{f}$.Therefore,$f = -30\, cm$.

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