A convex lens of focal length 10 cm is place | vedclass

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(A) Given: Focal length $f = +10 \ cm$,Image distance $v = +12 \ cm$ (since the image is real and formed on the wall on the other side of the lens).Us

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) Given: Focal length $f = +10 \ cm$,Image distance $v = +12 \ cm$ (since the image is real and formed on the wall on the other side of the lens).Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Substituting the values: $\frac{1}{12} - \frac{1}{u} = \frac{1}{10}$.Rearranging for $u$: $\frac{1}{u} = \frac{1}{12} - \frac{1}{10}$.Calculating the common denominator: $\frac{1}{u} = \frac{5 - 6}{60} = \frac{-1}{60}$.Thus,$u = -60 \ cm$.The negative sign indicates that the object should be placed at a distance of $60 \ cm$ in front of the lens.

$A$ convex lens of focal length $10 \ cm$ is placed at a distance of $12 \ cm$ from a wall. How far from the lens should an object be placed so as to form its real image on the wall (in $cm$)?

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