An object is placed at a distance of f/2 from | vedclass

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(A) Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Given that the object distance $u = -f/2$ (using sign convention).Substituting t

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At one of the foci,virtual and double its size

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(A) Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Given that the object distance $u = -f/2$ (using sign convention).Substituting the values: $\frac{1}{f} = \frac{1}{v} - \frac{1}{-f/2} = \frac{1}{v} + \frac{2}{f}$.Rearranging to solve for $v$: $\frac{1}{v} = \frac{1}{f} - \frac{2}{f} = -\frac{1}{f}$.Thus,$v = -f$. The negative sign indicates that the image is formed on the same side as the object,meaning it is virtual.The magnification $m = \frac{v}{u} = \frac{-f}{-f/2} = 2$.Since $m$ is positive and greater than $1$,the image is virtual,erect,and double the size of the object.

An object is placed at a distance of $f/2$ from a convex lens. The image will be

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