An object is placed at 15 , cm from a convex | vedclass

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(A) Given: Object distance $u = -15 \, cm$,Focal length $f = +20 \, cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Rearranging

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Virtual image is formed

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(A) Given: Object distance $u = -15 \, cm$,Focal length $f = +20 \, cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Rearranging for $v$: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} - \frac{1}{15}$.Taking the common denominator $(60)$: $\frac{1}{v} = \frac{3 - 4}{60} = -\frac{1}{60}$.Thus,$v = -60 \, cm$.The negative sign indicates that the image is formed on the same side as the object,which means it is a virtual image.

An object is placed at $15 \, cm$ from a convex lens of focal length $20 \, cm$. Which of the following is correct?

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