You are given two lenses,a converging lens wi | vedclass

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(A) For a converging lens (convex lens) with focal length $f = +10\, cm$,a virtual and magnified image is formed when the object is placed between the

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Placing the object $5\, cm$ from the converging lens.

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(A) For a converging lens (convex lens) with focal length $f = +10\, cm$,a virtual and magnified image is formed when the object is placed between the optical center and the focus (i.e.,$u In option $A$,the object is placed at $5\, cm$,which is less than the focal length $(5\, cm Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we have $\frac{1}{v} - \frac{1}{-5} = \frac{1}{10}$,which gives $\frac{1}{v} = \frac{1}{10} - \frac{1}{5} = -\frac{1}{10}$,so $v = -10\, cm$.The magnification $m = \frac{v}{u} = \frac{-10}{-5} = +2$. Since $|m| > 1$,the image is virtual,erect,and larger than the object.Diverging lenses always produce virtual,diminished images for real objects.

You are given two lenses,a converging lens with focal length $+10\, cm$ and a diverging lens with focal length $-20\, cm$. Which of the following would produce a virtual image that is larger than the object?

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