A lens forms a diminished and erect image of an object- The magnification is 1-4- Find ratio of distance between object and focus and focus and image

The correct option is C 8:1magnification is 14=vu v= v u= 4v 1v 1u=1f 1 v+14v=1f f= 4v3 ratio=u fv f=4v 4v3v 4v3=81 optio ...

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Standard XII

Physics

Power and Thin Lenses in Contact

A lens forms ...

Question

A lens forms a diminished and erect image of an object. The magnification is $\frac{1}{4}$ . Find ratio of distance between object and focus and focus and image

4:1

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1:4

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8:1

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2:1

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Solution

The correct option is C 8:1
magnification is $\frac{1}{4} = \frac{v}{u}$
$v = - v$
$u = - 4 v$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{- v} + \frac{1}{4 v} = \frac{1}{f}$
$f = \frac{- 4 v}{3}$
$r a t i o = \frac{u - f}{v - f} = \frac{4 v - \frac{4 v}{3}}{v - \frac{4 v}{3}} = \frac{8}{1}$
option $C$ is correct

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A lens forms a diminished and erect image of an object. The magnification is 14. Find ratio of distance between object and focus and focus and image

The correct option is C 8:1magnification is 14=vuv=−vu=−4v1v−1u=1f1−v+14v=1ff=−4v3ratio=u−fv−f=4v−4v3v−4v3=81option C is correct

A lens forms a diminished and erect image of an object. The magnification is $\frac{1}{4}$ . Find ratio of distance between object and focus and focus and image