Question

Any genius please solve it.

A concave mirror has focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also find the magnification produced by the lens.

Please step by step

Answer 1

hope this will help you

Answer 2

$\textbf{\huge{\gray{HEY THERE!}}}$

$\textbf{=>Here is ur Solution<=}$

$<b>A concave lens always form a virtual, erect image on the same side of the object.$

$image - distance = - 10 \: cm \\ focal \: length \: f \: = - 15cm \\ object - distance \: u = ?\\ since \frac{1}{v } = \frac{1}{u} - \frac{1}{f } \\ \frac{1}{u} = \frac{1}{ - 10} - \frac{1}{( - 15)} = \frac{1}{10} + \frac{1}{15} \\ \frac{1}{u} = \frac{ - 3 + 2}{30} = \frac{1}{ - 30} \\ or \: u = - 30 \: cm \\ \\ thus \: the \: object \: is \: 30 \: cm \: \\ magnification \: m = \frac{v}{u} \\ m = \frac{ - 10 \: cm}{ - 30 \: cm} = \frac{1}{3} = + 0.33$

The positive sign shows that the image is erect and virtual. The image is one-third of the size of the object.

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