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# The focal length of concave lens is 15cm at what distance object should be kept so that image formed will be 10cm away from lens. Find magnification of object.

Given:

Focal length of concave lens, $f$ = $-$15 cm $(focal\ length\ of\ a\ concave\ lens\ is\ always\ taken\ negative)$

Image distance, $v$ = $-$10 cm $(image\ distance\ is\ taken\ negative\ because\ image\ formed\ by\ a\ concave\ lens\ is\ on\ the\ left\ side\ of\ the\ lens)$

To find: Object distance $(u)$, and magnification $(m)$.

Solution:

From the lens formula, we know that-

$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$

Substituting the given values in the formula, we get-

$\frac {1}{(-10)}-\frac {1}{u}=\frac {1}{(-15)}$

$-\frac {1}{10}-\frac {1}{u}=-\frac {1}{15}$

$\frac {1}{u}=\frac {1}{15}-\frac {1}{10}$

$\frac {1}{u}=\frac {2-3}{30}$

$\frac {1}{u}=\frac {-1}{30}$

$u=-\frac {1}{30}$

$u=-30cm$

Thus, the object is at a distance of 30 cm from the concave lens, and the negative sign implies that it is on the left side of it.

Now,

Foo magnification we know that-

$m=\frac {v}{u}$

Substituting the given values in the formula, we get-

$m=\frac {-10}{-30}$

$m=\frac {1}{3}$

$m=0.33$

Thus, the magnification $m$, produced by the lens is 0.33 which is less than 1, therefore the image will be smaller than the object.

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