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# A converging lens of focal length 5 cm is placed at a distance of 20 cm from a screen. How far from the lens should an object be placed so as to form its real image on the screen?

** Given:**

Focal length, $f$ = 5 cm

Image distance, $v$ = Distance of the lens from the screen = $+$20 cm (for real image)

**To find:**Object distance, $u$.

Solution:

According to the lens formula, we know that:

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

Substituting the given values in the formula we get-

$\frac {1}{20}-\frac {1}{u}=\frac {1}{5}$

$\frac {1}{20}-\frac {1}{5}=\frac {1}{u}$

$\frac {1}{u}=\frac {1-4}{20}$

$\frac {1}{u}=\frac {-3}{20}$

$u=-\frac {20}{3}$

$u=-6.6cm$

Thus, the object $u$ should be placed at a distance of 6.6 cm from the lens, and the negative $(-)$ sign for object distance implies that the object is placed on the left side of the lens.

Hence, to form a real image, the object should be placed at a distance of 6.6 cm from the lens.

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