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# What is the position of image when an object is placed at a distance of 10 cm from a convex lens of focal length 10 cm?

**Given:**

Object distance, $u$ = $-$10 cm (negative sign shows that the object is placed on the left side of the lens)

Focal length, $f$ = $+$10 cm

**To find: **Position or distance of image $v$.

**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-

$\frac {1}{v}-\frac {1}{(-10)}=\frac {1}{10}$

$\frac {1}{v}+\frac {1}{10}=\frac {1}{10}$

$\frac {1}{v}=\frac {1}{10}-\frac {1}{10}$

$\frac {1}{v}=\frac {1-1}{10}$

$\frac {1}{v}=\frac {0}{10}$

$\frac {1}{v}=0$

$v=\frac {1}{0}$

$v=\infty$

Thus, the position or distance of the image will be at** infinity** $(\infty)$.

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