Question

An object placed at a distance of $16cm$ from a convex lens produces an image of magnification $m(m>1)$. If the object is moved towards the lens by $8cm$, then again an image of magnification $m$ is obtained. The numerical value of the focal length of the lens is:

• A. $12cm$
• B. $14cm$
• C. $18cm$
• D. $20cm$

Answer 1

The correct option is A $12cm$

For a convex lens, a magnified image is obtained for three positions of object:

i. When object lies between $f$ and $2f$

ii. When object is at $f$

iii. When object lies between $f$ and lens

Out of these positions, when object is at $f$, image will be formed at infinity and $m=\infty$.

For other two positions $m$ is not equal to $\infty$, therefore there is no other position for which $m=\infty$ and given that $m$ is equal for two positions of object i.e. object is not at $f$ because we have no other position for $m=\infty$.

Let in first position object lies between $f$ and $2f$.

Given: $u=-16cm$

From lens equation:

$v=\frac{uf}{u+f}$

$v=\frac{-16f}{-16+f}$

since $f<u<2f$, $v$ will be +ive ,

hence, $m=\frac{v}{-u}=\frac{\left(\frac{-16f}{-16+f}\right)}{-16}$ ...................eq1

Let in second position object lies between $f$ and lens.

Given: $u=-8cm$

From lens equation:

$v=\frac{uf}{u+f}$

$v=\frac{-8f}{-8+f}$

since $u<f$, $v$ will be -ive,

hence, $m=\frac{-v}{-u}=\frac{-\left(\frac{-8f}{-8+f}\right)}{-8}$ .........................eq2

By eq1 and eq2:

$\frac{\frac{-16f}{-16+f}}{-16}=\frac{-\left(\frac{-8f}{-8+f}\right)}{-8}$

$f=12cm$