Question

A $4cm$ tall object is placed perpendicular to the principal axis of a convex lens with a focal length of $24cm$. The distance of the object from the lens is $16cm$. Find the position, size, and nature of the image formed, using the lens formula.

Answer 1

Step 1: Given Data

Height of the object $h_o=4cm$

The focal length of the convex lens $f=24cm$

Distance of the object from the lens $u=-16cm$, negative because the object is on the left of the lens.

Let $v$ be the image distance and $h_i$ be the height of the image.

Step 2: Calculate Image Distance

According to the lens formula,

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

$\Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u}$

Upon substituting the values we get,

$\frac{1}{v}=\frac{1}{24}-\frac{1}{16}$

$\Rightarrow \frac{1}{v}=\frac{1}{8}\left(\frac{1}{3}-\frac{1}{2}\right)$

$\Rightarrow \frac{1}{v}=\frac{1}{8}\left(-\frac{1}{6}\right)$

$\Rightarrow \frac{1}{v}=-\frac{1}{48}$

$\therefore v=-48cm$

Step 3: Calculate Image Height

Magnification is given as,

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

$\Rightarrow h_i=\frac{v}{u}{h}_o$

Upon substituting the values we get,

$h_i=\frac{-48}{-16}\times 4$

$=12cm$

Hence, the position of the image is $48cm$ to the left of the lens, the size of the image is $12cm$, and the nature of the image formed is erect and magnified.