If an object of $7cm$height is placed at a distance of $12cm$from a convex lens of focal length$8cm$, find the position, nature and height of the image.

Step 1 - Given data

The height of the object $h=7cm$

The distance of the object from the optical center of the lens $u=-12cm$

The focal length of the lens $f=+8cm$ (The focal length of a convex lens is positive)

Step 2 - Finding the formula for the position of the image

The distance of the image from the optical center of the lens$\left(v\right)$ can be calculated using the lens formula given below

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \frac{1}{v}=\frac{1}{f}+\frac{1}{u} \\ \frac{1}{v}=\frac{u+f}{fu} \\ v=\frac{fu}{f+u}...\left(1\right) \end{gathered}$$

Step 3 - Finding the position and the nature of the image

Substitute the given value into the above formula.

$$\begin{gathered} v=\frac{fu}{f+u} \\ v=\frac{(8cm)\times (-12cm)}{(8cm)+(-12cm)} \\ v=-\frac{(8\times 12)}{(8-12)} \\ v=+24cm \end{gathered}$$

Since the image distance is positive, the image is formed on the right side of the convex lens at 24cm.

Step 4 - Finding the magnification

The magnification $\left(m\right)$can be calculated as

$$\begin{gathered} m=\frac{v}{u} \\ m=\frac{+24cm}{-12cm} \\ m=-\frac{24}{12} \\ m=-2 \end{gathered}$$

Since the magnitude of magnification is greater than 1, so the image formed is enlarged.

Step 5 - Finding the height of the image

The image height $(h\text{'})$can be calculated as

$$\begin{gathered} m=\frac{h\text{'}}{h} \\ h\text{'}=m\times h \\ h\text{'}=(-2)\times (7cm) \\ h\text{'}=-14cm \end{gathered}$$

Therefore, the height of the image is $14cm$ and the negative sign shows that the image is inverted and real.

Hence, the position of the image is $24cm$behind the lens, it is real and inverted and the height of the image is $14cm$(Enlarged).