Question

An object $5cm$ is placed at a distance of $20cm$ in front of a convex mirror of radius of curvature$30cm$. Find the position of the image, its nature, and size.

Answer 1

Step 1 - Givan data

Height of the object $h=5cm$

The distance of the object from the pole of the mirror $u=-20cm$ (Negative sign shows that the object is placed in the front of the mirror)

The radius of curvature $R=30cm$ (For convex mirror radius of curvature is positive)

Step 2 - Finding the focal length of the mirror

The focal length $\left(f\right)$ can be calculated as

$$\begin{gathered} f=\frac{R}{2} \\ f=\frac{30cm}{2} \\ f=15cm \end{gathered}$$

Step 3 - Finding the formula for the position of the image

The mirror equation is given by the expression

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

Where $v$ is the distance of the image from the pole of the mirror.

Rearrange the above equation.

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \\ \frac{1}{v}=\frac{u-f}{uf} \\ v=\frac{uf}{u-f} \end{gathered}$$

Step 4 - Finding the position of the image

Substitute the values into the above formula.

$$\begin{gathered} v=\frac{(-20cm)(15cm)}{(-20cm)-(15cm)} \\ v=\frac{-300}{-35} \\ v=+8.57cm \end{gathered}$$

Therefore, the image is formed behind the convex mirror at a distance $+8.57cm$ from the pole of the mirror.

Step 5 - Finding the nature of the object

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

$$\begin{gathered} m=-\frac{v}{u} \\ m=-\frac{(8.57cm)}{(-20cm)} \\ m=\frac{8.57}{20} \\ m=0.428 \end{gathered}$$

Therefore, the image is erect, virtual, and smaller in size by a factor of $0.428$.

Step 6 - Finding the size of the image

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

$$\begin{gathered} m=\frac{h\text{'}}{h} \\ h\text{'}=m\times h \\ h\text{'}=(0.428)\times (5.0cm) \\ h\text{'}=2.14cm \end{gathered}$$

Therefore, the height of the image is $2.14cm$.

Final answer- Hence, $v=+8.57cm$. The image is virtual and erect and the size of the image is $2.14cm$.