Question

An object of $5cm$ is placed at a distance of $15cm$ from a concave mirror of focal length $10cm$. Find the position, nature and height of the image.

Answer 1

Step 1 - Given data

Height of the object $h=5cm$

The distance of the object from the pole of the mirror $u=-15cm$ (Negative sign shows that the object is placed to the left of the mirror)

Focal length $f=-10cm$ (For concave mirror focal length is negative)

Step 2 - 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 3 - Finding the position of the image

Substitute the values into the above formula.

$$\begin{gathered} v=\frac{(-15cm)(-10cm)}{(-15cm)-(-10cm)} \\ v=\frac{150}{-5} \\ v=-30cm \end{gathered}$$

Therefore, the image is formed to the left of the mirror at a distance of $30cm$ from the pole of the mirror.

The negative sign shows that the image is formed on the left side of the mirror.

Step 4 - Finding the nature of the object

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

$$\begin{gathered} m=-\frac{v}{u} \\ m=-\frac{(-30cm)}{(-15cm)} \\ m=-\frac{30}{15} \\ m=-2 \end{gathered}$$

The negative magnification indicates that the image is real and inverted.

Step 5 - Finding the size of the object

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

$$\begin{gathered} m=\frac{h\text{'}}{h} \\ h\text{'}=m\times h \\ h\text{'}=(-2)\times (5cm) \\ h\text{'}=-10cm \end{gathered}$$

Therefore, the height of the image is $10cm$.

Final answer -

• Image is formed at $30cm$ to the left of the mirror.
• The nature of the image formed is real and inverted.
• The height of the image is $10cm.$