Question

An object of size $5.0cm$ is placed $30cm$ in front of a concave mirror of focal length$15cm$. At what distance should a screen be placed, so that a sharp, focussed image can be obtained? Find the size and nature of the image.

Answer 1

Step 1 - Given data

Height of the object $h=5.0cm$

The distance of the object from the pole of the mirror $u=-30cm$ (Negative sign shows that the object is placed in the front of the mirror)

Focal length $f=-15cm$ (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.

$$\begin{gathered} v=\frac{(-30cm)(-15cm)}{(-30cm)-(-15cm)} \\ v=\frac{+450}{-15} \\ v=-30cm \end{gathered}$$

Hence, the screen should be placed $30cm$ away in the front of the mirror.

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

Step 4 - Finding the size of the image

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

$$\begin{gathered} m=-\frac{v}{u} \\ m=-\frac{(-30cm)}{(-30cm)} \\ m=-\frac{30}{30} \\ m=-1 \end{gathered}$$

Therefore, the image is real, inverted, and has the same size as the object.

Final answer - For a sharp, focussed image, the screen should be placed $30cm$ away in the front of the mirror$(v=-30cm)$. The size of the image is the same as the object and the image is real and inverted.