Question

An object 4.0 cm in size, is placed 25.0 cm in front of a concave mirror of a focal length of 15.0 cm.

(i) At what distance from the mirror should a screen be placed in order to obtain a sharp image?

(ii) Find the size of the image.

(iii) Draw a ray diagram to show the formation of the image in this case.

Answer 1

Step 1:Given:

Height of the object, $h_1=4cm$

Distance of the object from the mirror, $u=-25cm$

The focal length of the mirror, $f=-15cm$

Step 2: (i) Finding the distance of the image (v) using the mirror formula:

We know that $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$$\begin{gathered} \Rightarrow \frac{1}{v}+\frac{1}{-25}=\frac{1}{-15} \\ \Rightarrow \frac{1}{v}=\frac{1}{-15}+\frac{1}{25}=\frac{(-25+15)}{375}=\frac{1}{-37.5}cm \\ \Rightarrow v=–37.5cm \end{gathered}$$

The image is formed object side, 37.5cm from the pole of the concave mirror.

Step 3: (ii) Finding the height of the image h2 using the formula:

We know that $\frac{H_2}{H_1}=\frac{-u}{v}$

Where $H_1$ is the size of the image .

Now, $\frac{H_2}{4cm}=\frac{-(-37.5cm)}{(-25cm)}$

$H_2=-4\times \frac{37.5}{25}=-6cm$

∴ size of image = 6cm

Step 4: (iii) The ray diagram showing the formation of the image in this case: