A object of 2.0cm high is placed 10cm from a converging mirror of focal length 5cm,what is the height of the image formed
Given:
Object height (h_o) = 2.0 cm
Object distance (u) = -10.0 cm (since the object is placed in front of the mirror, its distance is negative)
Focal length of the mirror (f) = +5.0 cm (since the mirror is converging, its focal length is positive)
Using the mirror formula:
\[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\]
Substitute the given values:
\[\frac{1}{5} = \frac{1}{-10} + \frac{1}{v}\]
Solving for v:
\[\frac{1}{5} = -\frac{1}{10} + \frac{1}{v}\]
\[\frac{1}{5} + \frac{1}{10} = \frac{1}{v}\]
\[\frac{2}{10} + \frac{1}{10} = \frac{1}{v}\]
\[\frac{3}{10} = \frac{1}{v}\]
\[v = \frac{10}{3} \text{ cm} = 3.33 \text{ cm}\]
Now, using the magnification formula:
Magnification (M) = \[\frac{Image height (h_i)}{Object height (h_o)}\] = \[\frac{-v}{u}\] = \[\frac{-3.33}{-10}\] = 0.3333
Therefore, the height of the image formed can be calculated as:
h_i = M * h_o = 0.3333 * 2.0 cm = 0.67 cm
The height of the image formed is 0.67 cm.