Answer:
Image distance = 23.6cm
Explanation:
Given
[tex]Focal\ Length\ (f) = 29cm[/tex]
[tex]Object\ Distance (u) = 13\ cm[/tex]
Required
Determine the image distance (v)
This can be calculated using:
[tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/tex]
[tex]\frac{1}{29} = \frac{1}{13} + \frac{1}{v}[/tex]
Collect Like Terms
[tex]\frac{1}{v} = \frac{1}{29} - \frac{1}{13}[/tex]
[tex]\frac{1}{v} = \frac{13 - 29}{29 * 13}[/tex]
[tex]\frac{1}{v} = \frac{-16}{377}[/tex]
[tex]v = \frac{-377}{16}[/tex]
[tex]v = -23.5625[/tex]
[tex]v = -23.6\ cm[/tex]
This implies that the image distance is 23.6cm and it is a virtual image.
