Answer:
[tex] 0.25[/tex] cm
Explanation:
[tex]h_{o}[/tex] = height of the object = 0.54 cm
[tex]d_{o}[/tex] = distance of the object = 8.5 cm
[tex]f[/tex] = focal length of the diverging lens = - 7.5 cm
[tex]d_{i}[/tex] = distance of the image = ?
Using the lens equation
[tex]\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}[/tex]
[tex]\frac{1}{8.5} + \frac{1}{d_{i}} = \frac{- 1}{7.5}[/tex]
[tex]d_{i} = - 3.98[/tex] cm
[tex]h_{i}[/tex] = height of the image
[tex]\frac{h_{i}}{0.54} = \frac{- (- 3.98)}{8.5}[/tex]
[tex]h_{i} = 0.25[/tex] cm
