Answer:
The object distance is [tex]D_{o}=15 cm[/tex]
Explanation:
According to the Descartes equation, you have:
[tex]\frac{1}{D_{o}}+\frac{1}{D_{i}} = \frac{1}{f}[/tex]
[tex]Do= ??[/tex] [object distance]
[tex]Di= -30cm[/tex] [negative image distance from being behind]
[tex]f= 30cm [/tex] [focal length]
Clearing in the Descartes equation:
[tex]D_{o}= \frac{D_{i} f}{(f-D_{i})} = \frac{30*30}{(30 - (-30))} = \frac{900}{60}} = 15[/tex]
The object distance is [tex]D_{o}=15[/tex] cm
