Given:
The object distance is,
[tex]u=15.0\text{ cm}[/tex]The focal length is,
[tex]f=8.00\text{ cm}[/tex]To find:
The image distance
Explanation:
The lens formula gives,
[tex]\frac{1}{v}-\frac{1}{u}=\frac{1}{f}[/tex]For the concave lens, the focal length is negative and the object distance is negative as per the sign convention. So we write,
[tex]\begin{gathered} \frac{1}{v}-\frac{1}{-15.0}=\frac{1}{-8.00} \\ \frac{1}{v}=-\frac{1}{8.00}-\frac{1}{15.0} \\ \frac{1}{v}=-\frac{23}{120} \\ v=-\frac{120}{23} \\ v=-5.22\text{ cm} \\ The\text{ }negative\text{ sign indicates the image is virtual and in front of the lens} \end{gathered}[/tex]Hence, the image is 5.22 cm in front of the lens.
