Given,
The image distance, v=17.22 cm
The focal length of the lens, f=3.05 cm
From the thin lens formula,
[tex]\begin{gathered} \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \\ \end{gathered}[/tex]Where u is the object distance.
On rearranging the above equation,
[tex]\begin{gathered} \frac{1}{u}=\frac{1}{f}-\frac{1}{v} \\ =\frac{v-f}{vf} \\ \Rightarrow u=\frac{vf}{v-f} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} u=\frac{17.22\times3.05}{17.22-3.05} \\ =3.71\text{ cm} \end{gathered}[/tex]Thus the object distance should be 3.71 cm
