We will have the following:
First, we will recall that:
[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}[/tex]That is:
[tex]\begin{gathered} \frac{1}{5.75}=\frac{1}{2.33}+\frac{1}{u}\Rightarrow\frac{1}{u}=-\frac{1368}{5359} \\ \\ \Rightarrow u=-\frac{5359}{1368}\Rightarrow u\approx-3.92 \end{gathered}[/tex]So, the image distance is approximately -3.92 cm.
