A 16 cm long pencil is placed as shown in figure. The central point C is 45 cm away from a 20 cm focal length lens and 15 cm above the optic axis. Find the length of image.

For point C $\frac{1}{v}-\frac{1}{-45}=\frac{1}{20}orv=36cm$
Using $M=\frac{I}{O}=\frac{v}{u}.y_c^{{}^{\prime }}=\frac{-36}{45}\times 15=-12.$ Thus co-ordinates of $C^{{}^{\prime }}$ are (36.-12)
For point B $$\begin{gathered} u=45+8\cos 45=45+5.656 \\ \frac{1}{v}=\frac{1}{20}-\frac{1}{50.66} \\ orv=\frac{50.66\times 20}{30.66}=33cm, \\ {y}_B^{{}^{\prime }}=\frac{-33}{50.66}\times \left(15-4\sqrt{2}\right) \end{gathered}$$
Thus coordinates of $B^{{}^{\prime }}$ are (33,-6.1)
For point A $$\begin{gathered} u=45-4\sqrt{2}=39.34cm \\ \frac{1}{v}=\frac{1}{20}-\frac{1}{39.34} \\ orv=\frac{39.34\times 20}{19.34}=40.70cm, \\ {y}_A^{{}^{\prime }}=\frac{-40.7}{39.34}\times \left(15-4\sqrt{2}\right)=21.4cm \end{gathered}$$
Thus coordinates of $A^{{}^{\prime }}$ are (40.7,-21.4)
Length of image $$\begin{gathered} A^{{}^{\prime }}{B}^{{}^{\prime }}=\sqrt{{\left(40.7-33\right)}^2+{\left(21.4-6.1\right)}^2} \\ =\sqrt{{7.7}^2+{15.3}^2}=17.1cm \end{gathered}$$.