Remember that the thin-lens equation relates the fcal length (f) of a tin lens with the distance from the lens to the ojbject d_o and the distance from the lens to the image, d_o.
According to he txext of the image, the focal length is equal to x cm and the distance from the lens to the object is x+20cm. Then, f and d_o are known and we are asked to find an expression fo d_i. Isolate d_i from the equation:
[tex]\begin{gathered} \frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o} \\ \\ \Rightarrow\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_o} \\ \\ \Rightarrow\frac{1}{d_i}=\frac{d_o-f}{fd_o} \\ \\ \Rightarrow d_i=\frac{fd_o}{d_o-f} \end{gathered}[/tex]
Replace f=x and d_o+x+20:
[tex]d_i=\frac{(x)(x+20)}{(x+20)-x}=\frac{x(x+20)}{x+20-x}=\frac{x(x+20)}{20}=\frac{1}{20}x^2+x[/tex]
The distance from the lens to the place where the image is form is given by the expression x(x+20)/20.
Therefore,the expression in simplest oform for how far from the lens should a screen be placed to put the image in focus, is:
[tex]\frac{1}{20}x^2+x[/tex]
