Question

If a thin converging lens of focal length 20 cm is placed at a definite position in between an object and a screen, an image is formed on the screen. If the lens is shifted through a distance of 20 cm towards the screen, again an image is formed on the screen. Determine the distance between the object and the screen and the magnification in each case.

Answer 1

Dear student
$$\begin{gathered} letthedis\tan cebetweentheobjectandthescreenisD.Letthedis\tan ceofobjectfromthelensisu.Dis\tan ceofimagefromthelensis(D-u). \\ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \frac{1}{D-u}-\frac{1}{-u}=\frac{1}{f} \\ \frac{1}{D-u}+\frac{1}{u}=\frac{1}{f} \\ \frac{1}{D-u}+\frac{1}{u}=\frac{1}{f} \\ \frac{u+D-u}{(D+u)\times u}=\frac{1}{f} \\ \frac{D}{(D+u)\times u}=\frac{1}{f} \\ \frac{D}{(D+u)\times u}=\frac{1}{f} \\ solving \\ {u}^2-Du+Df=0 \\ solvingquadraticeqn \\ {u}_1=\frac{D+\sqrt{D^2-4fD}}{2}and{u}_2=\frac{D+\sqrt{D^2-4fD}}{2} \\ Dis\tan cebywhichlensisshiftedtoformtheimageonthescreenis \\ shift=u_2-u_1=\frac{D+\sqrt{D^2-4fD}}{2}-\frac{D+\sqrt{D^2-4fD}}{2}=\sqrt{D^2-4fD} \\ d=\sqrt{D^2-4fD} \\ squaringbothsides \\ {d}^2=D^2-4fD \\ {20}^2=D^2-4\times 20\times D \\ 400=D^2-4\times 20\times D \\ {D}^2-80\times D-400=0 \\ D=\frac{80+\sqrt{8000}}{2}=84.72cmand\frac{80-\sqrt{8000}}{2}\to thisvalueisrejectedasitisnegative \\ regards \end{gathered}$$