Question

Question) The image of a film is to be cast on screen with a magnification 50 times its height. if focal length of the lens is7.5 cm, how far is the screen to be placed from the lens? under this condition what is the distance of the film from the lens?
[382.5 cm, 7.65 cm]

Answer 1

Dear student,

$$\begin{gathered} m=-50,f=7.5\sin cetheimageisrealthemagnificationshouldbe-ve. \\ m=\frac{v}{u} \\ v=-50u \\ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \frac{1}{-50u}-\frac{1}{u}=\frac{1}{7.5} \\ \frac{1}{u}\left[\frac{1}{-50}-1\right]=\frac{1}{7.5} \\ u=-7.65cmthedis\tan ceofthefilmfromthelensis7.65cm. \\ v=7.65*50=382.5cmisthethedis\tan ceofscreenfromthelens. \\ Regards \end{gathered}$$