# Solve the following A 1.0 -cm-high object is placed at a distance of 12 cm from a convex lens of focal length 16 cm.(a) Find the position of the image. (b) Is the image real or virtual? (c) Find the size of the image. (d) Is the image erect or inverted?

a) $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\ \Rightarrow \frac{1}{16}=\frac{-1}{12}+\frac{1}{v}\\ \Rightarrow \frac{1}{v}=\frac{1}{16}+\frac{1}{12}\\ \Rightarrow \frac{1}{v}=\frac{16+12}{16\times 12}\\ \Rightarrow v=\frac{16\times12}{16+12}\Rightarrow v=\frac{16\times 12}{28}=6.85cm$

b) real image

c) $m=\frac{v}{u}=\frac{h_i}{h_o}\Rightarrow \frac{6.85}{-12}=\frac{h_i}{1}\Rightarrow h_i=-0.5 cm$,

d) hi=-0.5 cm where -ve sign indicates that the image is inverted.

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