Question

An object 5cm in focal length is placed at a distance of 20cm in front of a convex mirror of focal length 15cm. Find the position, nature and size of the image.

Answer 1

Dear Student,
Applying sign conventions we can write the object height, ho = 5cm; object distance, u = - 20cm; focal length of the convex mirror,f = 15cm.
Using the mirror formula:
$$\begin{gathered} \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{15}-\frac{1}{-20}=\frac{1}{15}+\frac{1}{20}=\frac{4+3}{60}=\frac{7}{60} \\ Thus,v=\frac{60}{7}cm \\ Now,magnification,m=-\frac{v}{u}=-\frac{{\displaystyle \raisebox{1ex}{$60$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}}{-20}=\frac{3}{7} \\ Also,m=\frac{imageheight}{objectheight}=\frac{h_i}{5}=\frac{3}{7}=\frac{15}{7}=2.14cm \end{gathered}$$
Thus,
Since the magnification is positive, the image is erect and virtual .
Regards