Question

A $4.5\, cm$ needle is placed $12\, cm$ away from a convex mirror of focal length $15\, cm$. Give the location of the image and the magnification. What happens as the needle is moved farther from the mirror?

• A. Image goes on increasing
• B. lmage goes on decreasing
• C. Image will be unchanged
• D. Image may change or change

Answer 1

Answer: (B)

Given, focal length of convex mirror $f=+15 \,cm$
(Focal length of convex mirror is taken as positive)
Distance of object $u=-12 \,cm$
Size of object $O=4.5\,cm$
Using the mirror formula,
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$\frac{1}{15}=\frac{1}{v}-\frac{1}{12}$
$ \Rightarrow \frac{1}{v}=\frac{1}{15}+\frac{1}{12}$
$=\frac{4+5}{60}=\frac{9}{60}$
Distance of image from the mirror $v=6.7\, cm$ T
he positive sign shows that the image is formed behind the mirror.
Using the formula of magnification,
$m=-\frac{v}{u} =\frac{1}{O} $
$\frac{-6.7}{-12} =\frac{1}{4.5}$
Size of image $I=2.5\, cm$
As $I$ is positive, so image is erect and virtual.
Magnification $m$ is given by
$m=\frac{1}{O}=\frac{2.5}{4.5}=\frac{25}{45}=\frac{5}{9}$
As the needle moves away from the mirror, the image also moves away from the mirror (as $u \rightarrow \infty, v \rightarrow f)$ and the size of image goes on decreasing.