Question

A 2 cm tall object is placed perpendicular to the principal axis of a conave mirror of focal length 10 cm. The distance of object from the mirror is 15 cm. Find the position, nature and size of the image.

Answer 1

Given
Object distance, $u=-15cm$
Focal length, $f=-10cm$
Object height, $h=+2cm$
Image distance, $v=?$
Image height, $h^{{}^{\prime }}=?$

Using the mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}+\frac{1}{-15}=\frac{1}{-10}$
$v=-30cm$

Negative sign of $v$ shows that the image is formed at a distance of 30 cm on the left side of the mirror. Therefore, the image is real and inverted.

From the formula of magnification,
$m=-\frac{v}{u}=-\frac{-30}{-15}=-2$

Again, from the formula of magnification,
$m=\frac{h^{\prime }}{h}$
$\frac{h^{{}^{\prime }}}{2}=-2$
$h^{\prime }=-2\times 2=-4cm$

Negative sign with the magnification and height of the image shows that the image is inverted and real. Thus, a real image of height 4 cm is formed at a distance of 30 cm on the left side of the mirror. Image is inverted and twice the size of the object.