Question

An object of 5 cm is placed at a distance of 20 cm in front of a convex mirror with a radius of curvature of 30 cm. Find the position, nature, and size of the image.

Answer 1

Step (1) Given:

Object distance, $u=–20cm$

Object height, $h=5cm$

The radius of curvature, $R=30cm$

The radius of curvature $=2\times Focallength=30cm$

Step (2) Finding the focal length of the convex mirror:

We know that, $R=2f$

$f=15cm$

Step (3) Finding the distance of the image:

According to the mirror formula,

$$\begin{gathered} \begin{array}{l}\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\end{array} \\ \Rightarrow \begin{array}{l}\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\end{array} \\ \Rightarrow \begin{array}{l}\frac{1}{v}=\frac{1}{15}+\frac{1}{20}\end{array} \\ \begin{array}{l}=\frac{4+3}{60}\end{array} \\ =\frac{7}{60}=8.57cm \end{gathered}$$

The positive value of the image distance indicates that the image is formed behind the mirror

Step (4) Finding the magnification of the lens:

Magnification, $m=–\frac{Imagedis\tan ce}{objectdis\tan ce}=-\frac{8.57}{-20}=0.428$

The positive value of magnification indicates that the image is formed is virtual

When the value of the magnification is less than 1, then the image size is less than the object size.

Step (5) Finding the height of the image:

$$\begin{gathered} Magnification=\frac{Heightoftheimage}{Heightoftheobject}=\frac{h’}{h} \\ \Rightarrow h’=m\times h=0.428\times 5=21.4cm \end{gathered}$$

The positive value of image height indicates that the image formed is erect.

Hence, the image formed is erect, virtual, and smaller in size.