Question

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

Answer 1

Hint:This bis convex mirror and we know it always forms a virtual image. Virtual image is that image which if not formed by actual meeting of the rays but they appear to meet. Also, we know that virtual images are always erect.

Complete step by step answer:
Given convex mirror
Object distance, u= -20 cm
Height of the object, ho= 5 cm
Radius of curvature of convex mirror is always positive, R=30 cm
The relationship between radius of curvature and focal length is given as $$f={\displaystyle \frac{R}{2}}=15cm$$
Now using the mirror formula,
$$\begin{array}{rl}& {\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}\\ & {\displaystyle \frac{1}{v}}={\displaystyle \frac{1}{f}}-{\displaystyle \frac{1}{u}}\\ & v={\displaystyle \frac{fu}{u-f}}\end{array}$$
Putting the values, $$v={\displaystyle \frac{15\times (-20)}{-20-15}}=8.57cm$$
Thus, the image is formed at a distance of 8.57cm on the right side from the pole.
Now in order to find the magnification using the formula, $$m={\displaystyle \frac{-v}{u}}={\displaystyle \frac{-8.57}{-20}}=0.42$$
Since, the magnification is lesser than one, the image is diminished. Also, magnification is positive so the image is erect.
So, the answer is that the image is erect, diminished and is formed at a distance of 8.57cm on the right side from the pole.

Additional Information: Convex mirrors are often used in the hallways of buildings including stores, schools, hospitals, hotels and apartment buildings.

Note:While solving such problems we have to remember the sign convention. All the distances left to the pole are measured in negative and right to the pole are measured in positive