Question

An object 5.0 cm in length 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 of the image, its nature, and size.

Answer 1

Step1: Given data

The radius of curvature (R) of the mirror $=30cm$

Focal length, $f=\frac{R}{2}=\frac{30}{2}=15cm$

Distance of the object, $u=-20cm$

Height of the object, $h=5cm$

Step 2: Finding the distance of the image (v) using the formula

$$\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \\ \Rightarrow \frac{1}{v}=\frac{1}{15}-\frac{1}{-20} \\ \Rightarrow v=8.57cm \end{gathered}$$

Image is virtual and erect and formed behind the mirror.

Step 3: Finding the size of the image using the formula of magnification

$$\begin{gathered} m=\frac{-v}{u}=\frac{h_i}{h_o}=\frac{Heightofimage}{Heightofobject} \\ \frac{h_i}{5}=\frac{8.57}{20} \\ =0.428 \\ {h}_i=0.428x5=2.14cm \end{gathered}$$

Position of Image: Behind the mirror.

Nature of Image: Virtual and Erect.

Size of the Image: Diminished.