Question

An object is placed in front of a convex mirror of radius of curvature $40\mathrm{cm}$ at a distance of $10\mathrm{cm}$. Find the position and nature of the image.

Answer 1

Step 1: Given data:

Radius of curvature($\mathrm{R}$) of convex mirror = $+40\mathrm{cm}$

Object distance($\mathrm{u}$) = $-10\mathrm{cm}$

( Note: Sign convention:

1. All the distance to the left of the pole of a spherical mirror is negative($-$) and the distances taken to the right of the pole are positive($+$).
2. Thus, the sign of the radius of curvature($\mathrm{R}$) of the convex mirror is positive($+$) as it lies at the left of the pole.
3. Object distance($\mathrm{u}$)is negative($-$) as it is always considered to the left of the mirror.)

Step 2: formula used:

1. The radius of curvature($\mathrm{R}$) and the focal length($\mathrm{f}$) of a spherical mirror are related as;

$$\begin{gathered} \mathrm{Radius}\mathrm{of}\mathrm{curvature}\left(\mathrm{R}\right)=2\times \mathrm{Focal}\mathrm{length}\left(\mathrm{f}\right) \\ \mathrm{Or}, \\ \mathrm{Focal}\mathrm{length}\left(\mathrm{f}\right)=\frac{\mathrm{Radius}\mathrm{of}\mathrm{curvature}\left(\mathrm{R}\right)}{2} \end{gathered}$$

2. Object distance($\mathrm{u}$), image distance($\mathrm{v}$) and focal length($\mathrm{f}$) of a mirror are related as:

$\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$

3. Magnification of a mirror($\mathrm{m}$)= $-\frac{\mathrm{Image}\mathrm{distance}\left(\mathrm{v}\right)}{\mathrm{Object}\mathrm{distance}\left(\mathrm{u}\right)}$

(Note: Negative magnification suggests a real and inverted image whereas a positive magnification suggests a virtual and erect image.)

Step 3: Calculation of focal length($\mathrm{f}$) of the convex mirror:

$$\begin{gathered} \mathrm{f}=\frac{+40\mathrm{cm}}{2} \\ \mathrm{f}=+20\mathrm{cm} \end{gathered}$$

Step 4: Calculation of position and nature of image:

The position of the image or image distance($\mathrm{v}$) will be,

$$\begin{gathered} \frac{1}{\mathrm{v}}+\frac{1}{-10\mathrm{cm}}=\frac{1}{+20\mathrm{cm}} \\ \frac{1}{\mathrm{v}}-\frac{1}{10}=+\frac{1}{20} \\ \frac{1}{\mathrm{v}}=\frac{1}{20}+\frac{1}{10} \\ \frac{1}{\mathrm{v}}=\frac{1+2}{20} \\ \frac{1}{\mathrm{v}}=\frac{3}{20} \\ \mathrm{Or}, \\ \mathrm{v}=+\frac{20}{3}\mathrm{cm} \\ \mathrm{v}=+6.67\mathrm{cm} \end{gathered}$$

Now,

$$\begin{gathered} \mathrm{m}=-\frac{+6.67\mathrm{cm}}{-10\mathrm{cm}} \\ \mathrm{m}=+\frac{667}{1000} \\ \mathrm{m}=0.667\left(\mathrm{Positive}\right) \\ \mathrm{Which}\mathrm{suggests}\mathrm{a}\mathrm{virtual}\mathrm{and}\mathrm{erect}\mathrm{image}. \end{gathered}$$

Hence,

The image of the truck will be at $6.67\mathrm{cm}$ behind the mirror and it will be a virtual and erect image.