Question

The radius of curvature of a convex mirror used on a moving automobile is $2\mathrm{m}$. A truck is coming behind it at a constant distance of $3.5\mathrm{m}$. Calculate the position and nature of the image.

Answer 1

Step 1; Given data:

The radius of curvature of convex mirror($\mathrm{R}$) = $+2\mathrm{m}$

Distance of truck from the mirror($\mathrm{u}$) = $-3.5\mathrm{m}$

(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($\mathrm{R}$) and the focal length($\mathrm{f}$)of the convex mirror is positive(+) as both lie at the right 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 focal length of a convex mirror($\mathrm{f}$) = $\frac{\mathrm{Radius}\mathrm{of}\mathrm{curvature}\left(\mathrm{R}\right)}{2}$
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($\mathrm{m}$)= $-\frac{\mathrm{v}}{\mathrm{u}}$

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

Step 3: Calculation of position and nature of image:

$$\begin{gathered} \mathrm{Focal}\mathrm{length}\left(\mathrm{f}\right)=\frac{+2\mathrm{m}}{2} \\ \mathrm{f}=+1\mathrm{m} \end{gathered}$$

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

$$\begin{gathered} \frac{1}{\mathrm{v}}+\frac{1}{-3.5\mathrm{m}}=\frac{1}{+1\mathrm{m}} \\ \frac{1}{\mathrm{v}}-\frac{1}{3.5}=\frac{1}{1} \\ \frac{1}{\mathrm{v}}=\frac{1}{1}+\frac{1}{3.5} \\ \frac{1}{\mathrm{v}}=\frac{3.5+1}{3.5} \\ \frac{1}{\mathrm{v}}=\frac{4.5}{3.5} \\ \frac{1}{\mathrm{v}}=\frac{45}{35} \\ \frac{1}{\mathrm{v}}=\frac{9}{7} \\ \mathrm{Or}, \\ \mathrm{v}=+\frac{7}{9}\mathrm{m} \end{gathered}$$

Thus, the image will be formed at a distance of $\frac{7}{9}\mathrm{m}$ behind the mirror.

Now,

$$\begin{gathered} \mathrm{m}=-\frac{+{\displaystyle \frac{7}{9}}\mathrm{m}}{-3.5\mathrm{m}} \\ \mathrm{m}=\frac{7}{9\times 3.5} \\ \mathrm{m}=\frac{70}{9\times 35} \\ \mathrm{m}=\frac{2}{9}\left(\mathrm{Positive}\right) \end{gathered}$$

The image will be virtual and erect.

Hence,

The image of the truck will be at $\frac{7}{9}\mathrm{m}$ behind the mirror and it will be a virtual and erect image.