Question

A security mirror used in a big showroom has a radius of curvature $5m$. If a customer is standing at a distance of $20m$ from the cash counter, find the position, nature and size of the image formed in the security mirror.

Answer 1

Step 1: Given data

1. The radius of curvature of the mirror is $C=5m.$
2. The customer placed at a distance from the case counter is $u=20m.$

Step 2: Formula used

Focal length is defined by the form, $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$, where f is the focal length, u and v are the object distance and image distance.

Step 3: The radius of curvature and focal length

1. The radius of curvature of a mirror is the radius of the sphere, from which the mirror is created.
2. When a parallel beam of light rays falls on a mirror and converged or diverged from a point after reflection, this point is known as the focal point of the mirror. The distance of the focus from the center of a mirror is called the focal length.
3. For a mirror, the radius of curvature and focal length is related $f=\frac{C}{2}$, where, C is the radius of curvature and f is the focal length.

Step 4: Finding the image distance

It is clear that the mirror is a convex mirror. Because the convex mirror increases the range of vision. The focus of the convex mirror is placed behind the mirror and considering that the mirror is placed at the case counter. The customer is placed in front of the mirror, so the object distance is negative.

As we know, $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ or\frac{1}{2.5}=\frac{1}{v}+\frac{1}{-20}\left(Since,f=\frac{C}{2}\right) \\ or\frac{1}{v}=\frac{1}{2.5}+\frac{1}{20}=\frac{10}{25}+\frac{1}{20}=\frac{2}{5}+\frac{1}{20} \\ or\frac{1}{v}=\frac{8+1}{20}=\frac{9}{20} \\ orv=\frac{20}{9}=2.22 \\ orv=2.22 \end{gathered}$$

Step 5: Diagram: image formation on a convex mirror

Step 6: Size of the image

We know for a mirror, $linearmagnifcation=\frac{imagedis\tan ce}{objectdis\tan ce}=\frac{2.22}{20}=0.11$

In this case, $imagedis\tan ce<objectdis\tan ce.$

So, $linearmagnifcation<1.$

So, the image is shorter than the object.

Step 7: Nature of the image

We know that in a convex mirror all images are virtual, this is a property of a convex mirror. An image in a convex mirror and plane mirror is always upright.

Therefore, the image will be smaller than the actual size of the customer. The nature of the image is virtual, upright, and diminished.