A security mirror in a store is required to provide a wide field of view and always form diminished, upright images regardless of the position of objects. Which type of mirror and principle should be used for its design?

The correct option is: Convex mirror using the mirror formula for virtual, diminished images

1. Understanding the Requirements:
The security mirror needs to satisfy two main criteria:
- Provide a wide field of view to monitor a large area of the store.
- Always form diminished (smaller) and upright (erect) images, regardless of where the objects (like customers or items) are positioned.

2. Analyzing Mirror Types:
- Plane Mirror: Form images that are of the same size as the object. They do not offer a wider field of view than their physical size permits.
- Concave Mirror: Can form real and inverted images, or virtual and magnified images depending on the object's distance. They do not always form diminished, upright images.
- Convex Mirror: Diverges light rays incident on it. Because the rays diverge, they appear to meet behind the mirror, forming a virtual, upright, and diminished image. This divergence also allows the mirror to capture light from a much wider angle, providing a significantly larger field of view compared to plane or concave mirrors of the same size.

3. Applying the Mirror Formula:
We can mathematically verify these properties using the mirror formula:
$$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$$
and the magnification equation:
$$m=-\frac{v}{u}$$

For a convex mirror, the focal length f is always positive ($f>0$). The object distance u is always negative ($u<0$) because the object is placed in front of the mirror.
Rearranging the mirror formula to solve for the image distance v:
$$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$$
Since u is negative, let $u=-\left|u\right|$:
$$\frac{1}{v}=\frac{1}{f}-\left(\frac{1}{-\left|u\right|}\right)=\frac{1}{f}+\frac{1}{\left|u\right|}$$
Since both f and $\left|u\right|$ are positive, $\frac{1}{v}$ must be positive, which means v is always positive ($v>0$). A positive image distance confirms that the image is virtual and located behind the mirror.

Next, let's examine the magnification:
$$m=-\frac{v}{u}=-\frac{v}{-\left|u\right|}=\frac{v}{\left|u\right|}$$
Since $\frac{1}{v}=\frac{1}{f}+\frac{1}{\left|u\right|}$, it is clear that $\frac{1}{v}>\frac{1}{\left|u\right|}$, which implies that $v<\left|u\right|$.
Consequently, the absolute value of magnification is less than 1:
$$\left|m\right|=\frac{v}{\left|u\right|}<1$$
A magnification value between 0 and 1 mathematically guarantees that the image is upright (since m is positive) and diminished (since $m<1$). This behavior holds true for all possible object distances u.