Mirrors are considered as a polished surface that can reflect the light rays.

As shown in the figure all light rays emanating from a point source $P$ are reflected from a plane mirror so that the extensions of the reflected rays appear to come (or diverge) from point $P^{'}$. We call point $P$ an object point and point $P^{'}$ the corresponding image point. The rays do not actually pass through the mirror since the most mirrors in the market are opaque. In this case, no light rays don’t actually pass through the point $P^{'}$ (image point) and we call this image a virtual image. Therefore, the images that are formed on a plane mirror are virtual images.

To find the precise location of the images that are formed by a plane mirror, we must use at least two rays diverging from an object point $P$ as shown in the figure below. One of them incident normally on the mirror and the other strikes the mirror at angle of incidence $\theta$ and is reflected at an equal angle with the normal (due to the law of<img alt=">reflection: all rays striking any surface (polished or rough) are reflected at an angle from the normal equal to the incident angle). Now by extending these two reflected rays backward, they intersect at point $P^{'}$, at a distance $s^{'}$, behind the plane mirror. The distance $s^{'}$ which is the distance of the image from the mirror is called image distance. The location of the observer does not matter since if we draw the extension of all the reflected rays backward they appear to come from point $P^{'}$, confirming that $P^{'}$ is the image of $P$.

__Example__: **these two mirrors sit at angle of $120{}^\circ$ to one another. A ray of light is incident at $50{}^\circ$on the first mirror. What is the angle of refraction $\theta_r$, with respect to a perpendicular line to the second mirror, as shown?**

__Solution__:

By law of reflection, the angle of incidence equals the angle of reflection, the reflected ray from the first incident on the horizontal mirror is at angle $50{}^\circ$. In a triangle, the three interior angles always add to $180{}^\circ$, so the reflected ray of first mirror, which is served as the incident ray for second mirror, strike at angle of $70{}^\circ$ with respect to the perpendicular line of the second mirror. Again using the law of reflection, the reflected ray from the latter mirror is $\theta_r=70{}^\circ$.

For practical calculations, we must choose a sign convention that is applicable in all situations involving plane or spherical reflecting or refracting surfaces as follows:

**Sign rule for the object distance**: if the object is on the same side of the refracting or reflecting surface as the incoming light, then the object distance is positive $s>0$, otherwise, it is negative.**Sign rule for the image distance:**when the image is on the same side of the refracting or reflecting surface as the outgoing light, then the image distance is positive $s^{'}<0$, otherwise, it is negative.

Therefore, for a plane mirror, the object distance $s$ is equal to the image distance $s^{'}$ or $s=-s^{'}$.

- The image distance is equal to the object distance
- The image size is equal to the object size,$h_i=h_o$.
- Image is virtual (i.e. it is forms by extension of reflected rays).
- Image is upright.
- Image with respect to the object is lateral inversion.

The image formation of an extended object, which has finite size, is same as the point object. For example, consider an arrow oriented upright in front of a plane mirror. To construct the arrow’s image, it is sufficient to choose two rays from one of the points of the arrow such as its head (best choice) and reflect them from the mirror. It is obvious that the image of an extended object is extended. In any image forming situations we can define a useful quantity which is the ratio of image height $h^{'}$ to object height $h$ and is called the **lateral magnification** $m$

\[m=\frac{h^{'}}{h}\]

The lateral magnification for a plane mirror is unity since the image height of an object in the plane mirror is the same size as the real object. From this quantity we can deduce the orientation of object’s image. When $m>0$, we say that the image is erect or upright. In the plane mirrors we have always this case. But there are situations (such as in the spherical mirror) in which the lateral magnifications is negative $m<0$. In these cases, the image is in the opposite direction of the object or the image is inverted.

Plane Mirror,Object distance,Image distance,Lateral magnification,Magnification

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