Reflection On A Plane Mirror

Imagine standing in front of a perfectly still lake, where the water's surface acts like a mirror, reflecting your image with remarkable clarity. This natural phenomenon is akin to how a plane mirror functions. When light rays strike the smooth, flat surface of a plane mirror, they bounce back in a predictable manner, creating a clear and upright image of whatever is in front of it. This process, known as reflection, follows the law of reflection, which states that the angle of incidence (the angle at which the incoming light hits the mirror) is equal to the angle of reflection (the angle at which the light bounces off).

This Story also Contains

1. What is an Object?
2. What is Image?
3. Rotation of plane Mirror
4. The angle of Deviation
5. Solved Examples Based on Reflection On A Plane Mirror
6. Summary

In everyday life, plane mirrors are ubiquitous, from bathroom mirrors to periscopes in submarines. They provide accurate reflections that are essential for various tasks, such as grooming, enhancing room decor, and aiding in scientific observations. The simplicity and reliability of plane mirrors make them an indispensable tool in both everyday activities and advanced technological applications.

What is an Object?

Objects are sources of light rays that are incident on an optical element.

• Real object: An object is real if two or more incident rays actually emanate or seem to emanate from a point.
• Virtual object: An object is virtual when two incident rays seem to converge to that point.

What is Image?

An image is the point of convergence or apparent point of divergence of rays after they interact with a given optical element. An object provides rays that will be incident on an optical element. The optical element reflects or refracts the incident light rays which then meet at a point to form an image. As in the case of objects, images too can be real or virtual.

• Real Image: Real images are formed when the reflected or refracted rays actually meet or converge to a point. If a screen is placed at that
  point, a bright spot will be visible on the screen. Thus, a real image can be captured on a screen.
• Virtual image: an optical image formed from the apparent divergence of light rays from a point, as opposed to an image formed from their actual divergence

Image Formation by Plane Mirror

We have to see the rays coming from the object to see it. If the light first hits the mirror and then reflects with the same angle, the extensions of the reflected rays are focused at one point behind the mirror. We see the coming rays as if they are coming from behind the mirror. At a point, A' image of the point is formed and we call this image a virtual image The distance of the image to the mirror is equal to the distance of the object to the mirror.

a) For a point object

Distance of object from mirror = Distance of image from the mirror

(i) All the incident rays from a point object after reflection from a plane mirror will meet at a single point which is called an image.

(ii) The line joining a point object and its image is normal to the reflecting surface

b) For an extended object

The size of the image is the same as that of the object. An image of an extended object by a plane mirror is a virtual image. The image will be upright and laterally inverted.

Rotation of plane Mirror

When a plane mirror is rotated by an angle θ, the reflected ray rotates by an angle
2θ. This principle is crucial in optical instruments and applications where precise control of light direction is needed.

i) Mirror is rotated keeping ray of incidence fixed:

Consider that before the mirror is rotated the angle of incidence and recfection is $\theta$. When the mirror is rotated through and angle of say $\phi_{\text {in }}$ a clockwise direction, then the normal is also rotated by an angle $\phi$ and thus the new angle of incidence becomes $\theta+\phi$ and thus new angle of reflection will be to $\theta+\phi$ as shown in the below figure.

So the angle between the incident ray and new reflected ray is $2(\theta+\phi)$.
Let due to rotation of mirror new reflected ray get deflected by an angle $\delta$ in a clockwise direction with respect to the original reflected ray.

So So the angle between the incident ray and the new reflected ray is $2 \theta+\delta$..

i.e For fixed incident ray, When the mirror is rotated $\phi$ in a clockwise direction then reflected ray get deflected by $2 \phi_{\text {in the clockwise direction }}$

ii) Mirror is fixed, angle of the incident ray changed :

when the mirror is fixed and the angle of incidence is changed by an angle $\alpha$in an anti-clockwise direction, Then the angle of reflection will rotate by an angle $a$ in the clockwise direction as shown in the below figure.

The angle of Deviation

The angle of Deviation is the angle made by the reflected ray with the direction of the incident ray.

A ray of light (PO) that is incident onto the surface of a plane mirror is reflected as OR with the angle of incidence equal to the angle of reflection. Suppose that the ray had continued, through the mirror, in a straight line along OQ as shown in the below figure.

Then the angle made by the reflected ray (OR) with OQ is called the angle of Deviation $(\delta)$

As for the above figure

$\begin{aligned} & i+r+\delta=\pi \\ \Rightarrow & \delta=\pi-i-r \quad(\text { and using } i=r) \\ \Rightarrow & \delta=\pi-i-i=\pi-2 i\end{aligned}$

Number of images formed by two plane mirrors

Number of images formed by two plane mirrors:

The number of images formed by two adjacent plane mirrors depends on the angle between the mirror. If θ (in degrees) is the angle between the plane mirrors then the number of images are given by,

$n=\frac{360}{\theta}-1$

n = no. of images formed

Let us take a few cases(with some conditions) to understand it better -

Case 1: When two mirrors are placed parallel to each other

When the two mirrors are aligned at a 0-degree angle with each other (i.e., a parallel mirror system), there is an infinite number of images. The diagram below shows the multiple images for a parallel mirror system. Images I₁ and I₂ are primary images. Image I₁ is the image resulting from the reflection of the object O across mirror M₁ and image I₂ is the image resulting from the reflection of the object O across mirror M₂. Image I₃ is an image of image I₁, found by reflecting image I₁ across mirror M₂. Image I₄ is an image of image I₂; found by reflecting image I₂ across mirror M₁. This process could continue indefinitely, producing images of images for an infinite number of images extending to the right of mirror M₂ and to the left of mirror M₁. we can also calculate the number of images formed by using the formula

$n=\frac{360}{A}-1$

where $\theta$ is 0 degrees. So the number of images formed will be infinite.

Case 2: When two mirrors are placed perpendicular to each other

The number of images formed when two mirrors are placed at an angle theta to each other is given by:

$n=\frac{360}{\theta}-1$

So, here, we have the mirrors placed perpendicular to each other. So, $\theta$= 90 degree

$\begin{aligned} & n=\frac{360}{90}-1 \\ & n=4-1 \\ & n=3\end{aligned}$

Case 3: When two mirrors are placed at 120 degrees

Here, we have the mirrors placed at an angle of 120 degrees. and object is kept at angle bisector of two mirrors. So, $\theta$ = 120 degrees.

$\begin{aligned} & n=\frac{360}{120}-1 \\ & n=3-1 \\ & n=2\end{aligned}$

Also when the object is not kept at the angle bisector of two mirrors then the number of images formed by two mirrors can be calculated by the formula

$\begin{aligned} & n=\frac{360}{\theta} \\ & n=\frac{360}{120} \\ & n=3\end{aligned}$

• If the value of $\frac{360}{\theta}$ is even, then we will use the formula

$n=\frac{360}{\theta}-1$

But If the value of $\frac{360}{\theta}$ is odd, then we have two different cases

1.$n=\frac{360}{\theta}-1 \quad \ldots($ when the object is placed symmetrically $)$

2. $n=\frac{360}{\theta} \quad \ldots$ (when the object is placed asymmetrically)

Solved Examples Based on Reflection On A Plane Mirror

Example 1: A convex mirror is used to form an image of a real object. Point out the wrong statement.

1) The image lies between the pole and the focus

2) The image is diminished in size

3) The image is real

4) The image erect

Solution:

A source of light rays that are incident on an optical element. It may be a point object or an extended object. They are of two kinds real object & virtual object.

wherein

A convex mirror forms a real image only for a virtual object.

Example 2: Mark the wrong statement about a virtual image.

1) A virtual image can be photographed

2) A virtual image can be seen

3) A virtual image can be photographed by exposing a film at the location of the image.

4) A virtual image may be diminished or enlarged in size in comparison to an object.

Solution:

Point of convergence or apparent point of divergence of rays. The image can be real or virtual.

A virtual image can be photographed.

e.g. we can take photographs of our mirror image from our position

Hence, the answer is the option 3.

Example 3: There is a point object and a plane mirror. If a mirror is moved by 10cm away from the object then the image will move by?

1) 10cm

2) 20cm

3) 30cm

4) 40cm

Solution:

Image formation from a plane mirror

1) Distance of object from mirror = Distance of image from the mirror.

2) Line joining a point object and its image is normal to the reflecting surface

3) Size of the image is the same as that of the object.

4) For a real object the image is virtual and for a virtual object, the image is real.

After moving the mirror by 10cm the object distance = x+10cm

Image distance from m = x+ 10cm

$\therefore$ Movement of the image from m

= (x+20) -x = 20 cm

Hence, the answer is option (2).

Example 4: A point source of light S, placed at a distance of 60 cm in front of the centre of a plane mirror of width 50 cm, hangs vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance of 1.2 m from it (see in the figure). The distance between the extreme points where he can see the image of the light source in the mirror is _________cm.

1) 150

2) 300

3) 75

4) 120

Solution:

$
\begin{aligned}
& \tan \theta=\frac{25}{60}=\frac{x}{180} \\
& \mathrm{x}=75 \mathrm{~cm}
\end{aligned}
$
so distance between extreme point $=2 \mathrm{x}=2 \times \quad 75=150 \mathrm{~cm}$

Hence, the answer is option (1).

Example 5: A light ray is incident, at an incident angle $\theta_1$, on the system of two plane mirrors $\mathrm{M}_1$ and $\mathrm{M}_2$ having an inclination angle $75^{\circ}$ between them (as shown in figure). After reflecting from mirror $\mathrm{M}_1$ it gets reflected back by the mirror $\mathrm{M}_2$ with an angle of reflection $30^{\circ}$. The total deviation of the ray will be $\qquad$ degree.

1) 210

2) 310

3) 410

4) 510

Solution:

$
\begin{aligned}
& \delta \rightarrow \text { angle of deviation. } \\
& 60^{\circ}+75^{\circ}+90^{\circ}-\mathrm{x}=180 \\
& \mathrm{x}=45^{\circ}
\end{aligned}
$

Also, $2 \mathrm{x}+60^{\circ}+\mathrm{y}=180^{\circ}$
$
\begin{gathered}
\mathrm{y}=30^{\circ} \\
\delta=180^{\circ}+\mathrm{y}=210^{\circ}
\end{gathered}
$

The angle of deviation is $210^{\circ}$

Summary

Thinking about a plane mirror is one of the basic ideas in optics where light rays touch a smooth flat surface and bounce back. The angle of light striking the mirror (incidence angle) equals the angle, at which light leaves it; this is called the principle of reflection. Thus, for instance; we can conclude that when we use a plane mirror the picture produced is dematerialized, erect and equidistant with the subject.