Object And Image Velocity In Plane Mirror

Imagine running along the edge of a calm lake, where your reflection in the water mirrors your every step. In a plane mirror, the velocity of the image matches the object's velocity but in the opposite direction. If you move towards the mirror at a certain speed, your image moves towards you at the same speed; if you move away, your image does the same. This predictable relationship, dictated by the law of reflection, is essential for precise tasks like optical alignment and virtual reality environments. Just as your reflection in the lake keeps pace with you, the image in a plane mirror accurately mirrors your speed and direction.

This Story also Contains

1. The Relation Between the Velocity of the Object and Mirror in-Pane Mirror
2. Solved Examples Based on the Relation Between the Velocity of Object and Mirror in Plane Mirror
3. Example 1: If a plane mirror is approaching you with a speed of 10 cm/s, your image shall approach you with a speed of? Give your answer in cm/s.

The Relation Between the Velocity of the Object and Mirror in-Pane Mirror

In the case of a plane mirror, the distance of the object from the mirror is equal to the distance of the image from the mirror.

i.e. Distance of the Image formed in the mirror is the same as the distance of the object formed on the surface of the mirror.

Hence, from the mirror property:

$
x_{\mathrm{im}}=-x_{\mathrm{on}}, y_{\mathrm{im}}=y_{\mathrm{om}} \text { and } z_{\mathrm{im}}=z_{\mathrm{om}}
$

Here $x_{i m}$ means " x coordinate of the image with respect to mirror.
Differentiating w.r.t time, we get,
$
v_{(m m) x}=-v_{(\mathrm{om}) x} ; \quad v_{(\mathrm{im}) y}=v_{(\mathrm{om}) y} ; \quad v_{(\mathrm{im}) \mathrm{z}}=v_{(\mathrm{orn}) z}
$

Here,

$v_i=$ velocity of the image with respect to the ground.
$v_0=$ velocity of the object with respect to the ground.
$v_{\text {om }}=$ velocity of the object with respect to the mirror.
$v_{i m}=$ velocity of the object with respect to the mirror.
i.e $\vec{v}_{\text {om }}=\vec{v}_{\mathrm{o}}-\vec{v}_{\mathrm{m}} \quad$ and $\quad \vec{v}_{\mathrm{im}}=\vec{v}_{\mathrm{i}}-\vec{v}_{\mathrm{m}}$

For X -axis
$
\begin{aligned}
& v_{(\mathrm{m}) \mathrm{x}}=-v_{(\mathrm{om}) x} \\
& \Rightarrow \quad v_i-v_{\mathrm{m}}=-\left(v_{\mathrm{o}}-v_{\mathrm{m}}\right) \quad(\text { for } x \text {-axis })
\end{aligned}
$

l.e When the object moves with speed $v$ towards (or away) from the plane mirror then image also moves toward (or away) with speed $v$. But the relative speed of image w.r.t. the object is $2 v$.

For $y$-axis and $z$-axis
$
v_{(\mathrm{im}) y}=v_{(\mathrm{om}) y ;} \quad v_{(\mathrm{im}) \mathrm{z}}=v_{(\mathrm{om}) z}
$

| Relative velocity of image w.r.t. mirror | = | Relative velocity of object w.r.t. mirror |

But $\quad v_1-v_{\mathrm{m}}=\left(v_0-v_{\mathrm{m}}\right) \quad$ for $y$-and $z$-axis. or $\quad v_{\mathrm{i}}=v_{\mathrm{o}}$

Here, $v_i=$ velocity of the image with respect to the ground.
$v_0=$ velocity of the object with respect to the ground.

i.e. Velocity of the object is equal to the velocity of the image when the object is moving to parallel to the mirror surface.

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Solved Examples Based on the Relation Between the Velocity of Object and Mirror in Plane Mirror

Example 1: If a plane mirror is approaching you with a speed of 10 cm/s, your image shall approach you with a speed of? Give your answer in cm/s.

1) 10

2) 20

3) 30

4) 15

Solution:

If x is the distance between the object and the mirror, 2x is the distance between the object and its image because for a plane mirror object distance is equal to the image distance. So if the plane mirror is approaching the object with a speed of x/t, then the image will approach the object with a speed of 2x/t.If, $\frac{x}{t}=10 \mathrm{~cm} / \mathrm{s}$, then, $\frac{2 x}{t}=20 \mathrm{~cm} / \mathrm{s}$.

Hence the image will approach you at a speed of 20 cm/s.

Hence, the answer is the option 2.

Example 2: If an object moves towards a plane mirror with a speed v at an angle θ to the perpendicular to the plane of the mirror, find the relative velocity between the object and the image

1) v

2) 2v

3) 2vcosθ

4) 2vsinθ

Solution:

$
\begin{aligned}
& \vec{V}_{o, i}=\vec{V}_o-\vec{V}_i \\
& \vec{V}_{o, i}=v \cos \theta \vec{i}-(v \cos \theta(-\overrightarrow{-i})) \vec{V}_{o, i}=2 v \cos \theta \vec{i}
\end{aligned}
$

So magnitude will be $2 v \cos \theta$

Example 3: A plane mirror is placed at the origin parallel to the $y$-axis facing the positive x -axis. An object starts from $(2,0,0)$ with a velocity $(2 \hat{i}+2 \hat{j}) \mathrm{m} / \mathrm{s}$. The relative velocity of the image with respect to the object is along.

1) Positive x-axis

2) Negative x-axis

3) Positive y-axis

4) Negative y-axis

Solution:

Moving Object & Mirror

$
\vec{V}_i-\vec{V}_m=-\left(\vec{V}_o-\vec{V}_m\right)
$
wherein
$
\begin{aligned}
& \vec{V}_i=\text { Velocity of image } \\
& \vec{V}_m=\text { Velocity of mirror } \\
& \vec{V}_o=\text { Velocity of object }
\end{aligned}
$

$\begin{aligned} & \vec{V}_i=-2 \hat{i}+2 \hat{j} \\ & \Rightarrow \vec{V}_{i, o}=\vec{V}_i-\vec{V}_o=(-2 \hat{i}+2 \hat{j})-(2 \hat{i}+2 \hat{j})=-4 \hat{i}\end{aligned}$

$\therefore$ the velocity of the image w.r.t object is along the negative x-axis.

Example 4: Two plane mirrors $\mathrm{M}_1$ and $\mathrm{M}_2$ are at right angle to each other shown. A point source $P^{\prime}$ is placed at $a^{\prime}$ and $2 a^{\prime}$ meter away from $\mathrm{M}_1$ and $\mathrm{M}_2$ respectively. The longest distance between the images thus formed is: $($ Take $\sqrt{5}=2.3)$

1) $2.3 a$
2) $2 \sqrt{10} a$
3) $4.6 a$
4) $3 a$

Solution:

For two mirrors placed at the right angle to each other $\left(\theta=90^{\circ}\right)$ No. of images $=(n-1)=\left(\frac{360}{90}\right)-1$
No of images $=3$

$p_1, p_2 \& p_3$ are three images of p
Distance between image $p_2 \& p_3$ is the smallest i.e (2a)
Distance between image $p_1 \& p_3$ is 4 a
Distance between the image $p_1$ and $p_2$ is $\sqrt{(4 a)^2+(2 a)^2}=2 \sqrt{5} a=4 \cdot 6 a$ T

Hence, the answer is option (3).

Example 5: An object and a plane mirror are shown in the figure. The mirror is moved with velocity

V as shown. The velocity of the image is :

1) $2 V \sin \theta$
2) 2 V
3) $2 V \cos \theta$
4) none of these

Solution:

Moving Object & Mirror

$
\overrightarrow{V_i}-\overrightarrow{V_m}=-\left(\vec{V}_o-\vec{V}_m\right)
$
wherein
$
\begin{aligned}
& \vec{V}_i=\text { Velocity of image } \\
& \vec{V}_m=\text { Velocity of mirror } \\
& \vec{V}_o=\text { Velocity of object } \\
& \vec{V}_I m=\vec{V}_0 m \\
& \Rightarrow \vec{V}_I=\vec{V}_m=-\left(\overrightarrow{V_0}-\vec{V}_m\right) \\
& \vec{V}_I=V \sin \theta=-(0-V \sin \theta) \\
& V_I=2 V \sin \theta
\end{aligned}
$

Summary

The velocity of an object and its reflection in a plane mirror are closely related. When we move an object, we move its reflection also. The image’s velocity has the same magnitude as an object’s velocity, but the direction is opposite. Due to the fact that crossing their paths results in image reversal, we may say an image moves towards a mirror whenever an object does, suggesting a similar speed.