# Can the velocity of an image produced by a plane mirror be greater than that of light?

We are currently studying ray optics in school and it made me wonder if the velocity of an image produced by a plane mirror can theoretically be greater than that of light.

Using the relation $$v_i = 2 \ v_m - v_o$$

where
$$v_i$$ is the velocity of the image,
$$v_m$$ is the velocity of the mirror,
$$v_0$$ is the velocity of the object.

Using the formula, if the mirror is moving with a speed 80% the speed of light and the object is moving with a speed 10% the speed of light. the image is calculated to move with a speed 1.5 times the speed of light. Is this a contradiction to Special Relativity or is there some fine print for the relation that I don't know about. Perhaps a property of light that I am not aware about, or isn't normally taught in high school?

Why is there a 2 for that matter in front of $$v_m$$, why is $$v_0$$ preceded by a negative sign?

If an object is at the origin and the plane of the mirror is perpendicular to the x axis and located at $$x_m$$, then the reflection is located at $$2 x_m$$ by simple optical geometry.

Considering motion only along the x-axis, if the mirror is displaced by $$\Delta x$$ then the new location of the image is $$2 \times (x_m +\Delta x)$$. If this occurs in a time interval $$\Delta t$$ then the velocity of the reflected image is $$v_i = \frac{2 \Delta x}{ \Delta t} = 2 V_m.$$

The reflection moves in the opposite direction to the movement of the object, so $$v_i = -v_o$$.

• It's not clear how all these objects are moving with relation to one another since you didn't include a diagram from which these variables are defined. Why is there a 2 for that matter in front of $v_m$, why is $v_0$ preceded by a negative sign? Commented Apr 15 at 18:34
• I'm sorry I should have made it more clear. Here is the derivation: If an object is moving perpendicular to the plane of the mirror then the relative velocity of the object w.r.t mirror will be equal in magnitude but opposite in direction to the relative velocity of the image. v<sub>o/m</sub> = -v<sub>i/m</sub> v<sub>o</sub> - v<sub>m</sub> = - ( v<sub>i</sub> - v<sub>m</sub>) v<sub>i</sub> = 2v<sub>m</sub> - v<sub>o</sub> Commented Apr 16 at 2:00
• Edits should be made to the question itself and should employ the MathJaX typing, your comment is very complicated otherwise. Though the provided answer does do a good job of having answered your question anyways :) Commented Apr 16 at 2:14

I checked you equation ($$V_i = 2V_m - V_o$$) and it appears to be correct. (See caveat below.)

Can we say that if the mirror is moving with a speed 80% the speed of light and the object is moving with a speed 10% the speed of light. Would the image move with a speed 1.5 times the speed of light?

On the face of it we appear to have something that is exceeding the speed of light, but the image is a virtual image and not a tangible physical object. The virtual image is a mathematical geometric construction of where the light rays appear to be coming from, but if you look behind the mirror, there is nothing there. It joins a list of virtual objects that appear to exceed the speed of light, but these virtual objects are not physical and cannot be used to communicate at superluminal speeds.

This list includes:

1. The point between a long oblique wave and where it meets the shoreline.
2. Superluminal scissors
4. The leading edge of a jet from a neutron star or black hole seen at an oblique angle.

Details:

1. Imagine a 100 Km long straight wave that is approaching a straight shoreline that is almost parallel to it but not quite. It is possible that one end of the wave hits the shoreline a picosecond after the other end hits and the point of intersection with the beach exceeds the speed of light. The intersection point exists only as a mathematical entity and no molecules in the water wave exceed the speed of light. observers at each end of the shoreline cannot use this phenomena to meaningfully communicate with each superluminally.

2. Imagine sweeping a laser pointer across the surface of the Moon (while standing on the Earth). The dot on the Moon appears to move across the surface of the Moon faster than the speed of light (after an initial delay), but no physical entity actually moves across the surface. The dot at any instant comprises a set of reflecting photons and the dot at the successive instant, comprises of a different set of reflecting photons. See the link given by Pauline.

Caveat: The equation is valid for the actual position of the mirror, but if we take light travel times into account to work out what an observer would actually see, then things are different. To avoid things getting too complicated I am going to assume that the object and the observer remain at the origin and only the mirror moves (along the x axis). The reflective surface of the mirror is orthogonal to the x axis. Taking light travel time into account, the equation for the apparent velocity $$V_m'$$ of the mirror is:

$$V'_m = \frac{2 \Delta X}{\Delta T} = \frac{2 V_m }{1 + v_m/c}$$

where $$t_m$$ is the time that the mirror moves for and c is the speed of light. If we examine the case when the mirror is heading towards the object and observer at the origin at a velocity of -0.8c then the result is: $$V'_m = \frac{2 \times -0.8 }{1 - 0.8} = - 8 c$$

If $$V_m$$ is tends to c, $$V_m'$$ tends to infinite and we are talking about the velocity of a solid mirror and not a virtual image. So what is going on here?

The calculated velocity is the apparent velocity calculated according to where the light rays appeared to be coming from, from the point of view of the observer and not the actual velocity of the mirror. Again, this is not a real tangible object exceeding the speed of light. This is very similar in principle to the apparent superluminal motion described for the jets in (4) where the apparent velocity is exaggerated when an object is approaching an observer at high velocity.