If you look carefully into the animation I have attached, you will see, that at a screen that is placed equally distant to the centres of both the waves, an alternating maximum, zero, and minimum pattern is created. When we apply this concept of interference pattern to light waves, or any electromagnetic wave, we get the same result with a slight modification. With electromagnetic waves, on the screen, we get intensity patterns. Since the intensity of a wave is directly proportional to the square of the amplitude of the wave, the maxima and minima result in the same intensity. Remember that minima are the negative maxima and the square of -1 is just 1. Thus for EM waves, interference produces bright (max or min) and dark (zero displacement) patterns. We call these fringes.

**NOTE: For interference of two waves to produce fringes, they must be of the same frequency (or wavelength), else, the pattern will not be constant and won't be clearly visible.**

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It is a vector that doesn't follow vector laws.

A quantity that doesn't obey vector laws cannot be considered a vector.

Tension (or compression) in general is not a force but a pair of equal and opposite forces acting at a certain point. In problems in Physics and engineering, we often consider only one of the two forces in the tension pair and call it tension or tensile force. In this way, we can consider tension as a vector.

When you go by the layperson's definition of a tensor in Physics that says, "tensor is a physical quantity which has one magnitude and multiple directions.", you may consider tension as a tensor. However, tension does not fit the mathematical matrix in order to be a mathematical tensor. This is why the layperson's definition of a tensor isn't a valid one.

I hope my explanation helps.

]]>Here is the video of how a javelin flies.

]]>Although the question does not explicitly state it, we can assume the following:

The object is moving towards the mirror at a velocity $2\ \frac{m}{s}$ with respect to the ground and the mirror is moving towards the object at a velocity $3\ \frac{m}{s}$.

When all distances are measured from the centre of the mirror, let object distance be $u$ and image distance be $v$. We also have to follow the sign convention. However, to provide an easy and intuitive answer, let is also assume that the mirror is a plane mirror.

**For plane mirrors:**

It is well established that $v=u$ (ignoring the sign).

Thus the time rate of change of $v$ must be the same as the time rate of change of $u$.

$\frac{dv}{dt}=\frac{du}{dt}$.

To correctly find the value of $\frac{du}{dt}$, we have to consider the mirror to be at rest with us. Hence the object velocity now becomes relative to the mirror. Since the object and the mirror are moving towards each other, the magnitudes must be added.

Then, with respect to the mirror (and thus to us),

$\frac{dv}{dt}=\frac{du}{dt}=2+3=5\ \frac{m}{s}$.

This means that the image is also moving towards the mirror at a velocity $5\ \frac{m}{s}$, with respect to the mirror. Thus the image velocity with respect to the ground is $8\ \frac{m}{s}$. (Answer)

**Steps for spherical mirrors:**

- Use the mirror equation: $ \frac{1}{u}+ \frac{1}{v}= \frac{1}{f}$.
- Use coordinate sign convention.
- Assign positive directions along the axis of the mirror.
- Find the object and image distances and velocities WRT the mirror first.
- Add the image velocity WRT the mirror to the mirror velocity WRT the ground to find the image velocity WRT the ground.