Can We Explain Length Contraction in a Simpler Way? I have been studying relativity from a decent amount of time and all the books have a common thing.That is, They explain time dilation in a simple independent way but for length contraction they explain it using muons reaching earth or they use the lorentz transformation to directly derive it,Or use time dialation for mathematical derivation ,What I want is not a derivation.I want length contraction to pop up intuitively..this is what I mean(I will explain time Dialation)
TIME DILATION
Relativity is there because we accept that velocity of light($c$) is constant and won't change for different frames, This simply directly pops out time dilation
If we study the two mirrors thought experiment in which we measure $c$ in a frame at rest (wrt the observer) by reflecting light on the mirrors and then measure it from a observer moving wrt the mirrors and find out that the distance that light has to travel In the later is greater than the distance it has to travel in the former,so to keep $c$ a constant, we know $c=d/t$ ,so If d is  increased then t will also to keep $c$ constant and hence the time in the moving frame is greater than time in the rest frame..
Now I am unable to find an intuitive way like this to pop up length contraction directly just by saying velocity of light is constant,
This is what I have tried.. but not able to pop out length contraction
Let's say two mirrors separated by a distance $L$(now the mirrors are like this >> |  | and will move toward left or light along their length , now for a stationary observer, we measure the length through light going from mirror 1 to mirror 2 And $c=L/t$
Now doing the same for a observer moving with velocity $v$ wrt the mirror Now $c=(L' + vt')/t' - L'$ for contracted length and $t'$ for improper time, we know
$t'$ is greater than $t$ (I have included time dilation ) . can we conclude from here that $L'$ Is less than $L$?
 A: Here's a visual demonstration that length contraction is required if 
one assumes the Speed of Light Principle (so all light-signals at 45 degrees) and the Relativity Principle.
Each inertial observer is equipped with a Longitudinal light clock.
I use "rotated graph paper" because it is easy to draw light-signals...
in particular, it shows a coordinate system laid out by the ticks of the light clock at rest in this frame.

(When Bob's light clock is properly calibrated, the rotated graph paper
can be used to do calculations in relativity.)
THOUGHT EXPERIMENT:
Two inertial observers agree to send light signals to each other 2 minutes after separating. 
In this spacetime diagram, Alice is at rest, and Bob is moving (3/5)c with respect to Alice.
The Relativity Principle requires that they agree on when they receive the other observer's light signal...they should report the same elapsed time on their own wristwatches.
We make the naive assumption that 
the length of the [Bob's] moving clock is the same as 
the length of the [Alice's] stationary clock.
Since Alice receives a signal at 5 minutes on her clock, 
whereas Bob receives it at a different elapsed time: 3.2 minutes on his clock,
"absolute length" is not compatible with the Principles above.
Alice got Bob's signal too late and/or Bob got Alice's signal too early.
So, the ticks for Bob's clock are too large... they need to be smaller.
So, Bob's light clock must be shorter than Alice's light clock.
This suggests that there must be length-contraction.
(To get the actual length-contracted size [i.e. to derive the formula for length contraction], 
you must choose the length of Bob's light clock so that the reception times are equal. It turns out that this occurs when
their “light-clock diamonds” have equal areas.)

A: I read this in a book once. Let's do a thought experiment. You have a train which is say a 100 m long, and four people, say, Alice, Bob, George and Del. Now first, consider the train to be at rest. Let's say our group of people have one lasers, one detectors and two stopwatches. So, Alice stands at one end of the train and Bob stands at the other end. Now Alice fires the laser, parallel to the train, towards Bob. Bob holds the detector, and uses the stopwatch to measure the time it takes for the laser beam to travel from Alice to his detector. He measures the time to be 
$3.33 * 10^{-7} s$. Now he knowns that the length of the train should be $l = ct$, so he measures it and finds $(3 * 10^8 m/s) * (3.3 * 10^{-7}s) = 100 m$.
Now, let's say that the train starts moving, at a very high speed of 0.5c. Now, I make a assumption that the people outside the train can see the inside of the train. Let's say Alice and Bob are now in the train. They repeat the same experiment, and find that the train is 100 m long.
George and Del are outside the train, and George too is using his stopwatch to measure the time it takes for the beam to travel from Alice to Bob's detector. Alice's beam starts when the stopwatch shows $t=0$. But now the train is moving forward, so George sees Bob(who is at the end of the train), move towards the beam and catch up with it faster than if the train was at rest. Due to this observation, George's stopwatch shows him a smaller time interval between the release of the beam and it's detection. And so, doing the same calculation $l = ct$, George observes the train to be just around 86.6 meters long(see the figure).
Edit: If you want to totally generalize this, you can start by fixing Alice at the engine of the train and Bob at the rear of the train. Now, if the train starts moving right(in the figure it is shown moving left), then all you have to do is flip the train around in the x-direction. Doing this, we see that the same observation holds, and that the magnitude of the length contraction is the same. Because, the situation is symmetric in both the left and right directions.
Another way you get the value of the contracted length goes like this: Suppose you fix Alice at point $x_1$ and Bob at $x_2$. Now inside the train, the length will be measured by $l^{'}= ct = x_2^{'} - x_1^{'}$(assuming a certain choice of coordinates and convention). Now from outside the train we set $l = x_2 - x_1$. We know from the Lorentz transformation that $x_2^{'} = \gamma(x_2 - vt)$ and $x_1^{'} = \gamma(x_1 - vt)$, then after using $l^{'} = x_2^{'} - x_1^{1}$ and $l = x_2 - x_1$, we get: $$ l = \frac{l^{'}}{\gamma}$$
Thus, the $vt$ term does not factor into the final magnitude of length contraction. It only factors when we consider certain points, not lengths(the difference of two points).

(PS: Please ignore my bad drawing. And the typo. The ppears is 'appears')
So, that is why George and Del observe that the length of the train has contracted
