If two light rays start simultaneously in the space from exactly opposite ends in opposite direction that is separated by a distance of 600000 km in a way they meet at the mid point (300000 km from source), then:
How much time it will take to meet the front most photon of one ray to meet the front most photon of opposite ray and
What will be the speed of photon of the first ray relative to photon of other ray?
The question of what is the velocity of a photon relative to another photon does not make sense. Neither it does asking what is the velocity of anything relative to a photon. This is because in special relativity we only have the concept of a velocity defined for a massive observer, which is defined from the four-velocity $$ u^\mu = \frac{d x^\mu}{d\tau} $$ where $\tau$ is the proper time defined from the space-time interval as $ds^2=-d\tau^2$ (the sign depends on the convention). Then we identify the components of the four velocity as $u^\mu = (1,v_x,v_y,v_z)$ and the velocity as $v=\sqrt{(v_x^2+v_y^2+v_z^2)}$.
For massless particles the space-time interval is zero $ds^2=$ and then you cannot define the concept of a velocity in the usual sense. When we say that photons travel at $c$ is basically we are saying that photons are massless particles and actually the way to think about $c$ is as the "velocity" at which massless particles travel. I believe that the fact that we call $c$ the speed of light comes from the identification of $c$ as the phase velocity in the electromagnetic wave equation. But we have to be careful when we talk about particles.
As opposed to velocity we always have a definition of the four-momentum for any particle and that is why in particle physics we always talk about momenta and not velocities.
Also, when we talk about observers we usually assume that we are talking about massive observers and it is impossible to perform a boost which takes the reference frame of a massive observer into that of a massless (i.e we can't boost to velocity $c$). That is why we cannot ask this kind of question, as @UncleAl explains one cannot ride a photon.
Answering the first question, the time it will take is $l/c=1$ second and that will be the same for any massive observer.
Remember that you must always specify the inertial frame of the observer. Other than that, your question makes perfect sense.
The "closing" velocity of the two photons approaching each other will be 2c only in an inertial (stationary) observer at rest relative to the center point.
To an observer "riding" with one of the photons, (either one), the closing speed cannot be greater than c, but a Doppler shift will be seen toward the blue end of whatever spectrum these photons happen to be emitting.
The idea of an observer riding in this manner is not a fiction. Relative to an observer in a galaxy separated by sufficient distance (like 12 billion light years, for instance), all of the light from our galaxy, the milky way, will be red shifted sufficiently to clock a relative velocity that is very close (but never exceeding) the speed of light.
I understand that you ask this question not because you cannot calculate the results yourself, but because you believe they might be contrary to the axioms and conclusions of the SR theory. Am I correct?
Well, you can simplify your situation (first) to make it conceptually easier and actually experimentally verifiable:
If a rocket starts from Earth and goes with velocity $v$ directly toward a source of light (the Sun for example), which at the same time sends a ray of light directly toward the rocket, then:
In such case, the answers are rather straightforward:
So, going back to your original questions:
Well, obviously it would be the speed of light times two.
However, this can be misleading. Equations are all fine and dandy, but if you do not understand them, they are not of complete help. Imagine that you have the following...
1) A 300,000 km long spaceship which is at rest in space. 2) A clock is located at each opposite end of the spaceship, and these clocks are synchronized.
Therefore, if a burst of light is released from either end at precisely 12:00 noon, and the light is heading towards the opposite end, it will reach the opposite end when both clocks say 12:00 noon plus 1 second. Thus the light crossed 300,000 km's in a time period of 1 second, as expected.
However, if we launch the spaceship and set the spaceships forward velocity to 260,000 kps, then based upon the Time Dilation equation, the clocks onboard will be ticking at half speed. Also, based upon the Lorentz-Fitzgerald Contraction equation, the spaceship will have contracted to a new spatial length of 150,000 kms. And, if taking the Lorentz Transformation equations into account as well, we see that the clock at the rear of the spaceship is 0.866 of a second ahead of the clock at the front.
If we send a burst of light from the rear of the spaceship to the front, to an external observer, who is at rest in space, it appears to take 3.73 seconds for the light to complete the trip. To the observer, who is at rest in space, the light is only traveling 40,000 kps faster than the spaceship, thus 150,000 km length / 40,000 kps = 3.73 seconds.
Onboard however, clocks are ticking at half speed, thus they would measure 3.73 * 0.5 = 1.866 sec. However, the clock at the front is lagging behind the clock at the rear by 0.866 sec. Therefore, 0.866 is subtracted from that 1.866 sec measurement, thus the outcome is 1.866 - 0.866 = 1.000, meaning, when the light reaches the front clock, it will register 12:00 noon plus 1 second. Thus, to those on board, everything seems to be the same.
If we send a burst of light from the front of the spaceship to the rear, to an external observer, who is at rest in space, it appears to take only 0.268 of a second for the light to complete the trip. To the observer, who is at rest in space, the light travels across the 150,000 km spaceship at 560,000 kps ( 260,000 kps + 300,000 kps ), thus 150,000 km length / 560,000 kps = 0.268 of a second.
Onboard, however, clocks are ticking at half speed, thus they would measure 0.268 * 0.5 = 0.134. However, the clock at the rear is ahead of the clock at the front by 0.866 sec. Therefore, 0.866 is added to that 0.134 sec measurement, thus the outcome is 0.134 + 0.866 = 1.000, meaning, when the light reaches the rear clock, it will register 12:00 noon plus 1 second. Thus, to those on board, once again, everything seems to be the same.
Now if we send light from the front to the rear again, but the spaceship was moving almost at the speed of light itself, the length will have contracted to almost zero, the clocks will almost be at a stand still, and the clocks will be offset from each other by almost 1 second.
Thus the outcome, for example, may be....
0.0000001 sec.[time period ] + 0.9999999 sec. [clock offsets] = 1 second.
Thus one needs to understand that the measurement one takes is done with numerous measurement instruments, Thus as in the last example, even though the speed of the spaceship relative to the light is close to 600,000 kps, if someone in that spaceship measures the speed of that light, the outcome is still 300,000 kps.
Here, in this example, most of the measurement is determined by the 0.9999999 sec. clock offset rather than the ticking of time.
But when dealing with photons, and their point of view, we do not have these measurement instruments at hand.
