What will be the relative speed of a photon in a light ray to another photon of opposite direction light ray? 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?
 A: 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.
A: 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.
A: 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:


*

*How much time will it take for the front of the rocket to meet the first photon of the ray of light?

*What will be the velocity of this photon relative to the rocket?


In such case, the answers are rather straightforward:


*

*The time will be shorter than it would have been had the rocket stayed on Earth, and will be equal to the speed of light divided by the actual distance it had to travel (i.e. distance between the Sun and Earth less the distance traveled by the rocket)

*The velocity of the photon relative to the rocket will still be $c$, because that's what the famous Michelson-Morley experiment showed - regardless of the motion of the measuring equipment, $c$ remains constant.


So, going back to your original questions:


*

*The time it will take the photons to meet will be: $T=300000 km/c$

*Relative velocity (if it could be measured) would most likely equal $c$

