Does photon's energy increase when i emit it from a train and get it reflected back? Does photon's energy increase when i emit it from a train which is moving at comparable speed and get it reflected back by a mirror on a platform so that i can catch it again in the train?
  According to doppler effect the frequency of the photon for an observer standing near the mirror is greater to an observer in the train.So to the observer near the mirror, energy of the photon is greater than the energy observed by an observer in train.Similarly when it gets reflected back and reaches the train its frequency is larger than the initial frequency when observed by an observer in train .So obviously energy of photon is increased without doing any work....But it is against conservation of energy!!!!!!!!!!so can anyone help me with this problem ,please.....thanks in advance
 A: From the point of view of the observer in the train, the photon is reflected off a moving mirror. That reflection will increase the energy / momentum of the photon (it can't increase the speed, obviously). In the train frame of reference, the mirror is moving towards the observer at velocity $v$: if the outgoing photon has momentum $p$, the returning one will have momentum $$p' = p\frac{c+v}{c-v}$$
This is derived, for example, in "Reflection from a moving mirror—a simple derivation using the photon model of light", Gjurchinovski, European Journal of Physics, 34:1 (2012)
A: Floris is right, it does change exactly as he references from the paper. However, one does not need to look at any paper to figure it out -- it is a relatively simple physics problem if you think relativistically. I derive it below. And you do not need to involve momentum (though it's equivalent), but you can do it all from the way frequency changes from one inertial reference frame to another (i.e., special relativity, and also true in local inertial frames in gravity)
The idea is to consider the problem as two steps, and thus two changes from one reference frame to another. We'll do two transformations. It is important to do one at a time or you get confused.
We do them step by step, so it is clear there is absolutely NO MAGIC. Just straight SR transformations.   
FIRST TRANSFORMATION
Frame 1: the train rest frame (TRF). It is moving wrt the mirror, but in this frame we consider the train at rest, ie, we are standing on the train 
emitted frequency  = f
Frame 2: the mirror rest frame (MRF). In this frame we consider it at rest. 
Since the source of the emitted light is moving, there will be  a Doppler shift given by
$f_2$ = frequency in MRF = (1+$\beta$)$\gamma$f == (1+z)f 
where z is the blue shift, ie, $\Delta$f/f = (freq in MRF - f)/f, 
and $beta$ = v/c 
and the Lorentz factor $gamma$ = 1/$sqrt{(1-\beta^2)}$
as both are normally labeled. 
The equation for Doppler shift is exact in SR, for motion along the line of sight. See  https://en.wikipedia.org/wiki/Redshift where the equations are
1+z = (1+$\beta$)$\gamma$
and where I've take $\beta$ > 0 for the blue shift we would be seeing. (note: careful, Wiki takes the opposite, you can do it either way). 
SIMPLIFICATION
Note that 1+z = (1+$\beta$)$\gamma$
can be simplified to read
1+z = $sqrt{[(1+\beta)/(1-\beta)]}$
So 
$f_2$ = f (1+z)                                   (Eq. 1)
or 
$f_2$ = f $sqrt{[(1+\beta)/(1-\beta)]}$           (Eq. 2)
So that's a simple equation. We will reuse Eqs. 1 and 2 in the next steps
SECOND TRANSFORMATION
Frame 2: Mirror at rest, reflecting the light, ie, emitting light
Frequency for light reflected in Frame 2 = frequency in Frame 2 for incident light = frequency in MRF = $f_2$
Frame 3: Frame of the train when we receive the light back. In this frame the train (or us if you wish) see the source of the light, Frame 2, as approaching at velocity v
Thus at Frame 3 we see a frequency $f_3$ further blue shifted. 
We can apply the same 1+z factor we applied from Eq. 1 in Eq. 2 above, to get 
$f_3$ = $f_2$ $sqrt{[(1+\beta)/(1-\beta)]}$
and substituting for $f_2$ we get (with the two identical square roots multiplying) the received frequency at the train (RFT) 
$f_3$ = f $[(1+\beta)/(1-\beta)]$        (Eq.3)
That is, the received frequency at the train, $f_{RFT}$ is
$f_{RFT}$ =  f $[(1+\beta)/(1-\beta)]$      (Eq. 4)
Since momentum is proportional to energy and to frequency in SR, that is the same equation as in that paper referred by Floris.
Except that we derived it doing two consecutive Lorentz transformations where we used the equation for the Doppler shift in going from one frame to the next, twice.
Notice that this is the right way to do it. The mirror reflected the light, and we treated that in its rest frame. No need for other assumptions. The train received the light reflected, and it did not know that it was the light it emitted, it just saw light coming off the mirror. So it simply needed to make the transformation to get the right frequency [actually, obviously the train (or us at the train) can measure the frequency, we would not need to calculate it, and then use Eq. 4 to calculate our velocity with respect to the mirror]. 
CONSERVATION OF MOMENTUM AND ENERGY
Note, the paper does apply conservation of energy and momentum, at the mirror. It also states that one can obtain their slightly more complicated results (they do it for reflection at an angle, we could have also, not needed for the question at hand) simply using SR. It has to be so, and for the simple case covered in the question, it's very easy. Note also that we do not need to prove conservation of energy, we know that SR conserves energy if calculated right. Einstein and others proved that. Yes, the extra photon energy had to come from the mirror, it is also what changed its momentum by reversing direction. It's the only physical interaction.  
A: Energy balance in elastic scattering involving photons
Compton Scattering
First, consider the frame of reference $\mathcal{B}$ that the platform rests in.
A photon with the frequency $f^\mathcal{B}$ and momentum $p^\mathcal{B} = \frac{h}{c}f^\mathcal{B}$ will hit a mirror fixed to some resting object $\mathrm{M}$ of mass $m$. (You may imagine $\mathrm M$ to be the Earth.) The reflection will transfer a momentum of $S\approx 2p^\mathcal{B}$ to $\mathrm{M}$. After the collision the object $\mathrm{M}$ will have the velocity
$$
\Delta v \approx \frac{S}{m} \approx \frac{2h}{mc}f^\mathcal{B}
$$
and will have some kinetic energy. To account for that energy, the reflected photon will have a reduced frequency $f'{}^\mathcal{B}$. This phenomenon is called the Compton effect and is very small for a macroscopic object $\mathrm{M}$, but very well observable if $m$ is about the mass of an electron. 
This is not the energy transfer that you are interested in. It is only
$$
\Delta E \approx \frac{m}{2}(\Delta v)^2 \approx \frac{2h^2}{mc^2}f^\mathcal{B}
\approx \frac{10^{-83}}{m}f^\mathcal{B}
$$
which is very small - even compared to the energy of a customary photon ($\sim 10^{-19}\mathrm J$). Thus, in the frame of reference $\mathcal{B}$ there practically is no energy transfer and we may assume
$$f'{}^\mathcal{B}\approx f^\mathcal{B}$$
Reverse Compton scattering
In the frame of reference $\mathcal{A}$ the train is at rest. Note that the kinetic energy is not independent of the choice of reference frame. In $\mathcal{B}$ the object $\mathrm{M}$ is at rest and does not have any kinetic energy. In $\mathcal{A}$, $\mathrm{M}$ is moving with the velocity $v$ towards the observer, thus it has some kinetic energy $E_\text{kin}^\mathcal{B}\approx \frac{m}{2}v^2$.
Since the kinetic energy is not a linear function of $v$, but an approximately quadratic function, the change in energy by a boost $\Delta v$ depends on the energy that the object $\mathrm{M}$ has in the first place. (Note that $\Delta v$ is in opposite direction to $v$. Further note that $\Delta v$ is approximately the same in both frames of reference.)
$$
\Delta E_\text{kin}\approx \frac{m}{2}(v-\Delta v)^2-\frac{m}{2}v^2=\frac{m}{2}(-2v\Delta v + (\Delta v)^2)\approx -mv\Delta v\approx \frac{v}{c}2hf^\mathcal{B}
$$
Now I still assume that $\frac{m}{2}(\Delta v)^2$ is a very small and negligible change in kinetic energy, as I did above. However, 
$mv\Delta v$ cannot be neglected if $v$ is near the order of magnitude of the speed of light as your problem suggests. As a result, the kinetic energy of $\mathrm{M}$ is reduced. This effect is called reverse Compton effect.
Since trains don't move with, say, a tenth of the speed of light, the reverse Compton effect is not observed in trains, but it is observed with fast moving particles in particle accelerators.
In your example, the object $\mathrm{M}$ will loose some kinetic energy and the photon will gain energy in the frame of reference $\mathcal{A}$
Relativistic Doppler shift
The frequency of light in reference frames $\mathcal{A}$ and $\mathcal{B}$ are related by the following formulas:
For the photon moving from the train to the mirror
$$
f^\mathcal B = \sqrt{\frac{c + v}{c - v}}f^\mathcal A
$$
and for the photon moving from the mirror to the train
$$
f'{}^\mathcal A = \sqrt{\frac{c + v}{c - v}}f'{}^\mathcal B
$$
Combine that with $f'{}^\mathcal{B}\approx f^\mathcal{B}$ and conclude
$$
f'{}^\mathcal A \approx \frac{c + v}{c - v} f^\mathcal A \approx (1 + 2\frac{v}{c}) f^\mathcal A
$$
Therefore the change energy of the photon is
$$
\Delta E_\mathrm\gamma^\mathcal A = hf'{}^\mathcal A - hf^\mathcal A \approx h\cdot2\frac{v}{c}f^\mathcal A
$$
Summary
In $\mathcal{A}$ some kinetic energy of $\mathrm M$ is transferred to the photon. In $\mathcal B$ the changes in energy are negligible, but if $m$ is very small there is a transfer of energy from the photon to the mirror.
Similar transfers of energy happen in the train when emitting and absorbing the light. In $\mathcal A$ the beam of light will transfer kinetic energy from $\mathrm M$ to the train. In $\mathcal B$ the beam of light will transfer kinetic energy from the train to $\mathrm M$.
A: I got it. I thought the train launches a photon at right angle. It launches it into front.
In simple words you consider a photon which is jumping between two perfect mirrors, like in a light clock. If mirrors are rigidly fixed, proper frequency (let‘s say color) of photon doesn't change. If mirrors are not rigidly fixed, photon will transfer some energy to the mirrors and mirrors will start moving (receding) from each other.
Thus, photon will be more and more redshifted at every oscillation if mirrors receding, because mirrors will gain some of  it‘s energy.
If mirrors approach each other, even by inertia, they will transfer some of their kinetic energy to photon. Photon will slow them down a bit.  Photon will gain some energy and will be more and more blueish at every oscillation.
If someone "brings" mirrors closer by his hands, he has to do some work.
Interesting detail is the following: if a train launches a photon at right angle in his reference frame, photon will be blueshifted at the mirror on the platform. It will be redshifted on arrival back and will appear in the same frequency.  In this case we consider that observer and mirror are in motion relatively to each other at parallel patches and there is a gap between them.
Where is reciprocity of observations?
