Can the phase of a light wave be measured experimentally? The phase difference can be measured but is it possible to measure the instantaneous phase of the electric field in light wave at one position?
 A: Yes, of course it is possible to measure phase.   Holography works by
combining a source wave with a reflected-light wave and where the
phases match (zero degrees) the light intensity is high (and darkens the
film), and where the phases mismatch (180 degrees) the light intensity
is minimum, and the film stays transparent.   That hologram image, then,
captures the reflected-light phase (and reconstruction of the
image afterward is therefore possible by illuminating the film with
a copy of that original source wave).
A: Yes, for a classical field (e.g. a large photon number, coherent state), in principle the phase can be measured in the sense that one can in principle observe that the electric (magnetic) field at a given point oscillates sinusoidally with time and one can measure at what times the field is zero and when it reaches its maximums. 
If you don't believe this for light, simply imagine the experiment with microwaves, or radiowaves. Phase locked loops have been synchronizing oscillator circuits with incoming radio waves for the best part of a hundred years. 
So at optical frequencies, the difference is one of technology only, not physics. Optical phase locked loops now exist and realize coherent modulation schemes over optical fiber communication links.
If you mean by your question the phase of a quantum state of light, then, of course, the state is a ray in projective Hilbert space and is invariant with respect to multiplication by a global phase. No meaning can therefore be given to the absolute phase of a quantum state. Phase differences between the complex amplitudes of base states in superposition are meaningful, and these determine interference effects and the evolution of the quantum state. 
Classical and quantum phase are quite distinct, however, and don't simply merge through a "correspondence principle" in a large photon number limit. To understand the difference between them, and the lack of a "correspondence principle", consider a photon number coherent state $\psi(\alpha,\,t)$ of the quantum harmonic oscillator:
$$\begin{array}{lcl}\psi(\alpha,\,t) &=& e^{-i\,\frac{\hbar}{2}\,\omega_0\,t}\,\exp\left(\alpha\,e^{-i\,\hbar\,\omega_0\,t}\,a^\dagger - \alpha^\ast\,e^{+i\,\hbar\,\omega_0\,t}\,a\right)\,|0\rangle \\&=& e^{-i\,\frac{\hbar}{2}\,\omega_0\,t}\,e^{-\frac{|\alpha|^2}{2}}\,\sum\limits_{k=0}^\infty\,\frac{\alpha^k}{\sqrt{k!}}\,e^{-i\,k\,\hbar\,\omega_0\,t}\,|k\rangle\end{array}\tag{1}$$
which is a superposition of Fock (number) states beating together as their relative phases oscillate at different frequencies. Suppose we measure the photon number with the number observable $\hat{n}=a^\dagger\,a$; its mean $\langle\psi|\hat{n}|\psi\rangle$ is constant at $|\alpha|^2$ (indeed, so are all the moments $\langle\psi|\hat{n}^k|\psi\rangle;\,k\in\mathbb{N}$ and by calculating them, one can show that the photon number is Poisson distributed with mean $|\alpha|^2$). However, suppose we measure the position with the position observable $\hat{x}=  \sqrt{\frac{\hbar}{2}\frac{1}{m\omega}}\,(a^\dagger + a)$ or the momentum with observable $\hat {p}= i\sqrt{\frac{\hbar}{2}m\omega}\,(a^\dagger - a)$; the means of these measurements $\langle \psi|\hat{x}|\psi\rangle$, $\langle \psi|\hat{p}|\psi\rangle$ are phase-quadrature, time harmonic functions:
$$ \langle \hat{x}(t) \rangle = |\alpha|  \sqrt{\frac{2\,\hbar}{m\,\omega}} \cos (\arg(\alpha) - \omega_0 t)$$
$$ \langle \hat{p}(t) \rangle = |\alpha|  \sqrt{\frac{2\,m}{\hbar\,\omega_0}} \sin (\arg(\alpha) - \omega_0 t)\tag{2}$$
and the phase of these quantites, i.e. $\arg(\alpha)$, is the classical phase. Take heed that we can multiply the state $\psi$ in (1) by any wildly time-varying quantum phase $e^{i\,\varphi(t)}$ we like and the results of the above calculations will be unchanged; the quantum phase disappears under the multiplication of the state and its complex conjugate in the calculation $\langle \psi|\hat{X}^k|\psi\rangle$ of any moment of any observable $\hat{X}$. Note that the classical phase is perfectly well defined for any, arbitrarily small mean photon number coherent state (it's simply that for $\alpha$ very small, you'll need many measurements on the same state to measure it, whereas a single measurement will do for laser-sized values of $\alpha$).

For measurement of classical phase of light, see Emilio Pisanty's answer here describing the relevant experimental technology in the Physics SE thread Have we directly observed the electric component to EM waves?.
Also, one can indeed define an observable for the classical phase of a quantum state, although it's tricky; an influential quantum optics researcher has written a whole book on the subject: 
Stephen M. Barnett & John A. Vaccaro, "The Quantum Phase Operator: A Review"
(Stephen M. Barnett, along with Bruno Huttner, was one of the first people to work out a fully quantized theory of the electomagnetic field in a dielectric (in this paper here)).
A: No if you meant visible light. We can measure phase directly only when EM wave's frequency is lower than 400GHz. Because fastest transistor we made is about 400GHz. But visible light have frequency 400–790THz. We need 1000x faster transistor to measure phase of visible light.
