A plane wave (laser beam) with an optical power density 0.5 W/cm2 propagate in vacuum (n = 1). That beam travels into a new medium, optically denser than vacuum (n = 2). Is the optical power density still the same? (time-averaged Poynting vector) or the optical power density is multiply by refractive index of new medium?
Intensity depends on the refractive index of the medium, look it up in Saleh-Teich Photonics or any similar book. Attached the correct definition (intensity is the modulus of the time-averaged Poynting vector, E_0 the amplitude of the light-electric field) and the equation including the refractive index.
Cheers,
Martin
According to linear electrodynamics, on the boundary between two transparent media a normal component of the Poynting vector is continuous.This is just conservation of energy.
time average of Poynting vector gives the intensity is the quantity per unit time and area. when the light propagate in the different media, their velocity will be different. therefore the time of spending of wave in a given time in different media is different and it will be there in the form of refractive index in the wave propagation. so the intensity depends on the refractive index as dear Martin Fally already discussed. the intensity in the medium increases with increasing its refractive index.
As mentioned before, to my understanding, the intensity indeed increases with the refractive index. However, to my intuition, the reason lies in the spatial domain rather than the temporal domain. The minimimum spot size is proportional to inverse square of the refractive index
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