We know when light enters a refractive medium it's wavelength(l) is decreased. But it's frequency (f) remains unchanged. Light emits from the other side it gets back to its former wavelength. Does the phonemenon obey energy conservation?
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1Here's a hint: $E=hf$ – PM 2Ring Mar 15 '18 at 03:36
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Is this question a duplicate of this one : https://physics.stackexchange.com/questions/153904/how-does-light-speed-up-after-coming-out-of-a-glass-slab ? – dolun Mar 15 '18 at 07:17
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Energy conservation is maintained when you consider not only the transmitted but also the reflected light wave at the interfaces with different refractive indices.
freecharly
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If the question is about the change in wavelength of light when crossing a boundary between media with different refractive indices, and the relationship often cited between wavelength and photon energy, the answer is that the energy/wavelength relationship depends on the medium through which the photon is traveling. Inside a medium of refractive index greater than 1 (the index of the vacuum), a photon of any given wavelength has a smaller energy than a photon of the same wavelength has in a vacuum. The wavelength reduces inside the medium, just enough to give the photon the same energy as it has outside the medium.
So, even if the surfaces of the medium are given perfect antireflection coatings, energy is still conserved.
S. McGrew
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