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When a light microscope is used in the reflected mode there is a semi-transparent mirror angled at 45 degrees that reflects a portion of the light onto the sample. The sample in turn reflects the light coming off the beam and it goes straight through the 45 degrees angled semi-transparent mirror.

How is it that in this process we don't lose half the photons from the beam of the light source?

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You lose half the light twice, each time the light passes through the half-silvered mirror: Once heading toward the sample, and once again heading back. This leaves you with only one-quarter of the original light intensity reaching the eyepiece. This is why reflection-mode microscopes have to use very powerful bulbs to get enough illumination.

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  • $\begingroup$ That's what I also thought, but from the slide where I read this it says clearly: explain why we do NOT lose half of the photons. But if such a beamsplitter (I think that's what it is) is used, then of course some of the light of the light source will also be transmitted (and lost), even if it's at a 45 degree angle and so will some of the light reflected of the sample going through the beamsplitter. So could the question be wrong? should it say explain why we DO lose half of the photons? $\endgroup$ – Paperreader Aug 9 at 18:04
  • $\begingroup$ I have no idea. but in my previous career I had to make xenon flashtube light sources for microscopes like this and this made me very familiar with all the light loss mechanisms in them. $\endgroup$ – niels nielsen Aug 9 at 21:46
  • $\begingroup$ Ah okay, thanks for the answer, even if it confuses me even more. $\endgroup$ – Paperreader Aug 10 at 9:47
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In effect, you loose half the light intensity, but normally, that is still enough.

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