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Now instead a 3-mw laser of wavelength λ = 760 nm is directed into the interferometer (with the mirror m1 at the displacement calculated in part a). how much power is detected at d1?

User KiriSakow
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power received will always remain same and does not depends on the distance

it will be same as the power of source


P = 3 mW

now the energy released by one photon is given as


E = (hc)/(\lambda)


E = (6.6* 10^(-34)* 3 * 10^8)/(760 * 10^(-9))


E = 2.6 * 10^(-19) J

now let say N photons per second released


N * 2.6 * 10^(-19) = 3 * 10^(-3)


N = 1.15* 10^(16) per\: second

User Markstewie
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