Experience shows that a body irradiated with light with circular polarization acquires a torque. This happens because such a light possesses an angular momentum whose flow density in vacuum is equal to `M = I//omega`, where `I` is the intensity o flight, `omega` is the angualr oscillation frequacny. Suppose light with circualr polarization and wavelength `lambda = 0.70 mu m` falls normally on a uniform black disc of mass `m = 10 mg` which can freely rotate about its axis. How soon will its angualr velocity become equal to `omega_(0) = 1.0 rad//s` provided `I = 10 W//cm^(2)`?

Experience shows that a body irradiated with light with circular polar

1.	Experience shows that a body irradiated with light with circular polarization acquires a torque. This happens because such a light possesses an angular momentum whose flow density in vacuum is equal to `M = I//omega`, where `I` is the intensity o flight, `omega` is the angualr oscillation frequacny. Suppose light with circualr polarization and wavelength `lambda = 0.70 mu m` falls normally on a uniform black disc of mass `m = 10 mg` which can freely rotate about its axis. How soon will its angualr velocity become equal to `omega_(0) = 1.0 rad//s` provided `I = 10 W//cm^(2)`?
Answer» Let `r =` radius of the disc then its moment of inertia about its axis `= (1)/(2)mr^(2)` In time `t` the disc will acquire an angualr momentum `t.pi r^(2).(I)/(omega)` when circularly polarized light of intensity `I` falls on it. By conservation of angualr momentum this must equal `(1)/(2)mr^(2)omega_(0)` where `omega_(0) =` final angualr velocity. Equating `t = (m omega omega_(0))/(2pi I)` But `(omega)/(2pi) = v = (c)/(lambda)` so `t = (mc omega_(0))/(I lambda)` Subsituting the values of the various quantities we get `t = 11.9 hours`

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