The intensity of light from a continuously em | vedclass

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(B) Given:Wavelength $\lambda = 638 \,nm$Modulation frequency $f = 1 \,GHz = 1 	imes 10^{9} \,Hz$Speed of light $c = 3 	imes 10^{8} \,m/s$The pulse

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$30 \,cm$

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(B) Given:Wavelength $\lambda = 638 \,nm$Modulation frequency $f = 1 \,GHz = 1 	imes 10^{9} \,Hz$Speed of light $c = 3 	imes 10^{8} \,m/s$The pulse length (or the spatial extent of one pulse) is determined by the distance light travels in one period of the modulation frequency.The distance $D$ between two detectors such that they can see the same pulse simultaneously is equal to the wavelength of the modulation,which is given by:$D = \frac{c}{f}$Substituting the values:$D = \frac{3 	imes 10^{8} \,m/s}{1 	imes 10^{9} \,Hz} = 0.3 \,m$$D = 30 \,cm$Therefore,the farthest distance apart the two detectors can be placed is $30 \,cm$.

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