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Q) we have 1np=20log_(10)edB1^(mp)=8.68 dB(a) 1.38np//cm=1.38 xx8.68dB//cm=11.97(dB)/(Cm)Now suppose it has to travel x cm. for Intensity to reduce half.:. Peduction in Pntensity{:[Delta I=11.97(dB)/(cm)xx xcm],[Delta IdB=-11.97 x[[" The -ve sign "],[" curresponds to "]:}],[" decrease in "],[" Rntensity? "]:}Tow10 log (I)/(I_(0))=Delta I{:[:.quad-11.57 x=10 log (I)/(I_(0))],[" NoW "I=(D_(0))/(2)]:}{:[:.-11.97 x=10 log((R_(0))/(2I_(0)))=10 log((I_(0))/(2I_(0)))],[-11.97 x=1010 g((I_(0))/(2I_(0)))],[" or, "11.97 x=10 log 2], ... See the full answer