Q:

A ray of light with intensity lo falls vertically on the sur face of the sea and at ten meters deep its intensity is 10/3. Assume that the loss of light intensity is a process with a constant rate : a)Calculate the intensity of the light at 50 and 100 meters deep. b) At what depth will 1/100 of the initial intensity remain?

Accepted Solution

A:
Answer:intensity = [tex]\frac{Io}{15}[/tex] intensity = [tex]\frac{Io}{30}[/tex] depth = 333.33 mStep-by-step explanation:given data deep = 10 mintensity = Iointensity = Io/3to find outintensity of light at 50 m and 100 m and 1/100 of initial intensity remain ?solutionwe know here that intensity is inversely proportional to deep so intensity = k × [tex]\frac{1}{Deep}[/tex]      .................1here k is constantso we have given 10 m deep so[tex]\frac{Io}{3}[/tex]  = [tex]\frac{k}{10}[/tex]so k = Io × [tex]\frac{10}{3}[/tex]    ................2so from equation 1 when 100 m deep and 50 m deepintensity = k × [tex]\frac{1}{Deep}[/tex] intensity =  Io × [tex]\frac{10}{3}[/tex] × [tex]\frac{1}{50}[/tex] intensity = [tex]\frac{Io}{15}[/tex] andintensity =  Io × [tex]\frac{10}{3}[/tex] × [tex]\frac{1}{100}[/tex] intensity = [tex]\frac{Io}{30}[/tex] andat intensity Io/100intensity = k × [tex]\frac{1}{Deep}[/tex] [tex]\frac{Io}{100}[/tex]  =  Io × [tex]\frac{10}{3}[/tex] × [tex]\frac{1}{D}[/tex] D = 333.33 mso depth = 333.33 m