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?

Question

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Respuesta :

DeniceSandidge DeniceSandidge · Answer 1 · 2019-09-13T08:54:41+02:00

Answer:

intensity = [tex]\frac{Io}{15}[/tex]

intensity = [tex]\frac{Io}{30}[/tex]

depth = 333.33 m

Step-by-step explanation:

given data

deep = 10 m

intensity = Io

intensity = Io/3

to find out

intensity of light at 50 m and 100 m and 1/100 of initial intensity remain ?

solution

we know here that intensity is inversely proportional to deep so

intensity = k × [tex]\frac{1}{Deep}[/tex] .................1

here k is constant

so 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] ................2

so from equation 1 when 100 m deep and 50 m deep

intensity = k × [tex]\frac{1}{Deep}[/tex]

intensity = Io × [tex]\frac{10}{3}[/tex] × [tex]\frac{1}{50}[/tex]

intensity = [tex]\frac{Io}{15}[/tex]

and

intensity = Io × [tex]\frac{10}{3}[/tex] × [tex]\frac{1}{100}[/tex]

intensity = [tex]\frac{Io}{30}[/tex]

and

at intensity Io/100

intensity = 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 m

so depth = 333.33 m