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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?

User Lanrat
by
7.7k points

1 Answer

5 votes

Answer:

intensity =
(Io)/(15)

intensity =
(Io)/(30)

depth = 333.33 m

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 ×
(1)/(Deep) .................1

here k is constant

so we have given 10 m deep so


(Io)/(3) =
(k)/(10)

so k = Io ×
(10)/(3) ................2

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

intensity = k ×
(1)/(Deep)

intensity = Io ×
(10)/(3) ×
(1)/(50)

intensity =
(Io)/(15)

and

intensity = Io ×
(10)/(3) ×
(1)/(100)

intensity =
(Io)/(30)

and

at intensity Io/100

intensity = k ×
(1)/(Deep)


(Io)/(100) = Io ×
(10)/(3) ×
(1)/(D)

D = 333.33 m

so depth = 333.33 m

User Kristian Rafteseth
by
7.5k points

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