Question

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?

Answer 1

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