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

Answer:

intensity = $\frac{Io}{15}$

intensity = $\frac{Io}{30}$

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 × $\frac{1}{Deep}$      .................1

here k is constant

so we have given 10 m deep so

$\frac{Io}{3}$  = $\frac{k}{10}$

so k = Io × $\frac{10}{3}$    ................2

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

intensity = k × $\frac{1}{Deep}$

intensity =  Io × $\frac{10}{3}$ × $\frac{1}{50}$

intensity = $\frac{Io}{15}$

and

intensity =  Io × $\frac{10}{3}$ × $\frac{1}{100}$

intensity = $\frac{Io}{30}$

and

at intensity Io/100

intensity = k × $\frac{1}{Deep}$

$\frac{Io}{100}$  =  Io × $\frac{10}{3}$ × $\frac{1}{D}$

D = 333.33 m

so depth = 333.33 m