after passing therough a material t centimeters thick, the intensity I(t) of a light beam is given by I(t) =(4^-ct)Io, where Io is the initial intensity and c is a constant called the absorption factor. Ocean water absorbs light with an absorption factor of c= .0101. at what depth will a beam of light be reduced to 50% of its initial intensity? 2% of its initial intensity?
I will do the last part
to have only 2% of the intensity left,
l(t) = .02
.02 = 4^(-.0101t) (1)
ln .02 = -.0101t ln4
-.0101t = ln .02 /ln4 = -2.821928
t = -2.821928/-.0101 = 279.4
for the 50% question , set the left side equal to .5
To find the depth at which the light beam is reduced to 50% and 2% of its initial intensity, we can use the equation:
I(t) = (4^(-ct)) * Io
Given that Io is the initial intensity, we can represent 50% of Io as (0.5 * Io) and 2% of Io as (0.02 * Io).
Let's find the depth at which the light beam is reduced to 50% of its initial intensity:
0.5 * Io = (4^(-ct)) * Io
Now we can cancel out Io:
0.5 = 4^(-ct)
To remove the exponent, we can take the logarithm of both sides. Let's take the natural logarithm (ln) for simplicity:
ln(0.5) = -ct * ln(4)
Now, isolate t:
t = -ln(0.5) / (c * ln(4))
Substituting the value of c = 0.0101, we can calculate t:
t = -ln(0.5) / (0.0101 * ln(4))
Now, calculate the value of t to find the depth at which the light beam is reduced to 50% of its initial intensity.
Similarly, let's calculate the depth at which the light beam is reduced to 2% of its initial intensity:
0.02 * Io = (4^(-ct)) * Io
0.02 = 4^(-ct)
Again, take the logarithm of both sides:
ln(0.02) = -ct * ln(4)
Isolate t:
t = -ln(0.02) / (c * ln(4))
Substitute the value of c = 0.0101, and calculate t to find the depth at which the light beam is reduced to 2% of its initial intensity.
Please note that the calculations performed above are based on the given formula and assumptions.