Question 619340: The intensity of light is reduced by 2% for each metre that a diver descends below the surface of the water. At what depth is the intensity of light only 10% of that at the surface?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the intensity of light is reduced 2% for each meter that a diver descends below the surface.
if the intensity starts at 100, then after 1 meter, the intensity is 100 - .02*100 which is equivalent to 100 * (1-.02) which is equivalent to .98 * 100
the next meter reduces it another 2%.
you get (.98*100) - .02*(.98*100) which is equivalent to (.98*100) * (1-.02) which is equivalent to (.98*100) * .98 which is equivalent to 100 * (.98)^2
each additional meter reduces it by another addition to the exponent.
for 10 meters, the reduction is 100 * (.98)^10
the formula is, therefore:
reduction in visibility = 100% * .98^m
where m is equal to the number of meters.
you want to know how many meters to reduce the visibility to 10%.
the equation becomes:
10% = 100% * .98^m
you want to solve for m.
divide both sides of this equation by 100% to get:
10% / 100% = .98^m
this results in:
.1 = .98^m
take the log of both sides of this equation to get:
log(.1) = log(.98^m)
this becomes:
log(.1) = m*log(.98)
divide both sides of this equation by log(.98) to get:
m = log(.1) / log(.98)
use your calculator to solve for m to get:
m = 113.9740856
the visibility should be reduced to 10% at 113.9740856 meters.
100% * (.98)^(113.9740856) is equal to 10%
to determine if your formula is correct, use a much smaller number.
assume you want to know how much visibility you have after 3 meters.
your equation becomes:
100% * (.98)^3 which becomes 94.1192
you should only have 94.1192% visibility after 3 meters if the equation is correct.
start with 100% and reduce it by 2% to get 98
reduce 98 by 2% to get 96.04
reduce 96.04 by 2% to get 94.1192
looks like the formula is good.
your answer is 113.9740856 meters of depth reduces visibility to 10%.
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principle of logarithms used.
log(a^b) = b*log(a)
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