Light passing through water loses 30% of its intensity every metre of water depth. At what depth will the light intensity be half of what it is at the surface?

How is it 8.1? I don't know how.

Thanks

It isn't 8.1

1(.7)^x = .5
take log of both sides and use rules of logs
x log .7 = log .5
x = log.5/log.7 = 1.94

check:
after 1 m, int. = .7
after 2 m, int = .7(.7) = .49 , already less than half

my answer of 1.94 m is very reasonable.

Hmm the answer key says 8.1. I got the same answer you did.

To find the depth at which the light intensity is half of the surface intensity, we can use the concept of light attenuation. According to the given information, light passing through water loses 30% of its intensity every meter of water depth.

Let's assume the surface intensity of light passing through water is 100 units (it could be any measurement, but for simplicity, let's use units). We want to find the depth at which the intensity becomes half, so the final intensity we want to achieve is 50 units.

To determine the depth, we can set up an exponential equation using the attenuation factor of 30% (0.3) per meter. The equation is:
Final Intensity = Initial Intensity * (Attenuation Factor)^(Meters of Depth)

Plugging in the values:
50 = 100 * (0.7)^M

To solve for M (depth in meters), we can take the logarithm of both sides to remove the exponent:
log(50) = log(100) + log(0.7^M)

Using logarithmic properties, we can rewrite the equation as:
log(50) = log(100) + M * log(0.7)

Now, let's rearrange the equation to solve for M:
M * log(0.7) = log(50) - log(100)

Using the properties of logarithms again, we can simplify:
M * log(0.7) = log(50/100)
M * log(0.7) = log(0.5)

Now we just need to isolate M:
M = log(0.5) / log(0.7)

Using a calculator to evaluate the logarithms and perform the division, we find that M is approximately 8.1. Therefore, the light intensity will be half of what it is at the surface at a depth of approximately 8.1 meters.