Answer :
Certainly! Let's proceed step-by-step to find the light intensity, [tex]\( I \)[/tex], at a depth of 30 feet in Lake Heron using the given Beer-Lambert Law.
### Step-by-Step Solution:
1. Given Information:
- [tex]\( k = 0.035 \)[/tex] (constant for Lake Heron)
- [tex]\( J = 10 \)[/tex] lumens (light intensity at the surface)
- [tex]\( x = 30 \)[/tex] feet (depth in Lake Heron)
2. Beer-Lambert Law:
[tex]\[ \frac{-1}{k} \ln \left(\frac{I}{J}\right) = x \][/tex]
3. Rearranging the Equation:
First, we need to isolate [tex]\( \ln \left(\frac{I}{J}\right) \)[/tex].
[tex]\[ \frac{-1}{k} \ln \left(\frac{I}{J}\right) = x \][/tex]
Multiplying both sides by [tex]\( k \)[/tex]:
[tex]\[ - \ln \left(\frac{I}{J}\right) = kx \][/tex]
4. Eliminating the Negative Sign:
Multiply both sides by -1:
[tex]\[ \ln \left(\frac{I}{J}\right) = -kx \][/tex]
5. Exponentiate Both Sides:
To remove the natural logarithm, exponentiate both sides using base [tex]\( e \)[/tex]:
[tex]\[ \frac{I}{J} = e^{-kx} \][/tex]
6. Solve for [tex]\( I \)[/tex]:
Multiply both sides by [tex]\( J \)[/tex]:
[tex]\[ I = J \cdot e^{-kx} \][/tex]
7. Substitute Known Values:
Substitute [tex]\( J = 10 \)[/tex], [tex]\( k = 0.035 \)[/tex], and [tex]\( x = 30 \)[/tex] into the equation:
[tex]\[ I = 10 \cdot e^{-0.035 \cdot 30} \][/tex]
8. Calculate the Exponential Term:
[tex]\[ I = 10 \cdot e^{-1.05} \][/tex]
9. Final Calculation:
Compute [tex]\( e^{-1.05} \)[/tex]:
[tex]\[ I \approx 10 \cdot 0.3499377491111553 \][/tex]
10. Result:
[tex]\[ I \approx 3.499377491111553 \][/tex]
Therefore, the light intensity [tex]\( I \)[/tex] at a depth of 30 feet in Lake Heron is approximately [tex]\( 3.499377491111553 \)[/tex] lumens.
### Step-by-Step Solution:
1. Given Information:
- [tex]\( k = 0.035 \)[/tex] (constant for Lake Heron)
- [tex]\( J = 10 \)[/tex] lumens (light intensity at the surface)
- [tex]\( x = 30 \)[/tex] feet (depth in Lake Heron)
2. Beer-Lambert Law:
[tex]\[ \frac{-1}{k} \ln \left(\frac{I}{J}\right) = x \][/tex]
3. Rearranging the Equation:
First, we need to isolate [tex]\( \ln \left(\frac{I}{J}\right) \)[/tex].
[tex]\[ \frac{-1}{k} \ln \left(\frac{I}{J}\right) = x \][/tex]
Multiplying both sides by [tex]\( k \)[/tex]:
[tex]\[ - \ln \left(\frac{I}{J}\right) = kx \][/tex]
4. Eliminating the Negative Sign:
Multiply both sides by -1:
[tex]\[ \ln \left(\frac{I}{J}\right) = -kx \][/tex]
5. Exponentiate Both Sides:
To remove the natural logarithm, exponentiate both sides using base [tex]\( e \)[/tex]:
[tex]\[ \frac{I}{J} = e^{-kx} \][/tex]
6. Solve for [tex]\( I \)[/tex]:
Multiply both sides by [tex]\( J \)[/tex]:
[tex]\[ I = J \cdot e^{-kx} \][/tex]
7. Substitute Known Values:
Substitute [tex]\( J = 10 \)[/tex], [tex]\( k = 0.035 \)[/tex], and [tex]\( x = 30 \)[/tex] into the equation:
[tex]\[ I = 10 \cdot e^{-0.035 \cdot 30} \][/tex]
8. Calculate the Exponential Term:
[tex]\[ I = 10 \cdot e^{-1.05} \][/tex]
9. Final Calculation:
Compute [tex]\( e^{-1.05} \)[/tex]:
[tex]\[ I \approx 10 \cdot 0.3499377491111553 \][/tex]
10. Result:
[tex]\[ I \approx 3.499377491111553 \][/tex]
Therefore, the light intensity [tex]\( I \)[/tex] at a depth of 30 feet in Lake Heron is approximately [tex]\( 3.499377491111553 \)[/tex] lumens.