2. When light passes through a transparent medium, its intensity I at a depth d can be modelled using the equation where I0 is the initial intensity, and k is a constant.
If the intensity of sunlight is reduced to 54% of its original value after penetrating through a depth of 3.9 cm of filter glass, at what depth will the intensity be reduced to 19% of its original value?
To solve this problem, we can use the equation given for the intensity of light passing through a transparent medium:
I = I0 * e^(-k * d)
Where:
- I is the intensity at a depth d
- I0 is the initial intensity
- k is a constant
- e is the base of the natural logarithm (approximately equal to 2.71828)
- d is the depth
We are given that the intensity of sunlight is reduced to 54% of its original value (0.54 I0) after a depth of 3.9 cm. Plugging this information into the equation, we get:
0.54 I0 = I0 * e^(-k * 3.9)
To find the value of k, we can divide both sides of the equation by I0:
0.54 = e^(-k * 3.9)
To isolate k, we take the natural logarithm (ln) of both sides:
ln(0.54) = -k * 3.9
Now we can solve for k:
k = -ln(0.54) / 3.9
Next, we need to find the depth at which the intensity is reduced to 19% (0.19 I0) of its original value. Plugging this information into the equation, we get:
0.19 I0 = I0 * e^(-k * d)
Dividing both sides by I0:
0.19 = e^(-k * d)
Taking the natural logarithm (ln) of both sides:
ln(0.19) = -k * d
Now we can solve for d:
d = -ln(0.19) / k
Substituting the value we previously found for k, we can now compute d:
d = -ln(0.19) / (-ln(0.54) / 3.9)
Simplifying the expression further will give us the final result.