Step 1: Understanding the Concept:
The power (\(P\)) of a lens is a measure of its ability to converge or diverge light and is mathematically defined as the reciprocal of the focal length (\(f\)).
\[ P = \frac{1}{f} \]
When focal length is measured in meters, power is expressed in Diopters (D).
Since the relationship is inverse, an increase in the power of a lens necessarily implies a decrease in its focal length.
A "relative decrease" or "fractional change" refers to the ratio of the change in a quantity to its original value, i.e., \(\frac{\Delta f}{f}\).
Step 2: Key Formula or Approach:
From \(f = \frac{1}{P}\), we can use calculus for small changes. Differentiating with respect to \(P\):
\[ \frac{df}{dP} = -\frac{1}{P^2} \]
\[ df = -\frac{1}{P^2} dP \]
To find the relative change, divide by \(f\):
\[ \frac{df}{f} = \frac{-\frac{1}{P^2} dP}{\frac{1}{P}} = -\frac{dP}{P} \]
The negative sign indicates that as power increases, focal length decreases. For relative "decrease", we take the magnitude:
\[ \left| \frac{\Delta f}{f} \right| = \frac{\Delta P}{P} \]
Step 3: Detailed Explanation:
Given in the problem:
Initial optical power, \(P = 2.5 \text{ D}\)
Increase in power, \(\Delta P = 0.1 \text{ D}\)
We need to find the relative decrease in focal length, which is given by the formula derived above:
\[ \text{Relative decrease} = \frac{\Delta P}{P} \]
Substituting the given values:
\[ \text{Relative decrease} = \frac{0.1}{2.5} \]
To simplify this calculation, multiply both the numerator and the denominator by 10:
\[ \text{Relative decrease} = \frac{1}{25} \]
Now, convert the fraction into a decimal:
\[ \frac{1}{25} = 0.04 \]
This means the focal length decreases by 4% of its original value.
Step 4: Final Answer:
The relative decrease in the focal length is 0.04.