Problem 59 - Precalculus with Calculus Previews

Question

An empirical model published in 1952 by S. G. De Groot and \(J\). W. Gebhard in the Journal of the Optical Society of America relates the diameter \(d\) of the pupil of the eye (measured in millimeters, \(\mathrm{mm}\) ) to the luminance \(B\) of light source (measured in millilambert's, \(\mathrm{mL}\) ): $$ \log _{10} d=0.8558-0.000401\left(8.1+\log _{10} B\right)^{3} $$ (a) The average luminance of clear sky is approximately \(B=255 \mathrm{~mL}\). Use (13) to find the corresponding pupil diameter. (b) The luminance of the Sun varies from approximately \(B=190,000 \mathrm{~mL}\) at sunrise to \(B\) \(=51,000,000 \mathrm{~mL}\) at noon. Find the corresponding pupil diameters. (c) Find the luminance \(B\) corresponding to a pupil diameter of \(7 \mathrm{~mm}\).

Step-by-Step Solution

Answer

(a) \(d \approx 2.437\) mm; (b) Sunrise \(d \approx 0.788\) mm, Noon \(d \approx 0.187\) mm; (c) Calculation error with given data.

Step 1: Understand the formula

The formula given is \( \log_{10} d = 0.8558 - 0.000401 \left(8.1 + \log_{10} B\right)^3 \), where \( d \) is the pupil diameter in mm and \( B \) is the luminance in mL. Our goal is to find \( d \) or \( B \) given certain values.

Step 2: Part (a) - Substitute the average luminance

Substitute \( B = 255 \) mL into the given formula:\[ \log_{10} d = 0.8558 - 0.000401 \left(8.1 + \log_{10} 255\right)^3 \].Calculate \( \log_{10} 255 \approx 2.4065 \), then:\[ \log_{10} d = 0.8558 - 0.000401 \left(8.1 + 2.4065\right)^3 \].Evaluate the exponent:\( 8.1 + 2.4065 = 10.5065 \) and \( (10.5065)^3 \approx 1158.5 \), so:\[ \log_{10} d = 0.8558 - 0.000401 \times 1158.5 \approx 0.3865 \], thus \( d = 10^{0.3865} \approx 2.437\) mm.

Step 3: Part (b) - Sunrise luminance substitution

For \( B = 190,000 \) mL at sunrise, use:\[ \log_{10} d = 0.8558 - 0.000401 \left(8.1 + \log_{10} 190,000\right)^3 \].Calculate \( \log_{10} 190,000 \approx 5.2788 \), then:\[ \log_{10} d = 0.8558 - 0.000401 \left(8.1 + 5.2788\right)^3 \],\( 8.1 + 5.2788 = 13.3788 \) and \( (13.3788)^3 \approx 2391.0 \), so:\[ \log_{10} d = 0.8558 - 0.000401 \times 2391.0 \approx -0.1054 \], and \( d = 10^{-0.1054} \approx 0.788 \) mm.

Step 4: Part (b) - Noon luminance substitution

For \( B = 51,000,000 \) mL at noon, substitute:\[ \log_{10} d = 0.8558 - 0.000401 \left(8.1 + \log_{10} 51,000,000\right)^3 \].Calculate \( \log_{10} 51,000,000 \approx 7.7076 \), then:\[ \log_{10} d = 0.8558 - 0.000401 \left(8.1 + 7.7076\right)^3 \],\( 8.1 + 7.7076 = 15.8076 \) and \( (15.8076)^3 \approx 3947.4 \), so:\[ \log_{10} d = 0.8558 - 0.000401 \times 3947.4 \approx -0.7268 \], and \( d = 10^{-0.7268} \approx 0.187 \) mm.

Step 5: Part (c) - Finding luminance for given pupil diameter

To find \( B \) for \( d = 7 \) mm, rearrange the formula to fit \( B \):Starting from \( \log_{10} d = 0.8558 - 0.000401 (8.1 + \log_{10} B)^3 \),Substitute \( \log_{10} 7 \approx 0.8451 \), \[ 0.8451 = 0.8558 - 0.000401 (8.1 + \log_{10} B)^3 \],Solve for \( (8.1 + \log_{10} B)^3 \):\[ 0.000401 (8.1 + \log_{10} B)^3 = 0.8558 - 0.8451 \approx 0.0107 \],\[ (8.1 + \log_{10} B)^3 = \frac{0.0107}{0.000401} \approx 26.68 \],Taking cube root:\( 8.1 + \log_{10} B = (26.68)^{1/3} \approx 2.981 \), so \( \log_{10} B \approx 2.981 - 8.1 = -5.119 \),Thus, \( B = 10^{-5.119} \approx 7.59 \times 10^{-6} \) mL, which implies an error in given data or calculations.

Key Concepts

Empirical Model

Empirical models are established by observing and recording real-world data. In this exercise, an empirical model describes the relationship between the pupil diameter of the eye and the luminance of the light source. Such models are particularly useful in fields where theoretical models are hard to derive. They provide us with a practical approach to predict outcomes with reasonable accuracy.
The model by De Groot and Gebhard is based on data analysis and expresses pupil diameter (\(d\)) as a function of luminance (\(B\)) through a logarithmic equation. This approach indicates the non-linear nature of the relationship they discovered. The empirical model leverages the mathematical function that best fits the observed data, enabling scientists to make predictions about pupil size given different lighting conditions.
Understanding empirical models help students develop skills for analyzing and interpreting data, which is crucial for mathematical problem-solving and scientific research.

Logarithmic Functions

Logarithmic functions play a crucial role in this exercise. The pupil diameter is expressed in terms of the base-10 logarithm (\(\log_{10}\)) of the diameter, showing how logarithms can transform data.
Logarithmic functions are particularly useful because they can linearize relationships, making it easier to visualize and analyze varying magnitudes, like luminance. In contexts such as optical measurements, where there are large variabilities in the data (from moonlight to bright sunlight), using the logarithm can simplify computations.
In this exercise, students are tasked with inputting values into a logarithmic equation to derive results. Understanding these functions can significantly ease the process of dealing with exponential growth or reduction patterns in real-world scenarios, such as changes in light intensity affecting pupil size.

Problem-Solving in Mathematics

Mathematical problem-solving is at the core of this exercise. Students are given an equation and tasked with manipulating it to derive necessary values, such as pupil diameter or luminance. This requires a step-by-step approach:

• Understanding what each variable and constant represents in the equation.
• Substituting known values to find unknowns.
• Rearranging the equation where needed to isolate specific variables.
• Calculating logarithms and solving the resulting expressions.

Such exercises build critical thinking and numerical manipulation skills. It's important for students to practice identifying patterns, using mathematical tools (like logarithms), and interpreting results in a meaningful way. These skills are transferable across various branches of mathematics and its applications.

Optical Measurements

Optical measurements, such as pupil diameter and light luminance, are essential in the field of optics and vision science. Understanding how these measurements correlate can have practical implications for designing lighting systems or understanding human vision adaptations.
This exercise highlights how pupil diameter varies with light intensity—a fundamental concept in understanding how eyes react to different lighting conditions. This relationship allows for deeper insights into the automatic adjustments of the human eye, playing critical roles in design considerations for everything from camera technology to smart lighting systems.
By calculating the pupil's response to varying luminance levels, students can appreciate the optical measurements' complexity and their relevance in practical applications, bridging theoretical knowledge with real-world utility.