Problem 10 - Applied Physics

Question

If the distance from the lens in your eye to the retina is \(19.0 \mathrm{~mm}\), what is the focal length of the lens when reading a sign \(40.0 \mathrm{~cm}\) from the lens?

Step-by-Step Solution

Answer

The focal length of the lens is approximately 18.15 mm.

Step 1: Understanding the Problem

First, we need to identify what is given and what is unknown in this problem. We are given the distance from the lens to the image formed on the retina, which is the image distance, denoted as \(d_i = 19.0\, \mathrm{mm}\). We also know the distance from the lens to the object (the sign), which is the object distance, denoted as \(d_o = 40.0\, \mathrm{cm}\). Our task is to find the focal length \(f\) of the lens.

Step 2: Convert Units

To solve the problem, both distances should be converted to the same units. Convert the object distance from centimeters to millimeters: \(d_o = 40.0\, \mathrm{cm} = 400.0\, \mathrm{mm}\). Now, both distances are in millimeters: \(d_i = 19.0\, \mathrm{mm}\) and \(d_o = 400.0\, \mathrm{mm}\).

Step 3: Apply the Lens Formula

The thin lens formula is given by \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). Substitute the values for \(d_o\) and \(d_i\) into the formula: \[\frac{1}{f} = \frac{1}{400.0} + \frac{1}{19.0}\].

Step 4: Calculate Focal Length

Calculate the right side of the equation \(\frac{1}{f} = \frac{1}{400.0} + \frac{1}{19.0}\) to find \(f\). First, calculate \(\frac{1}{400.0} \approx 0.0025\) and \(\frac{1}{19.0} \approx 0.0526\). Add these values to get \(\frac{1}{f} \approx 0.0551\). Finally, find \(f\) by taking the reciprocal: \(f \approx \frac{1}{0.0551} \approx 18.15\, \mathrm{mm}\).

Key Concepts

Lens Formula

The lens formula is a fundamental concept in geometric optics, used to relate the focal length of a lens to the distances of the object and image. This formula is given by:
\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]
Where:

• \(f\) is the focal length of the lens.
• \(d_o\) is the object distance.
• \(d_i\) is the image distance.

To use the lens formula, it's crucial to ensure that all distances are in the same unit. This formula helps in solving problems that involve locating the image formation by lenses, making it a powerful tool for understanding how lenses function in various applications. By rearranging or manipulating this formula, one can find unknown parameters like focal length, given the other two distances.

Focal Length Calculation

Calculating the focal length of a lens involves using measurements of object and image distances. To find the focal length \(f\), you utilize the reciprocal relationship expressed in the lens formula:
\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]
Once you substitute the known values for \(d_o\) and \(d_i\), you perform the arithmetic operation to solve for \(\frac{1}{f}\). Then, to isolate \(f\), you take the reciprocal of the resulting sum.
Practically:

• Convert all measurements to a consistent unit if necessary (e.g., millimeters).
• Insert the values into the lens formula.
• Compute the sum of \(\frac{1}{d_o}\) and \(\frac{1}{d_i}\).
• Finally, find \(f\) by taking the reciprocal of the result.

Doing this correctly will yield the focal length, which tells you how strongly the lens converges or diverges light.

Thin Lens Equation

The thin lens equation is a simplified version used in lens optics. It assumes lenses with very small thickness compared to their radii of curvature. The thin lens equation is essentially the same as the lens formula:
\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]
This equation assumes that:

• The lenses are "thin," meaning that their thickness doesn't significantly affect how they bend light.
• Light rays close to the principal axis (paraxial rays) are mainly considered, ignoring aberrations.

Using the thin lens equation simplifies calculations in ideal conditions, common in educational problems and introductory physics. Understanding these conditions helps when expanding to more complex scenarios, such as real lenses with thickness and non-paraxial rays.

Unit Conversion in Physics

Unit conversion is vital in physics to ensure calculations are correct and consistent. Different parameters in problems often use varied units, so converting them to a common system is necessary.

• In the given exercise, distances were initially in different units: millimeters for image distance and centimeters for object distance.
• Converting all distances to millimeters helped standardize the measurements, making the application of the lens formula straightforward.

To convert from centimeters to millimeters, you multiply by 10 (since 1 cm = 10 mm). Such conversions help in avoiding errors and ensuring that equations are solved correctly.
In physics, mastering unit conversions enhances problem-solving skills and accurately represents real-world phenomena in calculations.