Problem 43 - Calculus - AP Edition

Question

The thin lens equation in physics is $$\frac{1}{s}+\frac{1}{S}=\frac{1}{f}$$ where \(s\) is the object distance from the lens, \(S\) is the image distance from the lens, and \(f\) is the focal length of the lens. Suppose that a certain lens has a focal length of \(6 \mathrm{cm}\) and that an object is moving toward the lens at the rate of \(2 \mathrm{cm} / \mathrm{s} .\) How fast is the image distance changing at the instant when the object is \(10 \mathrm{cm}\) from the lens? Is the image moving away from the lens or toward the lens?

Step-by-Step Solution

Answer

The image distance is increasing at 4.5 cm/s; the image moves away from the lens.

Step 1: Identify Given Variables

We are given the following values: focal length \( f = 6 \text{ cm} \), object distance \( s = 10 \text{ cm} \), and the rate at which the object distance is changing \( \frac{ds}{dt} = -2 \text{ cm/s} \). The negative sign indicates the object is moving towards the lens.

Step 2: Express the Change in Image Distance

We need to find \( \frac{dS}{dt} \), the rate of change of the image distance. Differentiating the thin lens equation \( \frac{1}{s} + \frac{1}{S} = \frac{1}{f} \) with respect to time \( t \) gives us:\[-\frac{1}{s^2} \frac{ds}{dt} - \frac{1}{S^2} \frac{dS}{dt} = 0\]

Step 3: Substitute Known Values

Substitute the values \( s = 10 \text{ cm}, \frac{ds}{dt} = -2 \text{ cm/s} \), and \( f = 6 \text{ cm} \) into the differentiated equation:\[-\frac{1}{10^2} (-2) - \frac{1}{S^2} \frac{dS}{dt} = 0\]

Step 4: Solve for Image Distance \( S \)

First, find \( S \) using the original lens equation, substituting \( s = 10 \text{ cm} \) and \( f = 6 \text{ cm} \):\[\frac{1}{S} = \frac{1}{6} - \frac{1}{10}\]Computing this gives: \[\frac{1}{S} = \frac{10 - 6}{60} = \frac{4}{60} = \frac{2}{30} = \frac{1}{15}\]So, \( S = 15 \text{ cm} \).

Step 5: Calculate \( \frac{dS}{dt} \)

Solve the differentiated equation with \( S = 15 \text{ cm} \):\[\frac{2}{100} - \frac{1}{S^2} \frac{dS}{dt} = 0\]Substitute \( S = 15 \text{ cm} \):\[\frac{2}{100} = \frac{1}{(15)^2} \frac{dS}{dt}\]\[\frac{2}{100} = \frac{1}{225} \frac{dS}{dt}\]Solving for \( \frac{dS}{dt} \):\[\frac{dS}{dt} = 2 \times \frac{225}{100} \]\[\frac{dS}{dt} = \frac{450}{100} = 4.5 \text{ cm/s}\]

Step 6: Interpret the Result

Since \( \frac{dS}{dt} = 4.5 \text{ cm/s} \) is positive, the image distance \( S \) is increasing. Therefore, the image is moving away from the lens.

Key Concepts

Thin Lens Equation

The thin lens equation is an important formula in optics, which establishes a relationship between the object distance (\(s\)), the image distance (\(S\)), and the focal length (\(f\)) of a lens. It is expressed as:

• \(\frac{1}{s} + \frac{1}{S} = \frac{1}{f}\)

Using this equation, you can determine any one of the three variables if the other two are known. This equation is fundamental in designing optical systems such as cameras, microscopes, and eyeglasses, helping to understand how lenses form images.
When a real object is placed at a certain distance from the lens, the equation helps in finding where the image will form. The focal length of the lens, a key characteristic, is constant and unique for each lens, reflecting its capacity to converge or diverge light.
For example, in our exercise, we know the focal length of the lens and the current object distance, which allows us to apply the thin lens equation to find the image distance.

Differentiation

Differentiation is a mathematical process used to determine the rate at which a function is changing at any given point. In the context of related rates problems, like those involving the thin lens equation, it allows us to understand how two or more related quantities change over time.
In our exercise scenario, the differentiation of the thin lens equation with respect to time yields a new equation:

• \[-\frac{1}{s^2} \frac{ds}{dt} - \frac{1}{S^2} \frac{dS}{dt} = 0\]

Here, \(\frac{ds}{dt}\) represents the rate at which the object distance changes, and \(\frac{dS}{dt}\) symbolizes the rate of change of the image distance. By solving this differentiated equation, we are able to determine how adjusting the object distance influences the movement of the image.
The pervasive application of differentiation in physics involves calculating changes in diverse scenarios, such as speed from position functions or current changes in electric circuits.

Physics Applications

The thin lens equation and its associated mathematics are cornerstones in various physics applications, specifically in optics. Understanding how lenses form images is crucial in designing and improving optical devices, such as telescopes, microscopes, and cameras.
These devices rely on precise calculations to optimize image placement, clarity, and focus, directly benefiting from the principles behind the thin lens equation.

• Telephoto lenses, used in cameras, depend heavily on the correct implementation of the lens formula to achieve desired zoom levels.
• Magnifying glasses employ the thin lens equation to determine the appropriate focal length to enlarge close-up objects.

Additionally, medical imaging equipment such as endoscopes and corrective lenses for vision deficiencies utilize these optical principles.
In this context, optical calculations extend beyond academic exercises and play a vital role in advancing technology and improving quality of life.

Focal Length

Focal length (\(f\)) is a critical concept in optics, defining the distance at which parallel rays of light converge or appear to diverge after passing through a lens. It is intrinsic to the lens and greatly influences image formation, governing the size and clarity of the images created.
The formula \(\frac{1}{s} + \frac{1}{S} = \frac{1}{f}\) integrates focal length to establish an interconnected relationship between the object distance and the image distance. A shorter focal length generally denotes stronger converging power and wider field of view, as seen in wide-angle lenses. Conversely, a longer focal length imparts zoom capabilities and narrower fields of view.
In our exercise, the known focal length of 6 cm serves as a constant helping to calculate the image distance when the object distance is also known.

• A precise understanding of focal length is crucial for photographers to select the right lens for specific shoots.
• It helps astronomers decide on the most effective lenses for observing distant celestial objects.

Inquiries into focal length are instrumental in both practical optics and theoretical studies of light behavior.