Problem 59 - University Physics with Modern Physics

Question

The focal length of the eyepiece of a certain microscope is 18.0 mm. The focal length of the objective is 8.00 mm. The distance between objective and eyepiece is 19.7 cm. The final image formed by the eyepiece is at infinity. Treat all lenses as thin. (a) What is the distance from the objective to the object being viewed? (b) What is the magnitude of the linear magnification produced by the objective? (c) What is the overall angular magnification of the microscope?

Step-by-Step Solution

Answer

(a) 8.48 mm. (b) -21.1. (c) -293.3.

Step 1: Understand the Problem

We are given a microscope with the eyepiece and objective having focal lengths of 18.0 mm and 8.00 mm, respectively. The lenses are 19.7 cm apart, with the final image at infinity. We need to find the object distance for the objective, the linear magnification, and the microscope's angular magnification.

Step 2: Relation for Lens Equation

We use the lens formula for the objective: \( \frac{1}{f_o} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f_o \) is the focal length of the objective (8.00 mm), \( d_o \) is the object distance, and \( d_i \) is the image distance. Since lenses are thin, \( d_i + d_e = 19.7 \text{ cm} \). Convert this to mm, giving 197 mm.

Step 3: Find the Image Distance for Objective

Because the final image is at infinity, the image formed by the objective must be at the focal point of the eyepiece, \( d_e = 18 \text{ mm} \). Then, \( d_i = 197 \text{ mm} - 18 \text{ mm} = 179 \text{ mm}. \) Substitute into the lens equation to find \( d_o \).

Step 4: Solve for Objective Object Distance

Using \( \frac{1}{8} = \frac{1}{d_o} + \frac{1}{179} \), solve for \( d_o \). Rearrange to get \( \frac{1}{d_o} = \frac{1}{8} - \frac{1}{179} \). Calculate to find \( d_o \approx 8.48 \text{ mm} \).

Step 5: Linear Magnification of the Objective

The magnification \( m_o \) of the objective is given by \( m_o = -\frac{d_i}{d_o} \). Substitute \( d_i = 179 \text{ mm} \) and \( d_o = 8.48 \text{ mm} \). Calculate to find \( m_o \).

Step 6: Overall Angular Magnification

The total magnification is the product of the objective's linear magnification and the eyepiece's angular magnification. The angular magnification of the eyepiece \( m_e = \frac{25 \, \text{cm}}{f_e} \), where \( 25 \, \text{cm} \) is the near point of human vision. Substitute \( f_e = 18 \text{ mm} = 1.8 \text{ cm} \). Calculate the final magnification \( m_{total} = m_o \times m_e \).

Step 7: Final Calculations

Calculate \( m_o = -\frac{179}{8.48} \approx -21.1 \). Then \( m_e = \frac{25}{1.8} \approx 13.9 \), giving \( m_{total} = (-21.1) \times 13.9 \approx -293.29 \).

Key Concepts

Focal Length

Focal length is a fundamental concept in optics that determines how strongly a lens converges or diverges light. It is the distance between the lens and its focus, where parallel light rays converge after passing through the lens.

• A shorter focal length means the lens is more powerful and converges light quickly.
• A longer focal length indicates a weaker lens that takes longer to bring light to a focus.

In the context of the provided exercise, the focal lengths play a crucial role in understanding how the microscope lenses function. The eyepiece has a focal length of 18.0 mm, while the objective's focal length is 8.00 mm. These measurements affect how magnification and image formation occur within the microscope.
The objective with a shorter focal length is capable of creating a highly magnified intermediate image, which the eyepiece further magnifies for the viewer.

Lens Formula

The lens formula is a key equation in optics, used to relate the object distance, image distance, and focal length of a lens. The standard form of this formula is:
\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]where:

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

In the original exercise, we use this formula to find the distance from the objective to the object being viewed. It's important to convert all measurements into consistent units, ensuring accurate calculations. For this problem, we derive that the image distance of the objective is effectively 179 mm, allowing us to calculate the object distance accurately as approximately 8.48 mm.

Magnification

Magnification in optics describes how much larger or smaller an image appears compared to the real object. It can be linear (for distances) or angular (for angles and visual perception).
Linear magnification for a lens, like the objective in a microscope, is given by:

• \( m = -\frac{d_i}{d_o} \)

In this equation, \( m \) represents magnification, \( d_i \) is the image distance, and \( d_o \) is the object distance. For the microscope's objective with the calculated distances, it becomes clear the magnification is about \(-21.1\), indicating the image is inverted and magnified 21.1 times.
The total magnification of the microscope combines the eyepiece's angular magnification with the objective's linear magnification, providing the viewer with a greatly enhanced view of the object.

Thin Lens Equation

In optics, the thin lens equation is a simplified model that assumes lenses have negligible thickness. This model allows for easier calculations, particularly in educational contexts or basic optical devices.
By treating lenses as thin, their focal lengths can be effectively modeled with the lens formula. Thin lens assumptions simplify complex systems, like microscopes, into manageable data points. This aids in calculating crucial measures, like object and image distances, and magnifications.
Within the provided exercise, assuming thin lenses helps find key distances and magnifications with minimal computational effort, using the relations derived from the lens formula.
While this model is quite effective for most calculations, it's important to remember that real lenses may have limited thickness, affecting precision in professional applications.