Problem 98 - Physics

Question

The near point of a naked eye is \(25 \mathrm{~cm}\). When placed at the near point and viewed by the naked eye, a tiny object would have an angular size of \(5.2 \times 10^{-5}\) rad. When viewed through a compound microscope, however, it has an angular size of \(-8.8 \times 10^{-3}\) rad. (The minus sign indicates that the image produced by the microscope is inverted.) The objective of the microscope has a focal length of \(2.6 \mathrm{~cm},\) and the distance between the objective and the eyepiece is \(16 \mathrm{~cm} .\) Find the focal length of the eyepiece.

Step-by-Step Solution

Answer

The focal length of the eyepiece is approximately 0.91 cm.

Step 1: Understand the Angular Magnification Formula

The angular magnification of a microscope is given by the formula \(M = \left( \frac{L}{f_o} \right) \left( \frac{25 \, \text{cm}}{f_e} \right)\) where \(L\) is the tube length of the microscope, \(f_o\) is the focal length of the objective, and \(f_e\) is the focal length of the eyepiece.

Step 2: Calculate the Total Angular Magnification

The total magnification \(M\) can also be calculated as the ratio of the angular size of the image when viewed through the microscope to that when viewed by the naked eye. Therefore, \(M = \frac{|\theta_m|}{|\theta_0|} = \frac{-8.8 \times 10^{-3}}{5.2 \times 10^{-5}}\). Calculate \(M\) using this ratio.

Step 3: Substitute Known Values into Magnification Equation

Substitute \(M\), \(L = 16\, \text{cm}\), and \(f_o = 2.6\, \text{cm}\) into the angular magnification formula: \( \frac{8.8 \times 10^{-3}}{5.2 \times 10^{-5}} = \left( \frac{16}{2.6} \right) \left( \frac{25}{f_e} \right)\).

Step 4: Solve for the Focal Length of the Eyepiece

First, calculate \( \frac{8.8 \times 10^{-3}}{5.2 \times 10^{-5}} \approx 169.231\). Next, simplify \( \frac{16}{2.6} \approx 6.154\). Substitute values to find \(f_e\): \(169.231 = 6.154 \times \frac{25}{f_e}\). Rearrange this to find \(f_e = \frac{6.154 \times 25}{169.231}\). Compute \(f_e\).

Step 5: Final Calculation

Calculate \(f_e = \frac{6.154 \times 25}{169.231} \approx 0.91\, \text{cm}\). Thus, the focal length of the eyepiece is approximately \(0.91 \text{ cm}\).

Key Concepts

Focal Length Calculation

When dealing with microscopes, calculating the focal length of components like the eyepiece is crucial for rendering the image focusing and magnification correctly. In our exercise, we determine the focal length of the eyepiece using given values for microscope components and formulas deriving from optics.

To start, the exercise provides several known values, such as the focal length of the microscope's objective lens, the tube length, and the angular sizes as observed. The formula used to achieve this is:\[ M = \left( \frac{L}{f_o} \right) \left( \frac{25 \, \text{cm}}{f_e} \right)\]where

• \(M\) is the total magnification,
• \(L\) is the tube length between lenses,
• \(f_o\) is the focal length of the objective,
• \(f_e\) is the focal length of the eyepiece.

After calculating the angular magnification, the next step entails substituting these values into the formula. Once this substitution is completed, we can rearrange to solve for \(f_e\). In our example, the focal length of the eyepiece was calculated as approximately \(0.91 \text{ cm}\). This value is essential in fine-tuning the clarity and detail of the image produced by the microscope.

Angular Magnification

Angular magnification is the process of enlarging the angular size of an image compared to what the naked eye would typically see. It describes how much larger or closer an object appears when viewed through an optical instrument like a microscope or telescope.

In the context of the exercise, angular magnification is key to understanding how the microscope creates a visually magnified image of the tiny object. The magnification \(M\) can be viewed as the relationship between the angular sizes observed with the microscope and the naked eye. This is computed using the formula:\[M = \frac{|\theta_m|}{|\theta_0|}\]where

• \(\theta_m\) is the angular size with the microscope, and
• \(\theta_0\) is the angular size with the naked eye.

Through this formula, our exercise identifies an angular magnification of 169.231, showing a significant enlargement of the image when viewed through the microscope. This measurement aids in further calculations, such as determining the microscope's eyepiece focal length.

Optical Instruments

Optical instruments like microscopes and telescopes are designed to assist human vision by enhancing the view of small or distant objects. They do this through a complex arrangement of lenses that focus light and increase the angular size of images.

A microscope, specifically, uses a system of lenses to produce a magnified, inverted image of small objects. The main components involved include:

• The objective lens, which is close to the specimen, collecting light and concentrating it to form a real image.
• The eyepiece lens, which further magnifies this image, making it possible to view comfortably by the eye.
• A support structure, such as a tube, which holds the lenses at a fixed distance apart, integral to determining the overall magnification.

In our exercise, the relationship between the focal lengths and their placement plays a crucial role in rendering the final magnified image. By calculating these components correctly, one ensures that the microscope achieves the desired level of detail and accuracy in ever-essential scientific observations.