Problem 36 - College Physics

Question

\(\bullet\) The focal length of the eyepiece of a certain microscope is 18.0 \(\mathrm{mm}\) . The focal length of the objective is 8.00 \(\mathrm{mm}\) . The distance between objective and eyepiece is 19.7 \(\mathrm{cm}\) . 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) 8.39 mm, (b) 21.34, (c) 296.3

Step 1: Understand the problem setup

The eyepiece and the objective lens form the microscope. We know their focal lengths (18.0 mm for the eyepiece and 8.00 mm for the objective) and the distance between them (19.7 cm). We're tasked with finding the distance from the objective to the object, and both linear and angular magnifications.

Step 2: Find distance from objective to image

Since the final image formed by the eyepiece is at infinity, the image from the objective acts as the object for the eyepiece and must be at the focal point of the eyepiece. Thus, the image distance from the objective \( v_o \) is 19.7 cm minus the focal length of the eyepiece. \[ v_o = 19.7 \, \text{cm} - 1.8 \, \text{cm} = 17.9 \, \text{cm} \]

Step 3: Use lens formula to find object distance

The lens formula is \( \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \), where \( f_o \) is the focal length of the objective, \( v_o \) is the image distance, and \( u_o \) is the object distance. \[ \frac{1}{8 \, \text{mm}} = \frac{1}{179 \, \text{mm}} - \frac{1}{u_o} \] Solving for \( u_o \), we get: \[ u_o = \frac{8 \times 179}{179 - 8} = 8.39 \, \text{mm} \]

Step 4: Calculate linear magnification

The magnification produced by a lens is given by the formula \( M_o = \frac{v_o}{u_o} \). Using \( v_o = 179 \, \text{mm} \) and \( u_o = 8.39 \, \text{mm} \): \[ M_o = \frac{179}{8.39} \approx 21.34 \].

Step 5: Calculate overall angular magnification

The angular magnification \( M_t \) of a microscope is given by the product of the magnifications of the objective \( M_o \) and the eyepiece \( M_e \). For the eyepiece: \( M_e = \frac{250}{f_e} \), where \( f_e \) is the focal length of the eyepiece in mm. \[ M_e = \frac{250}{18} \approx 13.89 \] Therefore, the overall magnification is: \[ M_t = M_o \times M_e = 21.34 \times 13.89 \approx 296.3 \]

Key Concepts

Focal Length

The focal length of a lens is a fundamental concept in understanding how microscopes and other optical instruments work. It is the distance between the lens and its focus, the point at which light rays converge to form a sharp image. In microscopes, both the eyepiece and the objective lens have their own focal lengths.

In the given problem, the focal length of the eyepiece is 18.0 mm and the objective lens has a focal length of 8.00 mm. These values determine how each lens bends light and contributes to image formation.

• Eyepiece lenses typically have longer focal lengths to allow for easier viewing of the image.
• Objective lenses often have shorter focal lengths to provide greater magnification as they are closest to the object.

Knowing the focal length helps to calculate the position of the image formed by the lenses, setting the stage for understanding other concepts like linear and angular magnification later on.

Linear Magnification

Linear magnification refers to how much larger or smaller an image is compared to the actual object being viewed through the microscope. It is a crucial part of understanding how microscopes enhance our view of tiny details.

The formula used to define linear magnification by a lens is: \[M_o = \frac{v_o}{u_o}\]Here, \(M_o\) represents the magnification, \(v_o\) is the image distance from the objective lens, and \(u_o\) is the object distance from the objective lens. In our exercise, these values are found to be 179 mm and 8.39 mm respectively, leading to a magnification of 21.34.

Linear magnification is important because it describes how much larger an object appears than its actual size, and is largely determined by the objective lens based on its focal length and position relative to the object. This magnification is then multiplied by the eyepiece magnification to get the total angular magnification.

Angular Magnification

Angular magnification provides insight into how much an image taken by the microscope appears to be enlarged to the human eye, which is a bit different from just how large the image itself is. It is particularly significant because it describes the extent to which an instrument can increase the angle under which the human eye views the object.

This concept is important in microscopes, as the total magnification, known as the angular magnification, is the product of the magnifications of both the objective and eyepiece lenses. The equation is: \[M_t = M_o \times M_e\]In the exercise, the magnification of the eyepiece \(M_e\) is calculated using its focal length as:\[M_e = \frac{250}{f_e}\]Inserting the focal length of 18 mm yields a magnification of approximately 13.89.

By multiplying the objective magnification \(M_o = 21.34\) by the eyepiece magnification \(M_e\), we obtain an overall angular magnification of about 296.3. This demonstrates how microscopes compound magnification from two lenses to significantly enlarge images, enhancing our ability to examine minute details.