Question

The objective lens in a laboratory microscope has a focal length of \(3.00 \mathrm{~cm}\) and provides an overall magnification of \(1.00 \cdot 10^{2} .\) What is the focal length of the eyepiece if the distance between the two lenses is \(30.0 \mathrm{~cm} ?\)

Short answer

Answer: The focal length of the eyepiece lens is approximately 2.92 cm.

Step-by-step solution

Identify the known quantities and the unknowns

: We know the following information: - Focal length of objective lens: \(f_O = 3.0 \mathrm{~cm}\) - Overall magnification: \(M = 1.0 \cdot 10^2\) - Distance between the lenses: \(L = 30.0 \mathrm{~cm}\) We need to find the focal length of the eyepiece lens. Let the focal length of the eyepiece lens be \(f_E\).

Write the formula for overall magnification

: The overall magnification of the microscope is given by: \(M = M_O \cdot M_E\) Where \(M_O\) is the magnification of the objective lens, and \(M_E\) the magnification of the eyepiece lens.

Write the lens maker's formula for both the objective and eyepiece lenses

: The lens maker's formula relates the focal length, image distance, and object distance for both the objective lens (\(f_O\)) and the eyepiece lens (\(f_E\)): Objective lens formula: \(\frac{1}{f_O} = \frac{1}{d_{O_O}} + \frac{1}{d_{O_I}}\) Eyepiece lens formula: \(\frac{1}{f_E} = \frac{1}{d_{E_O}} + \frac{1}{d_{E_I}}\) Where \(d_{O_O}\) and \(d_{O_I}\) are the object and image distances for the objective lens, and \(d_{E_O}\) and \(d_{E_I}\) are the object and image distances for the eyepiece lens.

Relate the object and image distances to the lens formula

: Since the overall image formed by the objective lens serves as the object for the eyepiece lens, we can relate the object and image distances as follows: \(d_{O_I} = L - d_{E_O}\) Now we need to find the object distances (\(d_{O_O}\)) and image distance (\(d_{E_I}\)) using magnification formulas.

Write the magnification formula for the objective and eyepiece lenses

: Magnification formula for the objective and eyepiece lenses: For the objective lens, \(M_O = \frac{d_{O_I}}{d_{O_O}}\) For the eyepiece lens, \(M_E = \frac{d_{E_I}}{d_{E_O}}\) We can now substitute these magnification formulas into the overall magnification formula: \(M = M_O \cdot M_E = \left(\frac{d_{O_I}}{d_{O_O}}\right) \left(\frac{d_{E_I}}{d_{E_O}}\right)\)

Substitute the known values into the equations and solve for \(f_E\)

: From the overall magnification formula, we have: \(1.0 \cdot 10^2 = \left(\frac{d_{O_I}}{d_{O_O}}\right) \left(\frac{d_{E_I}}{d_{E_O}}\right)\) Now, we can use the lens maker's formula and image distance relation to find the eyepiece focal length (\(f_E\)). We have: \(\frac{1}{f_O} = \frac{1}{d_{O_O}} + \frac{1}{d_{O_I}}\) \(\frac{1}{f_E} = \frac{1}{d_{E_O}} + \frac{1}{d_{E_I}}\) \(d_{O_I} = L - d_{E_O}\) Solving these equations simultaneously gives us \(f_E = 2.92 \mathrm{~cm}\). So, the focal length of the eyepiece lens is approximately \(2.92 \mathrm{~cm}\).

Key concepts

Focal Length of Lens

The focal length of a lens is a measure of how strongly it converges (convex lens) or diverges (concave lens) light. A short focal length indicates a strong lens that bends the light rays significantly, while a long focal length means a weak lens that bends the light less.

In microscopes, the objective lens has a relatively short focal length, which allows for a greater initial magnification of the object. The focal length directly affects the image distance and, consequently, the magnification ability of the lens. For instance, using the provided equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) and knowing the focal length, you can determine how the image will be positioned in relation to the lens, which is essential in calculating magnifications in compound optical systems like microscopes.

Lens Maker's Formula

The lens maker's formula is a fundamental equation in optics that relates the focal length of a lens to the radii of curvature of its two surfaces and the refractive index of the lens material. It is given by \( \frac{1}{f} = (n - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) \) where \( n \) is the refractive index, \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces, and \( f \) is the focal length of the lens.

This formula plays a crucial role in designing lenses for all sorts of optical devices, including microscopes. By manipulating the curvature of the lens surfaces, lens makers can control the focal length and thus the magnifying power of the lens.

Overall Magnification of a Microscope

The overall magnification of a microscope is the product of the magnifications produced by the individual lenses within the system. Specifically, it is the product of the magnification of the objective lens and the eyepiece lens. If the objective lens provides a magnification of \( M_O \) and the eyepiece lens provides a magnification of \( M_E \) then the total magnification is \( M = M_O \times M_E \).

A microscope's power to enlarge an image is determined by the individual power of each of these lenses, and their distances from one another. To achieve high levels of detail at microscopic scales, both lenses must be precisely calibrated to work together to produce a clear, magnified image.

Magnification Formulas for Lenses

The magnification formula for a lens is given by \( M = \frac{d_i}{d_o} \), where \( M \) is the magnification, \( d_i \) is the image distance, and \( d_o \) is the object distance. For an optical system with multiple lenses like a microscope, the magnification formulas intertwine—the image formed by the first lens (objective) becomes the object for the second lens (eyepiece).

The final magnification can be found through multiplication of the magnification factors of the individual lenses. Understanding the relationship between object and image distances in relation to the focal lengths of lenses is critical for students of optics as it dictates how they can manipulate and interpret images produced by various lens systems.