Question

A microscope has a \(2.0-\mathrm{cm}\) focal length eyepiece and a \(0.80-\mathrm{cm}\) objective lens. For a relaxed normal eye, calculate the position of the object if the distance between the lenses is \(16.2 \mathrm{~cm}\)

Short answer

Solution: To find the position of the object, follow the steps we provided in the solution, and after solving the system of equations, you will find the value of \(d_{oo}\), which gives the position of the object.

Step-by-step solution

Determine the lens formulas for the eyepiece and the objective lens

For a thin lens, the lens formula is given by: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\] where - \(f\) is the focal length of the lens - \(d_o\) is the object distance (distance between object and lens) - \(d_i\) is the image distance (distance between image and lens) We have separate lens formulas for the eyepiece and the objective lens: For the eyepiece: \[\frac{1}{f_e} = \frac{1}{d_{oe}} + \frac{1}{d_{ie}}\] For the objective lens: \[\frac{1}{f_o} = \frac{1}{d_{oo}} + \frac{1}{d_{io}}\]

Write the equation for the distance between the lenses

The distance between the lenses is given as 16.2 cm. This distance is the sum of the image distance for the objective lens and the object distance for the eyepiece: \[d_{io} + d_{oe} = 16.2\]

Express the object distance for the eyepiece in terms of the objective lens

From the equation in Step 2, we can express the object distance for the eyepiece in terms of the image distance of the objective lens: \[d_{oe} = 16.2 - d_{io}\]

Substitute the values of the focal lengths into the lens formulas

We are given the focal length of the eyepiece (\(f_e\)) as 2.0 cm and the focal length of the objective lens (\(f_o\)) as 0.80 cm. Substituting these values into the lens formulas from step 1, we get: For the eyepiece: \[\frac{1}{2.0} = \frac{1}{d_{oe}} + \frac{1}{d_{ie}}\] For the objective lens: \[\frac{1}{0.80} = \frac{1}{d_{oo}} + \frac{1}{d_{io}}\]

Use the equation in Step 3 to substitute into the eyepiece lens formula

Substitute the expression for \(d_{oe}\) from Step 3 into the eyepiece lens formula: \[\frac{1}{2.0} = \frac{1}{16.2 - d_{io}} + \frac{1}{d_{ie}}\]

Solve the system of equations

We have two equations to solve simultaneously: 1. \[\frac{1}{0.80} = \frac{1}{d_{oo}} + \frac{1}{d_{io}}\] 2. \[\frac{1}{2.0} = \frac{1}{16.2 - d_{io}} + \frac{1}{d_{ie}}\] Solve this system of equations for \(d_{oo}\), the object distance for the objective lens.

Calculate the position of the object

The object distance for the objective lens, \(d_{oo}\), gives the position of the object. Once you find the value by solving the equations above, the position of the object will be the obtained value.

Key concepts

Lens Formula

Understanding the lens formula is vital when working with optical instruments like microscopes. It is expressed as \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\] where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. This equation helps us in determining the relationship between how far an object is from the lens, the distance at which an image forms, and the lens's capability to focus light. When used correctly, the formula can reveal one of these distances if the other two are known.

Especially in complex systems like microscopes with multiple lenses, understanding how to apply this formula to each lens separately, and then together, allows for precise calculations of object and image positions, leading to a clearer understanding of magnification and image formation.

Focal Length

Focal length, denoted by \(f\), is a fundamental property of lenses that describes the distance over which initially parallel rays are brought to a focus. It is determined by the curvature and refractive index of the lens material. The focal length of a lens influences both the magnification and the field of view of the image produced.

Regarding the given microscope problem, the focal lengths of the eyepiece and the objective lens are primary factors. For instance, the shorter the focal length of the objective lens, the greater the magnification, which is why high-powered microscopes have objective lenses with minuscule focal lengths.

Object Distance

The object distance, \(d_o\), in the context of lens formulas, is the distance between the object being viewed and the lens. Its measurement is critical in calculating the size and placement of the resulting image. In a microscope, the object distance affects the clarity and detail of the observed specimen; hence, it's meticulously adjusted to achieve the right focus.

For microscopes with compound lenses, as seen in our example, the position of the object relative to the primary (objective) lens must be accurately calculated to ensure a well-focused and properly magnified image. The object distance has an inverse relationship with the image distance; if one increases, the other decreases, maintaining the balance as dictated by the lens formula.

Image Distance

Image distance, \(d_i\), is the distance from the lens to the image it forms. For real images, the distance is measured along the same direction as that of the incoming light, whereas for virtual images, it's in the opposite direction. In the realm of microscopy, this distance impacts the final magnification of the viewed specimen.

Calculating the image distance, particularly in a system with multiple lenses like the microscope in our exercise, requires understanding the interplay between the two lenses' properties. Both the image formed by the objective lens and how it serves as an object for the eyepiece are significant terms in determining the distance between the lenses and, ultimately, the total magnification power of the microscope.