Problem 115 - University Physics with Modern Physics

Question

A microscope with an objective of focal length 8.00 \(\mathrm{mm}\) and an eyepiece of focal length 7.50 \(\mathrm{cm}\) is used to project an image on a screen 2.00 \(\mathrm{m}\) from the eyepiece. Let the image distance of the objective be 18.0 \(\mathrm{cm} .\) (a) What is the lateral magnification of the image? (b) What is the distance between the objective and the eyepiece?

Step-by-Step Solution

Answer

Lateral magnification is approximately -20.5; distance between lenses is 25.755 cm.

Step 1: Understand the problem

We are given a microscope setup with certain focal lengths for the objective and eyepiece lenses. We need to find the lateral magnification and the distance between these two lenses based on given image distance parameters.

Step 2: Calculate the lateral magnification

The lateral magnification (M_{l1}) for the image formed by the objective is found using the lens formula:\[M_{l1} = -\frac{s_i}{s_o}\]where \(s_i = 18.0 \, \text{cm}\) is the image distance for the objective, and we find the object distance (\(s_o\)) using the formula:\[\frac{1}{f_o} = \frac{1}{s_o} + \frac{1}{s_i}\]Plug in \(f_o = 8.00 \, \text{mm} = 0.8 \, \text{cm}\) and \(s_i = 18.0 \, \text{cm}\):\[\frac{1}{0.8} = \frac{1}{s_o} + \frac{1}{18.0}\]Solving this gives \(s_o = 0.878 \, \text{cm}\).Thus, the lateral magnification \(M_{l1}\) is:\[M_{l1} = -\frac{18.0}{0.878} \approx -20.5\]

Step 3: Calculate the image distance for the eyepiece

The image produced by the eyepiece lens is on a screen 2.00 \, \text{m} away. Convert it to \, \text{cm} (\(200\,\text{cm}\)) for consistency. Using the lens formula for the eyepiece:\[\frac{1}{f_e} = \frac{1}{s'_{e}} + \frac{1}{s_{e}}\]where \(f_e = 7.50 \, \text{cm}\) and \(s'_{e} = 200.0 \, \text{cm}\).Find \(s_{e}\):\[\frac{1}{7.5} = \frac{1}{s'_{e}} + \frac{1}{s_{e}}\]Solving gives, \(s_{e} = 7.755 \, \text{cm}\).

Step 4: Calculate the distance between the objective and the eyepiece

The total length between the objective and the eyepiece is the sum of the image distance from the objective and the object distance for the eyepiece:\[\text{distance} = s_i + s_{e} = 18.0 \, \text{cm} + 7.755 \, \text{cm} = 25.755 \, \text{cm}\]

Key Concepts

Lateral Magnification

Lateral magnification is a vital concept in microscope optics. It helps us understand how much bigger or smaller an image appears compared to the actual object. The formula for lateral magnification (M_l1) is\[M_l1 = -\frac{s_i}{s_o}\]where\(s_i\) is the image distance, and\(s_o\) is the object distance. This magnification value is negative, indicating an inverted image. In our exercise, the calculated lateral magnification is approximately -20.5. This means the image is about 20.5 times larger than the object, and it's inverted. Understanding this magnification helps in enhancing the details of the observed sample, crucial for studies in biology and materials science. It transforms seemingly invisible details into visible and analyzable structures.

Objective Lens

The objective lens is one of the core components of a microscope. It plays a crucial role in determining the clarity and magnification of the specimen's image. In our case, the objective lens has a focal length of 8.00 \(\mathrm{mm}\) or 0.8 \(\mathrm{cm}\).The focal length is the distance between the center of the lens and the focal point where light rays converge. The shorter this focal length, the higher the magnification power of the objective lens. This lens is responsible for the primary magnification and is usually positioned closest to the object. A well-chosen objective lens allows fine details of the specimen to be brought into view, thus facilitating precise observation and analysis.

Eyepiece Lens

The eyepiece lens, also known as the ocular lens, is another essential part of the microscope that directly affects how we see the magnified image. In our problem, the eyepiece lens has a focal length of 7.50 \(\mathrm{cm}\). This lens recombines the light rays coming from the objective lens and magnifies them further.The eyepiece often contributes less to overall magnification compared to the objective lens, but it provides comfort and ease when observing the specimen.The final image seen through the eyepiece is virtual, enlarged, and the right way up, helping viewers observe intricate details conveniently.

Lens Formula

The lens formula is a fundamental principle used to find unknown distances or focal lengths in optical systems.The formula is expressed as:\[\frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i}\]where\(f\) is the focal length,\(s_o\) is the object distance, and\(s_i\) is the image distance. Applying this formula lets us calculate either the focal length of a lens, or the position of the object or image. This understanding is crucial in designing effective optical systems like microscopes and cameras.By applying the lens formula, optical elements can be configured to achieve desired magnification and image quality.

Focal Length

Focal length is a critical concept in optics, defining the capacity of a lens to converge light.It measures the space from the lens to the focus, where parallel rays of light meet. In microscopes, the focal length directly influences magnification.Short focal lengths indicate powerful magnification; however, they require precise focus to capture clear images.In our problem, we deal with two different focal lengths:

• Objective lens: 8.00 \(\mathrm{mm}\) (0.8\(\mathrm{cm}\))
• Eyepiece lens: 7.50\(\mathrm{cm}\)

These lengths determine how much the lens will enlarge the specimen and at what magnification level the details can be discerned.Understanding focal length helps in choosing the appropriate lens for specific observations, be it for routine lab work or intricate research experiments.