Problem 81
Question
The focal length of the objective and the eyepiece of a microscope are \(4 \mathrm{~mm}\) and \(25 \mathrm{~mm}\) respectively. If the final image is formed at infinity and the length of the tube is \(16 \mathrm{~cm}\), then the magnifying power of microscope will be (a) - \(337.5\) (b) \(-3.75\) (c) \(3.375\) (d) \(33.75\)
Step-by-Step Solution
Verified Answer
The magnifying power is -440 with potential revisits and clarification. Initial details encourage adjunction veccomp based checks.
1Step 1: Understanding the Formula
The magnifying power (or magnification) of a microscope is given by the formula \( M = -\dfrac{L}{f_o} \times \left(1 + \dfrac{D}{f_e}\right) \), where \( L \) is the tube length, \( f_o \) is the focal length of the objective lens, \( f_e \) is the focal length of the eyepiece, and \( D \) is the least distance of distinct vision, generally taken as \( 25 \mathrm{~cm} \). Here, \( L = 16 \mathrm{~cm} \), \( f_o = 4 \mathrm{~mm} = 0.4 \mathrm{~cm} \), and \( f_e = 25 \mathrm{~mm} = 2.5 \mathrm{~cm} \).
2Step 2: Calculating the First Part of the Formula
Calculate the first component of the formula \( -\dfrac{L}{f_o} = -\dfrac{16}{0.4} \). Carry out the division: \( -\dfrac{16}{0.4} = -40 \).
3Step 3: Calculating the Second Part of the Formula
Calculate the second part of the formula \( 1 + \dfrac{D}{f_e} = 1 + \dfrac{25}{2.5} \). Perform the division: \( \dfrac{25}{2.5} = 10 \). Thus, the term becomes \( 1 + 10 = 11 \).
4Step 4: Final Calculation for Magnification
Using the results from the previous steps, calculate the magnifying power: \( M = -40 \times 11 = -440 \).
5Step 5: Revisiting Possible Options
Re-evaluate and check the given options, as the result should match one of the presented options. The calculation should reflect the choices given. It appears there was a mistake and recalibration is needed, since the magnitude is too high and not aligned with multiple-choice answers.
6Step 6: Review and Calculate Correct Magnifying Power
Given calculation mistakes or misconception about result alignment, recalculate using correct formulation for infinity-based magnification: \( -\dfrac{L}{f_o} \times \dfrac{D}{f_e} \): \( -40 \times \dfrac{25}{2.5} = -40 \times 10 = -400 \). Check computational or mistake overlap initially. Equate variations or reasonable comprehension to nearest positive options.
Key Concepts
focal lengthmicroscope opticsleast distance of distinct vision
focal length
In the realm of microscope optics, the term focal length is pivotal. It describes the distance from the lens to the point where it brings light into perfect focus. This parameter is crucial in designing and understanding optical instruments, such as microscopes, which contain multiple lenses that work together to magnify images.
- **Objective Lens:** This is the lens closest to the specimen, and it usually has a very short focal length to allow for a high magnification. In our exercise, the focal length of the objective lens is given as 4 mm, or 0.4 cm.
- **Eyepiece Lens:** This lens is positioned closer to the viewer’s eye. It typically has a longer focal length compared to the objective lens, in this case, 25 mm or 2.5 cm.
The focal length influences how much you can magnify an object and still resolve detail. Shorter focal lengths can yield higher magnifications but require careful adjustments and precise positioning when capturing images.
Understanding these different focal lengths is essential when calculating the total magnifying power of the microscope, as it combines the optical strengths of both the objective and the eyepiece lenses.
- **Objective Lens:** This is the lens closest to the specimen, and it usually has a very short focal length to allow for a high magnification. In our exercise, the focal length of the objective lens is given as 4 mm, or 0.4 cm.
- **Eyepiece Lens:** This lens is positioned closer to the viewer’s eye. It typically has a longer focal length compared to the objective lens, in this case, 25 mm or 2.5 cm.
The focal length influences how much you can magnify an object and still resolve detail. Shorter focal lengths can yield higher magnifications but require careful adjustments and precise positioning when capturing images.
Understanding these different focal lengths is essential when calculating the total magnifying power of the microscope, as it combines the optical strengths of both the objective and the eyepiece lenses.
microscope optics
Microscope optics represent a sophisticated interplay of lenses and their alignments to enlarge tiny structures or organisms. In microscopes, magnification depends greatly on the optics, involving both the objective and eyepiece lenses. These lenses work together to form a highly magnified image of a small object.
- **Objective Lens System:** This lens is primarily responsible for creating a real, inverted, and magnified image of the object. It gathers light and brings it to focus at a certain tube length (distance from the objective lens to the intermediate image plane, here, 16 cm).
- **Eyepiece System:** The eyepiece, or ocular, then further magnifies this image formed by the objective. When the final image is focused at infinity, the eyepiece allows for a relaxed viewing experience because the eye does not need to accommodate for near viewing distances.
The combined magnification in microscopy is calculated using each lens's contribution, reflecting the system's overall capability to enlarge the view.
Notably, in microscope optics, a change in the focal lengths of either the objective or the eyepiece lens will proportionately affect the total magnification power, emphasizing the importance of precise optical design and alignment in microscopes.
- **Objective Lens System:** This lens is primarily responsible for creating a real, inverted, and magnified image of the object. It gathers light and brings it to focus at a certain tube length (distance from the objective lens to the intermediate image plane, here, 16 cm).
- **Eyepiece System:** The eyepiece, or ocular, then further magnifies this image formed by the objective. When the final image is focused at infinity, the eyepiece allows for a relaxed viewing experience because the eye does not need to accommodate for near viewing distances.
The combined magnification in microscopy is calculated using each lens's contribution, reflecting the system's overall capability to enlarge the view.
Notably, in microscope optics, a change in the focal lengths of either the objective or the eyepiece lens will proportionately affect the total magnification power, emphasizing the importance of precise optical design and alignment in microscopes.
least distance of distinct vision
The least distance of distinct vision is a critical concept in optics, particularly in understanding human visual perception and its interaction with instruments like microscopes. This distance is essentially the closest point at which the human eye can focus on an object without straining, and it's conventionally taken as 25 cm for a normal eye.
- **Relevance in Calculations:** In microscope magnification, the least distance of distinct vision (often denoted as "D") is integral to the magnification formula. For the typical human observer, this standard value simplifies calculations and provides a uniform basis for comparison across optical systems.
- **Use in Instruments:** When designing optical devices, engineers account for "D" to ensure images appear sharp when viewed through eyepieces. It's leveraged in equations to help adjust the magnification to achieve clarity and comfort.
Understanding this concept is crucial when dealing with microscopes, as it assists in determining the precise magnifying power needed to render an image clear and useful without eye strain.
It serves as a bridge between the mechanical functions of the microscope and the natural limitations of human vision.
- **Relevance in Calculations:** In microscope magnification, the least distance of distinct vision (often denoted as "D") is integral to the magnification formula. For the typical human observer, this standard value simplifies calculations and provides a uniform basis for comparison across optical systems.
- **Use in Instruments:** When designing optical devices, engineers account for "D" to ensure images appear sharp when viewed through eyepieces. It's leveraged in equations to help adjust the magnification to achieve clarity and comfort.
Understanding this concept is crucial when dealing with microscopes, as it assists in determining the precise magnifying power needed to render an image clear and useful without eye strain.
It serves as a bridge between the mechanical functions of the microscope and the natural limitations of human vision.
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