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Resolution of a Microscope. The image formed by a microscope objective with a focal length of 5.00 \(\mathrm{mm}\) is 160 \(\mathrm{mm}\) from its second focal point. The eyepiece has a focal length of 26.0 \(\mathrm{mm}\) . (a) What is the angular magnification of the microscope? (b) The unaided eye can distinguish two points at its nicroscope? (b) The unaided eye can distinguish two points at its near point as separate if they are about 0.10 \(\mathrm{mm}\) apart. What is the minimum separation between two points that can be observed (or resolved) through this microscope?

Short Answer

Expert verified
(a) The angular magnification is approximately 298.22. (b) The minimum resolvable separation is approximately 0.000336 mm.

Step by step solution

01

Calculate the Objective Magnification

The total magnification of a microscope is given by the product of the magnification of the objective lens and the magnification of the eyepiece. Start by finding the magnification provided by the objective lens. The magnification of the objective lens is calculated using the formula \( M_{objective} = \frac{L - f_{objective}}{f_{objective}} \), where \( L \) is the distance from the image to the second focal point of the objective lens (160mm) and \( f_{objective} \) is the focal length of the objective lens (5.00mm).\[ M_{objective} = \frac{160\, \text{mm} - 5\, \text{mm}}{5\, \text{mm}} = 31 \]
02

Calculate the Eyepiece Magnification

Now, calculate the magnification of the eyepiece. The eyepiece magnification is given by \( M_{eyepiece} = \frac{25\, \text{cm}}{f_{eyepiece}} \), with 25 cm being the normal near point distance for a human eye, and \( f_{eyepiece} \) the focal length of the eyepiece (26mm).Since we are dealing with different units here (cm and mm), convert 25 cm to 250mm.\[ M_{eyepiece} = \frac{250\, \text{mm}}{26\, \text{mm}} \approx 9.62 \]
03

Calculate the Total Angular Magnification

The total angular magnification of the microscope is the product of the magnifications of the objective and eyepiece lenses. Combine the results from the previous steps:\[ M_{total} = M_{objective} \times M_{eyepiece} = 31 \times 9.62 \approx 298.22 \]
04

Determine Minimum Resolvable Separation

The unaided eye can distinguish points 0.10 mm apart at its near point. When using the microscope, the minimum separation \( d_{min} \) that can be resolved is given by: \( d_{min} = \frac{0.10\, \text{mm}}{M_{total}} \).\[ d_{min} = \frac{0.10\, \text{mm}}{298.22} \approx 0.000336\, \text{mm} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Magnification
Angular magnification refers to how much larger or smaller an object appears when viewed through an optical instrument compared to the naked eye. In the context of microscopes, angular magnification is the result of the combined magnifying power from two key elements: the objective lens and the eyepiece lens. When you view an object with the naked eye, it forms an image on the retina based on the angle of light entering the eye. A microscope enhances this by converging light from the object through lenses, increasing the perceived angular size, making tiny details visible. Key points about angular magnification: - It provides a measure of how much larger an image appears compared to the actual size of the object. - The total angular magnification of a microscope is the product of the magnifications of the objective and eyepiece lenses. - Adjusting either lens can affect the overall magnification of the microscope.
Objective Lens
The objective lens is the primary lens in a microscope that determines the initial magnification and resolution of the observed specimen. It is placed close to the object, capturing light rays and focusing them to create a real image. The focal length of the objective lens is a crucial factor that affects magnification. In the given exercise, the focal length is 5.00 mm. Here's how the objective lens affects microscope usage:- **Magnification Calculation**: The magnification by the objective lens is calculated as the ratio of the distance from the image to the second focal point of the objective lens to the lens's focal length.- **Image Quality**: A shorter focal length typically results in higher magnification but can also impact image clarity and field of view.Using the formula:\[ M_{objective} = \frac{L - f_{objective}}{f_{objective}} \] we calculated that the objective lens magnifies the image 31 times.
Eyepiece Magnification
The eyepiece magnification plays an essential role in the final step of image magnification in a microscope. It enlarges the image formed by the objective lens, allowing further exploration of details by the human eye. The eyepiece magnification is determined by the focal length of the eyepiece lens, which, in this exercise, is 26.0 mm. Here's what to know about eyepiece magnification:- **Magnification Formula**: Calculated using the formula \( M_{eyepiece} = \frac{25}{f_{eyepiece}} \), where 25 is the near point distance in centimeters converted to millimeters for consistency. This gives us an eyepiece magnification of approximately 9.62.- **User Experience**: A higher eyepiece magnification results in greater enlargement of the primary image created by the objective lens, enhancing detail visibility without significantly affecting the field of view. By multiplying the eyepiece magnification with that of the objective lens, one can determine the total angular magnification of the microscope.
Resolvable Separation
Resolvable separation refers to the smallest distance between two points that can be distinguished as separate entities by the optical system. It is an essential indicator of the resolving power of a microscope. For the unaided human eye, this distance is about 0.10 mm. Under a microscope, this separation reduces significantly, allowing for analysis at a much finer scale. Here's how resolvable separation is determined in microscopes:- **Resolution Limit**: The ability to resolve points closer than the unaided eye's limit relies on the microscope's total magnification. - **Minimum Resolvable Distance Formula**: Calculated using \( d_{min} = \frac{0.10}{M_{total}} \), with \( M_{total} \) being the combined magnification. In this exercise, it results in a minimum separation of approximately 0.000336 mm.Thus, the microscope dramatically enhances the resolving power, enabling the observation of finer details otherwise imperceptible to the naked eye.

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Most popular questions from this chapter

A camera with a 90 -mm-focal-length lens is focused on an object 1.30 \(\mathrm{m}\) from the lens. To refocus on an object 6.50 \(\mathrm{m}\) from the lens, by how much must the distance between the lens and the film be changed? To refocus on the more distant object, is the lens moved toward or away from the film?

Three thin lenses, each with a focal length of \(40.0 \mathrm{cm},\) are aligned on a common axis; adjacent lenses are separated by 52.0 \(\mathrm{cm} .\) Find the position of the image of a small object on the axis, 80.0 \(\mathrm{cm}\) to the left of the first lens.

An object to the left of a lens is imaged by the lens on a screen 30.0 \(\mathrm{cm}\) to the right of the lens. When the lens is moved 4.00 \(\mathrm{cm}\) to the right, the screen must be moved 4.00 \(\mathrm{cm}\) to the left to refocus the image. Determine the focal length of the lens.

A tank whose bottom is a mirror is filled with water to a depth of 20.0 \(\mathrm{cm} .\) A small fish floats motionless 7.0 \(\mathrm{cm}\) under the surface of the water. (a) What is the apparent depth of the fish when viewed at normal incidence? (b) What is the apparent depth of the image of the fish when viewed at normal incidence?

A converging lens with a focal length of 12.0 \(\mathrm{cm}\) forms a virtual image 8.00 \(\mathrm{mm}\) tall, 17.0 \(\mathrm{cm}\) to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.

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