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Chapter 25: Problem 27

A microscope has an objective lens with a focal length of 12.0 \(\mathrm{mm} .\) A small object is placed \(0.8 \mathrm{~mm}\) beyond the focal point of the objective lens. (a) At what distance from the objective lens does a real image of the object form? (b) What is the magnification of the real image? (c) If an eyepiece with a focal length of \(2.5 \mathrm{~cm}\) is used, with a final image at infinity, what will be the overall angular magnification of the object?

Short Answer

Expert verified
(a) 192 mm; (b) -15; (c) -150.

Step by step solution

01

Understanding the Problem

We need to determine three things: 1. The image distance (where the real image forms) using the lens formula. 2. The magnification produced by the objective lens. 3. The overall angular magnification including the eyepiece.
02

Calculate Image Distance Using Lens Formula

The focal length of the objective lens is given as 12.0 mm and the object is placed 0.8 mm beyond the focal point, which means the object distance \(d_o = 12 + 0.8 = 12.8\) mm.Using the lens formula:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Substitute the known values:\[ \frac{1}{12} = \frac{1}{12.8} + \frac{1}{d_i} \]Solving for \(d_i\), we find:\[ \frac{1}{d_i} = \frac{1}{12} - \frac{1}{12.8} = \frac{0.8}{153.6} \]\( d_i \approx 192 \) mm.
03

Calculate Magnification of Objective Lens

The magnification \(m\) produced by a lens is given by the formula:\[ m = -\frac{d_i}{d_o} \]Substitute the calculated \(d_i\) and known \(d_o\):\[ m = -\frac{192}{12.8} = -15 \]The magnification of the objective lens is -15, indicating an inverted image.
04

Calculate Overall Angular Magnification with Eyepiece

The angular magnification \(M\) of the microscope is given by the product of the magnifications of the objective lens \(m_o\) and the eyepiece lens \(m_e\).The magnification of the eyepiece is given by:\[ m_e = \frac{250}{f_e} \]Where 250 mm (or 25 cm) is the "standard" near point of the human eye. Given \(f_e = 2.5\) cm (or 25 mm), we have:\[ m_e = \frac{250}{25} = 10 \]Therefore, the overall angular magnification \(M\) is:\[ M = m_o \cdot m_e = (-15) \times 10 = -150 \]
05

Interpret the Results

The negative sign in the magnifications indicates that the images are inverted. Therefore: (a) The real image forms at a distance of 192 mm from the objective lens. (b) The magnification of the real image is -15. (c) The overall angular magnification of the microscope is -150.

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Key Concepts

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

Objective Lens
In a microscope, the **Objective Lens** is one of the critical components. It is the lens closest to the object being viewed and the one that primarily contributes to the magnification of the specimen. The objective lens is responsible for creating an enlarged and real image of the object. The focal length of the objective lens is crucial as it determines the image's size and position. In our example, the focal length is 12.0 mm, and the object is positioned slightly beyond this focal point. This specific positioning plays a significant role in calculating where the real image will form. Understanding the working of an objective lens requires recognizing how it affects light rays. Light from the object enters the objective lens, converges at a point where a real image is formed, which can be further magnified by other lenses in the system, such as the eyepiece.
Lens Formula
The **Lens Formula** is an essential physics principle used to determine the relationship between the object distance, image distance, and the focal length of a lens. The formula is mathematically expressed as:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Where:
  • \( f \) is the focal length of the lens.
  • \( d_o \) is the object distance from the lens.
  • \( d_i \) is the image distance from the lens.
In the example provided, knowing that the object is placed 0.8 mm beyond the focal length, we can calculate the image distance where the real image forms. By substituting the given values into the lens formula, we find the image distance to be approximately 192 mm. The lens formula enables us to solve crucial problems in optics, providing insights into how images are formed by lenses.
Angular Magnification
**Angular Magnification** in the context of microscopes refers to how much larger an object appears when viewed through the microscope compared to the naked eye. It is often represented as the product of the magnifications of the objective and eyepiece lenses.The formula for calculating the total angular magnification \( M \) is:\[ M = m_o \cdot m_e \]Where:
  • \( m_o \) is the magnification by the objective lens.
  • \( m_e \) is the magnification by the eyepiece lens calculated by \( \frac{250}{f_e} \).
In our example, the objective lens magnifies by -15 and the eyepiece by 10, resulting in an overall magnification of -150. This indicates that the final image is 150 times larger than the actual object. Even though the negative sign shows the image is inverted, it remains crucial for understanding how effective the microscope is in enhancing the clarity and visibility of tiny objects. Angular magnification helps you appreciate why microscopes are invaluable tools in various fields, from biology to material science.

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Most popular questions from this chapter

Zoom lens. A zoom lens is a lens that varies in focal length. The zoom lens on a certain digital camera varies in focal length from \(6.50 \mathrm{~mm}\) to \(19.5 \mathrm{~mm}\). This camera is focused on an object \(2.00 \mathrm{~m}\) tall that is \(1.50 \mathrm{~m}\) from the camera. Find the distance between the lens and the photo sensors and the height of the image (a) when the zoom is set to \(6.50 \mathrm{~mm}\) focal length and (b) when it is at \(19.5 \mathrm{~mm}\). (c) Which is the telephoto focal length, \(6.50 \mathrm{~mm}\) or \(19.5 \mathrm{~mm} ?\)

Measurements on the cornea of a person's eye reveal that the magnitude of the front surface radius of curvature is \(7.80 \mathrm{~mm},\) while the magnitude of the rear surface radius of curvature is \(7.30 \mathrm{~mm}\) (see Figure 25.18 ), and that the index of refraction of the cornea is 1.38 . If the cornea were simply a thin lens in air, what would be its focal length and its power in diopters? What type of lens would it be?

In one form of cataract surgery, the person's natural lens, which has become cloudy, is replaced by an artificial lens. The refracting properties of the replacement lens can be chosen so that the person's eye focuses on distant objects. But there is no accommodation, and glasses or contact lenses are needed for close vision. What is the power, in diopters, of the corrective contact lenses that will enable a person who has had such surgery to focus on the page of a book at a distance of \(24 \mathrm{~cm} ?\)

To determine whether a frog can judge distance by means of the amount its lens must move to focus on an object, researchers covered one eye with an opaque material. An insect was placed in front of the frog, and the distance that the frog snapped its tongue out to catch the insect was measured with high-speed video. The experiment was repeated with a contact lens over the eye to determine whether the frog could correctly judge the distance under these conditions. If such an experiment is performed twice, once with a lens of power -9 diopters and once with a lens of power -15 diopters, in which case does the frog have to focus at a shorter distance, and why? A. With the -9 -diopter lens; because the lenses are diverging, the lens with the longer focal length creates an image that is closer to the frog. B. With the -15 -diopter lens; because the lenses are diverging, the lens with the shorter focal length creates an image that is closer to the frog. C. With the -9 -diopter lens; because the lenses are converging, the lens with the longer focal length creates a larger real image. D. With the -15 -diopter lens; because the lenses are converging, the lens with the shorter focal length creates a larger real image.

The objective lens and the eyepiece of a microscope are \(16.5 \mathrm{~cm}\) apart. The objective lens has a magnification of \(62 \times,\) and the eyepiece has a magnification of \(10 \times .\) Assume that the image of the objective lies very close to the focal point of the eyepiece. Calculate (a) the overall magnification of the microscope and (b) the focal length of each lens.
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