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The focal length of the eyepiece of a certain microscope is 18.0 \(\mathrm{mm}\) . The focal length of the objective is 8.00 \(\mathrm{mm}\) . The distance between objective and eyepiece is 19.7 \(\mathrm{cm} .\) The final image formed by the eyepiece is at infinity. Treat all lenses as thin. (a) What is the distance from the objective to the object being viewed? (b) What is the magnitude of the linear magnification produced by the objective? (c) What is the overall angular magnification of the microscope?

Short Answer

Expert verified
(a) 7.73 mm, (b) 23.38, (c) 324.46

Step by step solution

01

Understanding the Setup

The microscope consists of two lenses: the objective lens and the eyepiece lens. The total tube length (distance between eyepiece and objective) is given as 19.7 cm, which is equivalent to 197 mm to match the units of the focal lengths.
02

Calculate the Image Distance for Objective

For the eyepiece to focus the image at infinity, the intermediate image formed by the objective lens must be at the focal point of the eyepiece. Given the total tube length, we can calculate the image distance from the objective by subtracting the eyepiece focal length from the total length. Thus, the image distance (\( v_o \)) is \( 197 \text{ mm} - 18 \text{ mm} = 179 \text{ mm} \).
03

Using Lens Formula for Objective

For the objective lens, use the lens formula \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \) where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. For the objective, \( f_o = 8 \text{ mm} \) and \( v_o = 179 \text{ mm} \). Substitute into the equation to find \( u \): \[ \frac{1}{u_o} = \frac{1}{f_o} - \frac{1}{v_o} = \frac{1}{8} - \frac{1}{179}. \]Calculate \( u_o \) to find the object distance from the objective.
04

Calculate Magnification by Objective

Linear magnification by the objective is given by \( m_o = \frac{v_o}{u_o} \). Use the values of \( v_o \) and \( u_o \) calculated in the previous steps to find the magnification produced by the objective lens.
05

Calculate Overall Angular Magnification

Overall angular magnification \( M \) of the microscope is \( M = m_o \times m_e \). For the eyepiece, when the final image is formed at infinity, angular magnification is \( m_e = \frac{25}{f_e} \), where \( f_e \) is the focal length of the eyepiece in centimeters. Substitute \( m_o \) and \( m_e \) into this formula to find \( M \).
06

Calculate and Convert Eyepiece Magnification

Given that \( f_e = 18 \text{ mm} = 1.8 \text{ cm} \), substituting into the angular magnification formula for the eyepiece gives \( m_e = \frac{25}{1.8} \). Calculate this value to determine the eyepiece magnification.
07

Calculate the Final Distance and Magnifications

Compute the distances and magnifications using the step-wise calculations:(a) The object distance \( u_o \) from the objective based on the lens formula.(b) The linear magnification \( m_o \) from the objective.(c) The overall angular magnification \( M \) of the microscope using the derived expressions.

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Key Concepts

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

Focal Length
In optics, the focal length of a lens is the distance between the lens and the point where all incoming parallel light rays converge to a single point called the focus. It is a crucial measurement that determines how much a lens converges or diverges light. In our microscope example, we have two lenses: the objective with a focal length of 8 mm and the eyepiece with a focal length of 18 mm.

Focal length provides insight into the optical power of a lens. A shorter focal length means greater lens power, allowing for stronger magnification. In a microscope, the objective lens, with its short focal length, plays a critical role in enlarging small objects for detailed viewing.

Understanding focal length is vital as it affects how light behaves through lenses and how images are formed. The focal length of each lens in a microscope influences how it magnifies and resolves detail.
Linear Magnification
Linear magnification refers to how much larger (or smaller) an image is compared to the actual object. This concept is especially important when using an objective lens in a microscope. Linear magnification is calculated as the ratio of the image distance to the object distance, expressed in the formula:

\( m_o = \frac{v_o}{u_o} \).

In this formula, \( v_o \) represents the image distance (the distance from the lens to the image), and \( u_o \) is the object distance (the distance from the lens to the object). For instance, if the image distance is 179 mm, and the object distance is calculated using the lens formula, this ratio tells us how much the object is magnified by the objective lens.

Linear magnification is an intrinsic property of lenses and is useful in determining how powerful the lens is in making small objects visible in significant detail through a microscope. It helps in examining the fine details of specimens at an elevated size.
Angular Magnification
Angular magnification describes how an optical system like a telescope or microscope increases the angular size of the image compared to the object, as seen from the eye's perspective. It is crucial in microscope optics to find out how much bigger an object appears when using both the objective and eyepiece lenses.

Angular magnification is calculated by the formula:

\( M = m_o \times m_e \),where \( m_o \) is the linear magnification by the objective, and \( m_e \) is the magnification by the eyepiece.

For the eyepiece, specifically when the image is at infinity, its magnification is given by:

\( m_e = \frac{25}{f_e} \),where \( f_e \) is the focal length of the eyepiece. This highlights its power to enlarge the angular size of the image.

Understanding angular magnification is essential in designing microscopes that deliver a sufficiently large, clear image of tiny objects, making it easier to study their details at higher resolutions.
Lens Formula
The lens formula is a simple yet powerful equation connecting three key elements in optics: focal length \( f \), image distance \( v \), and object distance \( u \). It's expressed as:

\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \].

This relationship is fundamental in predicting how lenses form images and their characteristics. By knowing any two of these quantities, the third can be effortlessly found using this formula.

In our microscope example, the lens formula helps determine the object distance \( u_o \) from the objective lens when \( f \) (8 mm) and \( v_o \) (179 mm) are given. Solving this equation offers insights into where the object must be for a sharp image formation at a given distance.

The lens formula serves as a bridge to understanding how to position an object and lens to achieve desired optical outcomes, making it an indispensable tool in optics and lens design.

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Most popular questions from this chapter

A convex spherical mirror with a focal length of magnitude 24.0 \(\mathrm{cm}\) is placed 20.0 \(\mathrm{cm}\) to the left of a plane mirror. An object 0.250 \(\mathrm{cm}\) tall is placed midway between the surface of the plane mirror and the vertex of the spherical mirror. The spherical mirror forms multiple images of the object. Where are the two images of the object formed by the spherical mirror that are closest to the spherical mirror, and how tall is each image?

Figure 34.41 shows photographs of the same scene taken with the same camera with lenses of different focal length. If the object is 200 \(\mathrm{m}\) from the lens, what is the magnitude of the lateral magnification for a lens of focal length (a) \(28 \mathrm{mm} ;\) (b) 105 \(\mathrm{mm}\) ; (c) 300 \(\mathrm{mm}\) ?

A converging lens with a focal length of 7.00 \(\mathrm{cm}\) forms an image of a \(4.00-\mathrm{mm}\)-tall real object that is to the left of the lens. The image is 1.30 \(\mathrm{cm}\) tall and erect. Where are the object and image located? Is the image real or virtual?

Focal Length of a Zoom Lens. Figure P34.113 shows a simple version of a zoom lens. The converging lens has focal length \(f_{1},\) and the diverging lens has focal length \(f_{2}=-\left|f_{2}\right|\) The two lenses are separated by a variable distance \(d\) that is always less than \(f_{1} .\) Also, the magnitude of the focal length of the diverging lens satisfies the inequality \(\left|f_{2}\right|>\left(f_{1}-d\right) .\) To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius \(r_{0}\) entering the converging lens. (a) Show that the radius of the ray bundle decreases to \(r_{0}^{\prime}=r_{0}\left(f_{1}-d\right) / f_{1}\) at the point that it enters the diverging lens. (b) Show that the final image \(I^{\prime}\) is formed a distance \(s_{2}^{\prime}=\left|f_{2}\right|\left(f_{1}-d\right) /\left(\left|f_{2}\right|-f_{1}+d\right)\) to the right of the diverging lens. (c) If the rays that emerge from the diverging lens and reach the final image point are extended backward to the left of the diverging lens, they will eventually expand to the original radius \(r_{0}\) at some point \(Q .\) The distance from the final image \(I^{\prime}\) to the point \(Q\) is the effective focal length \(f\) of the lens combination; if the combination were replaced by a single lens of focal length \(f\) placed at \(Q,\) parallel rays would still be brought to a focus at \(I^{\prime}\) . Show that the effective focal length is given by \(f=f_{1}\left|f_{2}\right| /\left(\left|f_{2}\right|-f_{1}+d\right) .\) (d) If \(f_{1}=12.0 \mathrm{cm}\) \(f_{2}=-18.0 \mathrm{cm},\) and the separation \(d\) is adjustable between 0 and \(4.0 \mathrm{cm},\) find the maximum and minimum focal lengths of the combination. What value of \(d\) gives \(f=30.0 \mathrm{cm} ?\)

Contact Lenses. Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is 45.0 \(\mathrm{cm}\) from his eyes instead of the usual 25.0 \(\mathrm{cm} .(\mathrm{a})\) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal length lens is needed and what is its power in diopters?

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