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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?

Short Answer

Expert verified
(a) The lateral magnification is -21.5. (b) The distance between the objective and the eyepiece is 174.5 cm.

Step by step solution

01

Calculate the Lateral Magnification of the Objective

For a lens, the lateral magnification \( m \) is given by the formula \( m = -\frac{s_i}{s_o} \), where \( s_i \) is the image distance and \( s_o \) is the object distance. We know that \( s_i = 18.0 \ \text{cm} \). We need to find \( s_o \) using the lens formula: \( \frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i} \). Here, \( f = 8.00 \ \text{mm} = 0.8 \ \text{cm} \). Substitute the values and solve for \( s_o \).
02

Calculate the Object Distance for the Objective

Using the lens formula \( \frac{1}{0.8} = \frac{1}{s_o} + \frac{1}{18} \), solve for \( \frac{1}{s_o} \):\[ \frac{1}{s_o} = \frac{1}{0.8} - \frac{1}{18} = 1.25 - 0.0556 = 1.1944 \ \text{cm}^{-1} \]So, \( s_o = \frac{1}{1.1944} \approx 0.837 \ \text{cm} \).
03

Calculate the Lateral Magnification of the Objective

Now substitute \( s_i = 18.0 \ \text{cm} \) and \( s_o = 0.837 \ \text{cm} \) into the magnification formula:\[ m = -\frac{18}{0.837} \approx -21.5 \].
04

Calculate the Overall Distance for the Eyepiece Setup

The image is projected 2.00 \text{m} from the eyepiece. Convert to cm: 2.00 m = 200 cm. The image from the objective acts as the object for the eyepiece. We'll call this distance \( d_1 \). This distance plus the length \( L \) between the lenses and the image distance for the eyepiece \( s_{ie} \) must sum to 200 cm. Thus, \( s_{ie} = 200 - (L + 18) \).
05

Calculate the Distance the Eyepiece

Using the lens maker's formula for the eyepiece \( \frac{1}{f_e} = \frac{1}{s_{oe}} + \frac{1}{s_{ie}} \), with \( f_e = 7.50 \ \text{cm} \), we solve for \( s_{ie} \) and hence find \( L \). If \( s_{ie} = 200 - (L + 18) \), solve \( \frac{1}{7.5} = 0 + \frac{1}{s_{ie}} \) to find \( s_{ie} = 7.5 \). Now, we find \( L = 200 - 18 - 7.5 = 174.5 \ \text{cm} \).

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Key Concepts

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

Lateral Magnification
Lateral magnification is a key concept when using lenses, especially in microscopes. It describes how much larger (or smaller) an image appears compared to the object itself. It is calculated using the formula:
  • \( m = -\frac{s_i}{s_o} \)
Where \( s_i \) is the image distance from the lens, and \( s_o \) is the object distance from the lens.
In this particular problem, the image distance \( s_i \) is 18 cm, while \( s_o \) is calculated using the lens formula. The resulting lateral magnification turns out to be approximately -21.5. Negative sign indicates that the image is inverted compared to the object.
Understanding lateral magnification helps in determining how effectively a microscope can enlarge small details of an object. This measure is crucial for examining minute structures that are not visible to the naked eye.
Lens Formula
The lens formula is an essential tool in optics that relates the focal length of a lens to the object and image distances. It is expressed as:
  • \( \frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i} \)
Here, \( f \) represents the focal length of the lens, \( s_o \) is the object distance, and \( s_i \) is the image distance.
Using this formula, we can solve for any unknown quantity if others are known. For example, if you have the focal length and the image distance, you can find the object distance easily. In the given problem, this formula helps calculate the object distance to find out the lateral magnification. It serves as a strong foundation for understanding not only microscopes but other optical instruments as well.
It underlines how light paths are influenced by lenses, contributing to image formation and manipulation.
Objective Lens
The objective lens in a microscope is probably the most important component. It is responsible for the initial magnification and image formation of the specimen being examined. In a compound microscope, the objective lens collects light from the specimen and creates a real, inverted image.
Objective lenses have different focal lengths, influencing their magnifying power. Shorter focal lengths increase magnification power, as is evident in our exercise where the focal length is 8.00 mm (0.8 cm). This lens, due to its closeness to the object, captures more details and provides a clearer image.
Understanding how the objective lens works and its relationship with other components like the eyepiece is critical for effectively using a microscope. It highlights the significance of combining optics to enhance our view of microscopic entities.
Eyepiece Lens
The eyepiece lens is the lens into which you actually look when using a microscope. It works in conjunction with the objective lens to further magnify the image formed by the objective lens. Typically, the eyepiece lens has a lower magnifying power compared to the objective lens.
In our microscope example, the eyepiece has a focal length of 7.50 cm, which is larger than that of the objective lens. This setup modifies the initial image to be suitable for human vision, enhancing overall magnification. The eyepiece lens allows fine adjustments to the focus for comfort and clarity.
The overall distance setup allows the final enlarged image to be projected onto a screen at a set distance of 2 meters. By understanding the role of the eyepiece, users can adjust the magnification and focus to best meet their viewing needs. It bridges the gap between initial image formation and the final projection.

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Most popular questions from this chapter

Rear-View Mirror. A mirror on the passenger side of your car is convex and has a radius of curvature with magnitude 18.0 \(\mathrm{cm}\) (a) Another car is seen in this side mirror and is 13.0 \(\mathrm{m}\) behind the mirror. If this car is 1.5 \(\mathrm{m}\) tall, what is the height of the image? (b) The mirror has a warning attached that objects viewed in it are closer than they appear. Why is this so?

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