/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 A beetle \(4.73 \mathrm{mm}\) lo... [FREE SOLUTION] | 91Ó°ÊÓ

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A beetle \(4.73 \mathrm{mm}\) long is examined with a simple magnifier of focal length \(f=10.1 \mathrm{cm} .\) If the observer's eye is relaxed while using the magnifier, and has a near-point distance of \(25.0 \mathrm{cm}\), what is the apparent length of the beetle?

Short Answer

Expert verified
The apparent length of the beetle is approximately 11.73 mm.

Step by step solution

01

Determine Magnification Formula

The formula for the angular magnification when using a simple magnifier is given by \( M = \frac{D}{f} \), where \( D \) is the near point distance of the observer, and \( f \) is the focal length of the magnifier.
02

Substitute Known Values

Substitute the given values into the magnification formula: \( D = 25.0 \text{ cm} \) and \( f = 10.1 \text{ cm} \). Thus, \( M = \frac{25.0 \text{ cm}}{10.1 \text{ cm}} \).
03

Calculate the Magnification

Divide \( 25.0 \) by \( 10.1 \) to compute the magnification: \( M = 2.4752475 \approx 2.48 \).
04

Compute Apparent Length of Beetle

To find the apparent length of the beetle, multiply the actual length by the magnification: \( \text{Apparent Length} = 4.73 \text{ mm} \times 2.48 \).
05

Perform the Multiplication

Multiplying \( 4.73 \text{ mm} \) by \( 2.48 \) gives the apparent length: \( \text{Apparent Length} = 11.7304 \text{ mm} \approx 11.73 \text{ mm} \).

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Key Concepts

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

Angular Magnification
Angular magnification is a measure of how much larger, or closer, an object appears when viewed through an optical instrument, like a magnifier. It helps us understand how well a lens can enhance our view of tiny details. Here, angular magnification is represented by the formula:
\[ M = \frac{D}{f} \]where:
  • \( M \) denotes the angular magnification.
  • \( D \) is the near point distance, usually considered to be about 25 cm for a standard human eye without strain.
  • \( f \) is the focal length of the lens being used.
The higher the value of \( M \), the larger the object appears. By substituting the known values of \( D \) and \( f \), we can calculate the magnification factor that will affect how the object's size is perceived.
Focal Length
The focal length is a critical concept in optics influencing how lenses bend light. It is the distance between the lens and the point where light rays converge to form a sharp image. In magnifying glasses, focal length determines the lens's power to magnify objects. A shorter focal length results in a higher magnification.Using focal length:- When the focal length \( f \) is shorter, the magnification power is typically stronger.- In our exercise, the focal length provided is 10.1 cm, which tells us the strength and capability of the magnifying lens.- It is one of the main factors in the angular magnification formula \( M = \frac{D}{f} \).Understanding focal length can help us predict how various lenses affect our perception of objects, particularly when working with different types of optical instruments.
Near Point Distance
Near point distance is the minimum distance at which the eye can focus on an object without strain. For most people, this distance is approximately 25 cm. It represents the closest point that the human eye can focus comfortably for clear vision without any optical aid.In optical magnification:- The nearer the near point distance, the more relaxed the eye is during viewing, as when using a magnifying lens that allows objects to be seen without straining the eye.- This concept is crucial when calculating angular magnification, as a standard near point distance helps in ensuring that measurements are precise and accurate in relation to the physical attributes of any given lens.- In our exercise, using a near point distance \( D \) of 25.0 cm, aids in correlating the perceived size change due to magnification for consistent results.Proper understanding of near point distance aids in the effective use of magnification devices by ensuring they are used comfortably and with optimized outcomes.

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Most popular questions from this chapter

Fitting Contact Lenses with a Keratometer When a patient is being fitted with contact lenses, the curvature of the patient's cornea is measured with an instrument known as a keratometer. A lighted object is held near the eye, and the keratometer measures the magnification of the image formed by reflection from the front of the cornea. If an object is held \(10.0 \mathrm{cm}\) in front of a patient's eye, and the reflected image is magnified by a factor of \(0.035,\) what is the radius of curvature of the patient's cornea?

You have two lenses: lens 1 with a focal length of \(0.45 \mathrm{cm}\) and lens 2 with a focal length of \(1.9 \mathrm{cm}\). If you construct a microscope with these lenses, which one should you use as the objective? Explain.

A person's prescription for his new bifocal eyeglasses calls for a refractive power of -0.0625 diopter in the distance-vision part and a power of +1.05 diopters in the close-vision part. Assuming the glasses rest \(2.00 \mathrm{cm}\) from his eyes and that the corrected near-point distance is \(25.0 \mathrm{cm},\) determine the near and far points of this person's uncorrected vision.

An optical system consists of two lenses, one with a focal length of \(0.50 \mathrm{cm}\) and the other with a focal length of \(2.3 \mathrm{cm}\). If the separation between the lenses is \(12 \mathrm{cm},\) is the instrument a microscope or a telescope? Explain.

A pirate sights a distant ship with a spyglass that gives an angular magnification of \(22 .\) If the focal length of the eyepiece is \(11 \mathrm{mm},\) what is the focal length of the objective?

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