/*! 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 frog can see an insect clearly... [FREE SOLUTION] | 91Ó°ÊÓ

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A frog can see an insect clearly at a distance of \(10 \mathrm{~cm}\). At that point the effective distance from the lens to the retina is \(8 \mathrm{~mm} .\) If the insect moves \(5 \mathrm{~cm}\) farther from the frog, by how much and in which direction does the lens of the frog's eye have to move to keep the insect in focus? A. \(0.02 \mathrm{~cm},\) toward the retina B. \(0.02 \mathrm{~cm}\), away from the retina C. \(0.06 \mathrm{~cm}\), toward the retina D. \(0.06 \mathrm{~cm}\), away from the retina

Short Answer

Expert verified
The lens should move 0.12 cm away from the retina.

Step by step solution

01

Understand the Problem

When the insect is initially at a distance of 10 cm, the image forms on the frog's retina at an effective lens distance of 8 mm (0.8 cm). The task is to find how much and in which direction the lens must move to keep the insect in focus if it moves 5 cm farther away.
02

Model the Situation with Lens Equation

We use the lens formula: \( \frac{1}{v} = \frac{1}{u} + \frac{1}{f} \), where \(v\) is the image distance from the lens, \(u\) is the object distance, and \(f\) is the focal length of the lens. Initially, \(u = 10\) cm and \(v = 0.8\) cm are known. We can rearrange to find the focal length:\[ \frac{1}{f} = \frac{1}{0.8} - \frac{1}{10} = \frac{10 - 0.8}{8} = \frac{9.2}{8} \]Thus, \( f = \frac{8}{9.2} \approx 0.8696 \) cm.
03

Determine New Object Distance

The insect moves 5 cm farther away, making the new object distance \(u' = 15\) cm.
04

Calculate New Image Distance

Using the focal length calculated earlier and new object distance, use the lens formula again to find the new image distance \(v'\):\[ \frac{1}{v'} = \frac{1}{f} - \frac{1}{u'} = \frac{1}{0.8696} - \frac{1}{15} \]Calculate \( \frac{1}{0.8696} \approx 1.1505 \) and \( \frac{1}{15} \approx 0.0667 \). Thus \( \frac{1}{v'} = 1.1505 - 0.0667 \approx 1.0838 \), so \( v' = \frac{1}{1.0838} \approx 0.9223 \) cm.
05

Determine the Lens Movement

Initially, the image distance \(v\) is 0.8 cm. The new image distance \(v'\) is approximately 0.9223 cm. The lens must move to position the new image 0.9223 cm from the lens. Therefore, the lens moves by:\[ v' - v = 0.9223 - 0.8 = 0.1223 \text{ cm} \approx 0.12 \text{ cm} \]Since \(v' > v\), the lens moves away from the retina.

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Key Concepts

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

Lens Equation
The lens equation is a fundamental concept in optical physics that helps in understanding how lenses form images. This equation can be expressed as:
  • \( \frac{1}{v} = \frac{1}{u} + \frac{1}{f} \)
Here, \(v\) is the image distance from the lens to where the image is formed, \(u\) is the object distance from the lens, and \(f\) is the focal length of the lens.
This equation allows us to calculate one of these values if the other two are known. By using this formula, we can predict how a lens will need to adjust to focus on objects at different distances.
When an object moves, this equation helps in determining how the lens needs to adjust either by changing its position or by altering some other property like curvature to maintain focus.
Focal Length Calculation
Focal length is an intrinsic property of a lens, which determines how strongly it converges or diverges light. It is represented by \(f\). You can calculate it using the lens equation once you know the image and object distances.
To find the focal length \( f \) when the initial object distance \( u \) and image distance \( v \) are known, you can rearrange the lens equation:
  • \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
  • Replace \(v\) and \(u\) with their values
  • Solve for \(f\)
This process ensures you know how much the lens bends light, which is crucial for applications in photography, vision correction, and in this case, the vision of a frog.
Image Distance
Image distance is the distance from the lens to where the image is formed. It is denoted by \(v\). The lens equation plays a crucial role in finding this when the object distance \(u\) and the focal length \(f\) are known.
By rearranging the lens formula, you get:
  • \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \)
Solving this equation lets you determine exactly where the image will be formed relative to the lens. This is important for devices like cameras and the human eye because it dictates whether the image is in focus or blurry.
For the frog's eye, keeping the image on the retina is crucial for clear vision, so the lens must adjust as the object distance \(u\) changes.
Object Distance
Object distance, represented by \(u\), is the distance between the lens and the object being viewed. It is one of the key components of the lens equation alongside image distance and focal length.
As objects move closer or farther from a lens (such as a frog's eye), the lens must adjust to keep the object in focus. The change in \(u\) directly affects the calculations and adjustments within the lens equation.
  • When an object is moved farther from the lens, \(u\) increases
  • The lens equation then helps determine the needed adjustment in \(v\) for focusing
Understanding object distance is critical because it sets the starting point for determining other parameters in the optical setup, which are crucial for maintaining sharp, clear images.

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Most popular questions from this chapter

BIO A person can see clearly up close but cannot focus on objects beyond \(75.0 \mathrm{~cm} .\) She opts for contact lenses to correct her vision. (a) Is she nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct her vision? (c) What focallength contact lens is needed, and what is its power in diopters?

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?

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 \(25.0 \mathrm{~cm}\). (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?

Given that frogs are nearsighted in air, what is most likely to be true about their vision in water? A. They are even more nearsighted; because water has a higher index of refraction than air, a frog's ability to focus light increases in water. B. They are less nearsighted because the cornea is less effective at refracting light in water than in air. C. Their vision is no different because only structures that are internal to the eye can affect the eye's ability to focus. D. The images projected on the retina are no longer inverted because the eye in water functions as a diverging lens rather than a converging lens.

A refracting telescope has an objective lens of focal length 16.0 in. and eyepieces of focal lengths \(15 \mathrm{~mm}, 22 \mathrm{~mm}, 35 \mathrm{~mm},\) and \(85 \mathrm{~mm}\) What are the largest and smallest angular magnifications you can achieve with this instrument?

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