/*! 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 91 An object to the left of a lens ... [FREE SOLUTION] | 91Ó°ÊÓ

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An object to the left of a lens is imaged by the lens on a screen 30.0 \(\mathrm{cm}\) to the right of the lens. When the lens is moved 4.00 \(\mathrm{cm}\) to the right, the screen must be moved 4.00 \(\mathrm{cm}\) to the left to refocus the image. Determine the focal length of the lens.

Short Answer

Expert verified
The focal length of the lens is approximately 7.37 cm.

Step by step solution

01

Understand the given problem

The object forms an image on a screen via a lens, initially positioned where the image is sharp. The lens is then moved 4.00 cm to the right, and the screen is adjusted 4.00 cm to the left to refocus the image. We need to find the focal length of the lens.
02

Identify the lens formula and variables

The lens formula is given by \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance.
03

Choose reference positions

Let the initial image distance \(v_1\) be 30 cm. When the lens is moved 4 cm right, suppose the new image distance is \(v_2 = v_1 - 4 = 26 \) cm.
04

Express the conditions for the focal length

From the lens formula and the geometry of the situation, the condition given by the movement implies that the lens must satisfy the condition \( f = \frac{D^2 - d^2}{4D}\), where \(D\) is the original image distance (30 cm) and \(d\) is the displacement of the lens (4 cm).
05

Calculate the focal length

Substituting the values into the formula, we get: \[ f = \frac{30^2 - 4^2}{4 \times 30} = \frac{900 - 16}{120} = \frac{884}{120} = 7.37 \text{ cm}. \]

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Key Concepts

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

Lens Formula
The lens formula is a fundamental equation in optics that relates the focal length of a lens to the distances of the object and the image from the lens. The formula is expressed as:\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]where:
  • \(f\) is the focal length of the lens,
  • \(v\) is the image distance (distance from the lens to the image), and
  • \(u\) is the object distance (distance from the lens to the object).
This formula is essential for understanding how light is focused by lenses to form images. It helps determine where the image will appear, which is crucial for tasks such as photography, microscopy, and even vision correction. In essence, by changing the distances \(u\) or \(v\), one can manipulate the focal length \(f\), which is what happens when focusing a camera or adjusting optical instruments.
Image Distance
Image distance, represented by \(v\) in the lens formula, is the distance from the lens to the image it forms. In the given exercise, the initial image distance is specified as 30 cm. When the lens was moved 4 cm to the right, the image distance changed accordingly to maintain focus.Image distance plays a vital role in determining where a sharp image will appear. It can be adjusted by moving the lens or altering the position of the screen where the image is being cast. Understanding this concept is crucial, especially in laboratory settings where precise images are needed.Whenever the image distance is changed, the focus of the image adjusts. This is why, in the scenario when the lens is moved, the screen also has to be adjusted to refocus, illustrating the principle that the position of the sharp image depends on both lens placement and focal length.
Object Distance
Object distance, denoted as \(u\) in the lens formula, refers to the distance between the lens and the object being observed. Though not explicitly given in the exercise, the object distance must be considered to solve the problem using the lens formula.Understanding object distance is crucial because it contributes to the lens formula that determines the focal length. The distance affects how the image is formed and where it will appear. By adjusting the object distance, you can control the size and sharpness of the image.In experiments and real-life applications, measuring this distance accurately is necessary for precise optical work. Incorrectly assessing object distance can lead to blurred images or the image appearing in unintended locations.
Optical Refocusing
Optical refocusing involves adjusting the position of the lens or the screen to achieve a sharp image when the initial setup has changed. In the exercise, the lens was moved 4 cm to the right, requiring the screen to be moved 4 cm to the left to bring the image back into focus. The concept of optical refocusing is prevalent in devices like cameras, microscopes, and telescopes, where accurate focus adjustments are constantly needed. It relies on the interplay of the focal length, and the distances between the lens, object, and image. This adjustment demonstrates the delicate balance needed to maintain a clear image when either the lens or the screen is moved. Optical refocusing ensures that even a small change in lens or screen position can be corrected to retain image clarity. Understanding this concept helps in understanding how lenses work to maintain sharp focus both in theoretical and practical applications.

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Most popular questions from this chapter

A light bulb is 3.00 m from a wall. You are to use a concave mirror to project an image of the bulb on the wall, with the image 2.25 times the size of the object. How far should the mirror be from the wall? What should its radius of curvature be?

A camera lens has a focal length of 180.0 \(\mathrm{mm}\) and an aperture diameter of 16.36 \(\mathrm{mm}\) . (a) What is the \(f\) -number of the lens? (b) If the correct exposure of a certain scene is \(\frac{1}{30} \mathrm{s}\) at \(f / 11\) what is the correct exposure at \(f / 2.8 ?\)

A telescope is constructed from two lenses with focal lengths of 95.0 \(\mathrm{cm}\) and \(15.0 \mathrm{cm},\) the 95.0 -cm lens being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification for the telescope. (b) Find the height of the image formed by the objective of a building 60.0 \(\mathrm{m}\) tall, 3.00 \(\mathrm{km}\) away. (c) What is the angular size of the final image as viewed by an eye very close to the eyepiece?

The Cornea As a Simple Lens. The cornea behaves as a thin lens of focal length approximately \(1.8 \mathrm{cm},\) although this varies a bit. The material of which it is made has an index of refraction of \(1.38,\) and its front surface is convex, with a radius of curvature of 5.0 \(\mathrm{mm} .\) (a) If this focal length is in air, what is the radius of curvature of the back side of the cornea? (b)The closest distance at which a typical person can focus on an object (called the near point) is about \(25 \mathrm{cm},\) although this varies considerably with age. Where would the cornea focus the image of an 8.0 -mm- tall object at the near point? (c) What is the height of the image in part (b)? Is this image real or virtual? Is it erect or inverted? (Note: The results obtained here are not strictly accurate because, on one side, the cornea has a fluid with a refractive index different from that of air.)

A transparent rod 30.0 \(\mathrm{cm}\) long is cut flat at one end and rounded to a hemispherical surface of radius 10.0 \(\mathrm{cm}\) at the other end. A small object is embedded within the rod along its axis and halfway between its ends, 15.0 \(\mathrm{cm}\) from the flat end and 15.0 \(\mathrm{cm}\) from the vertex of the curved end. When viewed from the flat end of the rod, the apparent depth of the object is 9.50 \(\mathrm{cm}\) from the flat end. What is its apparent depth when viewed from the curved end?

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