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An object is \(20.0 \mathrm{~cm}\) from a converging lens, and the image falls on a screen. When the object is moved \(4.00 \mathrm{~cm}\) closer to the lens, the screen must be moved \(2.70 \mathrm{~cm}\) farther away from the lens to register a sharp image. Determine the focal length of the lens.

Short Answer

Expert verified
The focal length of the lens is approximately 10.4 cm.

Step by step solution

01

Understand the Lens Formula

The lens formula relates the object distance \(d_o\), image distance \(d_i\), and focal length \(f\) of a lens as follows: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). We'll utilize this formula to solve the problem.
02

Initial Object and Image Distance Setup

Initially, the object is \(20.0\, \text{cm}\) from the lens. Let's denote the initial image distance as \(d_{i1}\). Hence, \( \frac{1}{f} = \frac{1}{20.0} + \frac{1}{d_{i1}} \) (Equation 1).
03

Adjust Object and Image Distances

The object is moved \(4.00\, \text{cm}\) closer, so the new object distance \(d_{o2}\) is \(16.0\, \text{cm}\). The screen is moved \(2.70\, \text{cm}\) further back, making the new image distance \(d_{i2} = d_{i1} + 2.70\, \text{cm}\). Now we have another equation: \( \frac{1}{f} = \frac{1}{16.0} + \frac{1}{d_{i2}} \) (Equation 2).
04

Solve the System of Equations

We have two equations:1. \( \frac{1}{f} = \frac{1}{20.0} + \frac{1}{d_{i1}} \)2. \( \frac{1}{f} = \frac{1}{16.0} + \frac{1}{d_{i1} + 2.70} \).Since both expressions equal \(\frac{1}{f}\), set them equal:\( \frac{1}{20.0} + \frac{1}{d_{i1}} = \frac{1}{16.0} + \frac{1}{d_{i1} + 2.70} \). Let’s solve this equation for \(d_{i1}\).
05

Simplify and Solve

Rearrange the equation:\( \frac{1}{20.0} - \frac{1}{16.0} = \frac{1}{d_{i1} + 2.70} - \frac{1}{d_{i1}} \).Calculate each side:\( \frac{1}{20.0} - \frac{1}{16.0} = -0.0125 \).Thus, \[ -0.0125 = \frac{-(2.70)}{(d_{i1})(d_{i1}+2.70)} \].Resolve for \(d_{i1}\):Multiply both sides by \((d_{i1})(d_{i1}+2.70)\) and solve the quadratic equation to find \(d_{i1} = 21.6\, \text{cm}\).
06

Find Focal Length

Substitute \(d_{i1} = 21.6 \text{ cm}\) back into Equation 1:\[ \frac{1}{f} = \frac{1}{20.0} + \frac{1}{21.6} \].Calculate:\[ \frac{1}{f} = 0.05 + 0.0463 = 0.0963 \].Thus, \(f = \frac{1}{0.0963} \approx 10.4 \text{ cm}\).

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Key Concepts

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

Converging Lens
A converging lens, often referred to as a convex lens, is designed to converge rays of light that are initially parallel. This means that, when light passes through a converging lens, it is bent in such a way that the rays meet at a single point known as the focal point. This principal feature makes converging lenses essential in various optical applications such as cameras, magnifying glasses, and even eyeglasses for correcting farsightedness.

The shape of convexus on both sides of the lens aids in this convergence. The thickest part at the center helps focus light at the focal point, bringing objects into sharp detail. It is essential to understand how a converging lens works to grasp how images are formed and how focal length plays a crucial role in this process.
Lens Formula
The lens formula is a crucial equation that connects the object distance (\(d_o\)), image distance (\(d_i\)), and focal length (\(f\)) of a lens. Mathematically, it is expressed as:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]This equation allows us to find any of these three parameters if the other two are known. It plays an essential part in optical physics and is a fundamental concept for solving problems involving lens systems.

Using this formula, one can predict where the image will form based on where the object is placed relative to the lens. It helps in calculating the sharpness and size of the image formed by adjusting the object and image distances accordingly.
Object and Image Distance
The object distance (\(d_o\)) and image distance (\(d_i\)) are essential metrics in understanding how lenses work. The object distance is the distance from the object to the lens's center, and the image distance is from the lens's center to where the image is formed. These distances are pivotal when using the lens formula to calculate the focal length.

In exercises such as the problem given, altering either of these distances affects how the lens creates the image. Moving the object closer or further changes the image position, necessitating adjustments to maintain focus. Understanding this interaction helps in practical applications like photography or vision correction, where precise focusing is required.
Quadratic Equation Solution
Often, determining the exact values for lens equations involves solving quadratic equations. In the context of lens problems, when we derive equations involving squared terms from the lens formula, we end up with a quadratic equation.

Solving a quadratic involves rearranging terms to fit the standard form \(ax^2 + bx + c = 0\) and then applying the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula provides solutions for the unknown variable, which, in our lens problem, indicates the object or image distance at given conditions.

Such techniques help us derive precise and well-defined solutions essential for tasks like creating sharp images through a lens, thus showing the importance of mathematical tools in real-world applications.

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Most popular questions from this chapter

A farsighted person can read printing as close as \(25.0 \mathrm{~cm}\) when she wears contacts that have a focal length of \(45.4 \mathrm{~cm}\). One day, however, she forgets her contacts and uses a magnifying glass, as in Figure \(26-40 b\). It has a maximum angular magnification of 7.50 for a young person with a normal near point of \(25.0 \mathrm{~cm}\). What is the maximum angular magnification that the magnifying glass can provide for her?

To focus a camera on objects at different distances, the converging lens is moved toward or away from the film, so a sharp im age always falls on the film. A camera with a telephoto lens \((f=200.0 \mathrm{~mm})\) is to be focused on an object located first at a distance of \(3.5 \mathrm{~m}\) and then at \(50.0 \mathrm{~m}\). Over what distance must the lens be movable?

Red light \((n=1.520)\) and violet light \((n=1.538)\) traveling in air are incident on a slab of crown glass. Both colors enter the glass at the same angle of refraction. The red light has an angle of incidence of \(30.00^{\circ} .\) What is the angle of incidence of the violet light?

A beaker has a height of \(30.0 \mathrm{~cm}\). The lower half of the beaker is filled with water, and the upper half is filled with oil \((n=1.48)\). To a person looking down into the beaker from above, what is the apparent depth of the bottom?

The contacts wom by a farsighted person allow her to see objects clearly that are as close as \(25.0 \mathrm{~cm}\), even though her uncorrected near point is \(79.0 \mathrm{~cm}\) from her eyes. When she is looking at a poster, the contacts form an image of the poster at a distance of \(217 \mathrm{~cm}\) from her eyes. (a) How far away is the poster actually located? (b) If the poster is \(0.350 \mathrm{~m}\) tall, how tall is the image formed by the contacts?

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