/*! 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 41 The relationship between the dis... [FREE SOLUTION] | 91Ó°ÊÓ

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The relationship between the distance of an object to a lens \(d_{\mathrm{o}}\), the distance to the image \(d_{\mathrm{i}}\), and the focal length \(f\) of the lens is given by $$\frac{f\left(d_{\mathrm{i}}+d_{\mathrm{o}}\right)}{d_{\mathrm{i}} d_{\mathrm{o}}}=1$$ Use partial-fraction decomposition to write the lens law in terms of sums of fractions. What does each term represent?

Short Answer

Expert verified
The lens formula can be decomposed as \( \frac{f}{d_{\mathrm{i}}} + \frac{f}{d_{\mathrm{o}}} = 1 \).

Step by step solution

01

Recognize Given Formula and Objective

We're given the lens formula: \( \frac{f(d_{\mathrm{i}} + d_{\mathrm{o}})}{d_{\mathrm{i}} d_{\mathrm{o}}} = 1 \). The task is to express this as a sum of fractions using partial-fraction decomposition. This will help us interpret each term's physical significance.
02

Rewrite the Formula

Rewrite the formula in a form ready for partial fraction decomposition. Begin by expanding the numerator: \( f(d_{\mathrm{i}} + d_{\mathrm{o}}) = f d_{\mathrm{i}} + f d_{\mathrm{o}} \). So the formula becomes \( \frac{f d_{\mathrm{i}} + f d_{\mathrm{o}}}{d_{\mathrm{i}} d_{\mathrm{o}}} = 1 \).
03

Set Up Partial Fraction Decomposition

We decompose \( \frac{f d_{\mathrm{i}} + f d_{\mathrm{o}}}{d_{\mathrm{i}} d_{\mathrm{o}}} \) as a sum of two fractions: \( \frac{f d_{\mathrm{i}} + f d_{\mathrm{o}}}{d_{\mathrm{i}} d_{\mathrm{o}}} = \frac{A}{d_{\mathrm{i}}} + \frac{B}{d_{\mathrm{o}}} \), where \(A\) and \(B\) are constants to be determined.
04

Calculate Constants Using Comparison

Multiply both sides by \(d_{\mathrm{i}} d_{\mathrm{o}}\) to clear the denominators: \( f d_{\mathrm{i}} + f d_{\mathrm{o}} = A d_{\mathrm{o}} + B d_{\mathrm{i}} \). By comparing coefficients, equate terms to solve for \(A\) and \(B\). This gives us two equations: \( A = f \) and \( B = f \).
05

Write the Decomposed Formula

Insert the values of \(A\) and \(B\) back into the decomposition: \( \frac{f d_{\mathrm{i}} + f d_{\mathrm{o}}}{d_{\mathrm{i}} d_{\mathrm{o}}} = \frac{f}{d_{\mathrm{i}}} + \frac{f}{d_{\mathrm{o}}} \).
06

Interpret Each Term

Each term in the result represents the contribution to the lens formula from the image side and the object side respectively. Specifically, \( \frac{f}{d_{\mathrm{i}}} \) represents the influence of the image distance, and \( \frac{f}{d_{\mathrm{o}}} \) represents the influence of the object distance.

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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 that links the object distance, image distance, and the focal length of a lens. It is used in optics to determine the position of an image formed by a lens. The generalized form of the lens equation is \( \frac{1}{f} = \frac{1}{d_i} + \frac{1}{d_o} \), where \( f \) is the focal length, \( d_i \) is the image distance, and \( d_o \) is the object distance. Understanding this equation helps predict how a lens will behave when forming an image.

In practical applications, this formula allows us to calculate one of these three variables if the other two are known. This is especially useful in designing optical systems like cameras, microscopes, and spectacles. The equation emphasizes the reciprocal nature of distances involved in image formation through a lens.
Focal Length
The focal length of a lens is a crucial parameter in optics that describes how strongly the lens converges or diverges light. It is the distance between the lens and the point where parallel rays of light converge (focus) after passing through the lens. A shorter focal length means a more powerful lens, as it bends light more sharply.

Focal length determines the magnification and field of view of a lens. In the context of the lens formula, the focal length \( f \) is incorporated to express fundamental relationships in the lens equation. It influences how an object's distance compresses or stretches, thereby impacting the size and location of the image formed. For lenses doing everyday tasks such as photography, understanding focal length assists in achieving the desired image composition and perspective.
Image Distance
Image distance, denoted as \( d_i \), represents the distance from a lens to the image it forms. In the lens formula \( \frac{1}{f} = \frac{1}{d_i} + \frac{1}{d_o} \), the image distance is a crucial variable that helps describe the image's location relative to the lens. When a lens focuses light, the light rays converge at this point, forming a clear image.

Changes to the image distance affect the image size and sharpness. Playful experiments like adjusting the position of a lens relative to a screen can help one observe the shifts in image clarity and size. By incorporating the concept of partial fraction decomposition into the lens formula, understanding the image distance's role becomes clearer, as represented in the decomposed expression \( \frac{f}{d_i} \), highlighting its influence on image formation.
Object Distance
Object distance \( d_o \) signifies the distance from the lens to the object being viewed. In the lens formula, it is a fundamental variable used to predict the nature of the image produced when the object is placed at a specific distance from the lens. Along with image distance and focal length, it is part of the essential relationship depicted by the formula \( \frac{1}{f} = \frac{1}{d_i} + \frac{1}{d_o} \).

Changing the object distance modifies the image attributes such as size, type (upright/inverted), and clarity. The term \( \frac{f}{d_o} \) in the partial fraction decomposition signifies the contribution of object distance to the overall lens formula. For practical optics problems, such as using a projector or a magnifying glass, object distance is central to adjusting how the image appears.

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