/*! 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 105 (a) For a lens with focal length... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) For a lens with focal length \(f,\) find the smallest distance possible between the object and its real image. (b) Graph the distance between the object and the real image as a function of the distance of the object from the lens. Does your graph agree with the result you found in part (a)?

Short Answer

Expert verified
Hence, the smallest distance between the object and its real image for a lens of given focal length \(f\) is \(2f\). It was verified graphically by plotting the relationship between object distance \(p\) and image distance \(q\) which indeed agreed with the predicted minimum value from the mathematical derivation, hence confirming the solution.

Step by step solution

01

Identify the Relevant Equation

The equation that links the object distance (\(p\)), the image distance (\(q\)), and the focal length (\(f\)) of the lens is known as the thin lens equation: \(1/f = 1/p + 1/q\)
02

Derive the expression

Rearrange the lens equation to show the q in terms of p and f: \(q = p*f / (p - f)\). Since the condition is that the object height and the image height us equal, h = h'. So, the magnification m = -q/p = -1.
03

Find Minimum Distance

According to the magnification, -1 = -q/p, q = p. Substitute q = p to the equation: \(q = p*f / (p - f) = p\). Thus, p = 2f. This is the smallest distance possible between the object and its real image. The factor is double because the image distance is negative for real images.
04

Graphing the Relationship

Plot the graph of q versus p with p on the x-axis and q on the y-axis, where q = f*p / p - f. This will yield a hyperbolic graph. Confirming that when p = 2f, q = -2f is the point of the minimum distance between the object and its real image.

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Key Concepts

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

Focal Length
The focal length of a lens (denoted as \(f\)) is a critical measure that describes how strongly the lens converges or diverges light. This is the distance over which initially collimated rays are brought to a focus. For converging lenses, the focal point is where light rays parallel to the principal axis converge after passing through the lens.

This property determines the lens's effectiveness in focusing light and is a key parameter in the thin lens equation, given by \(\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\), where \(p\) is the object distance and \(q\) is the image distance.

When using lenses in photography or optics, a shorter focal length means a wider field of view and a larger image produced at a given distance. Conversely, a longer focal length means a narrower field of view but greater magnification. The focal length is an intrinsic property of each lens, determined by the curvature of the lens surfaces and the refractive index of the lens material.
Object Distance
Object distance \(p\) is the distance from the object to the lens. This is one of the main variables in calculating image formation using the thin lens equation. A crucial part of any optics experiment is to measure this distance accurately, as it directly impacts the resulting image.

The relationship between object distance, image distance \(q\), and focal length \(f\) in the lens equation is shown by \(\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\). For real images, the object distance is always positive.

As the object is moved further from the lens, the object distance increases, affecting factors such as magnification and the position of the image formed. It is important to understand how changing \(p\) while keeping \(f\) constant can affect \(q\) and thereby the image we see through the lens.
Image Distance
Image distance \(q\) refers to the distance from the lens to the image formed. It is another critical component in the thin lens equation, alongside object distance \(p\) and focal length \(f\). The resulting value can determine whether the image is formed in front of or behind the lens.

A positive \(q\) denotes real images, which can be projected on a screen. Meanwhile, a negative \(q\) indicates virtual images, which are seen through the lens but cannot be projected.

When working with the thin lens equation \(\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\), adjusting \(p\) will influence \(q\). In scenarios like our exercise, when \(p = 2f\), we find \(q = -2f\), which verifies the smallest possible distance between the object and the lens for a real image.
Magnification
Magnification is a measure of how much larger or smaller the image is compared to the object itself. In terms of lenses, it’s given by the ratio of image distance \(q\) to object distance \(p\), and also can be expressed as \(m = \frac{-q}{p}\). Magnification provides insight into how the size of the image changes as you change \(p\) or \(q\).

In our exercise, the magnification becomes \(-1\), meaning the image and object are of equal size but inverted. This particular situation arises when \(q = p\), emphasizing the instance of minimum distance between object and image when \(p = 2f\).

Understanding magnification is crucial in applications like photography and microscopy, where detail and clarity depend on being able to control and predict image size relative to the original object.
Graphing in Optics
Graphing is an essential tool in optics for visualizing relationships like those described by the thin lens equation. In our exercise, graphing the relationship between object distance \(p\) and image distance \(q\) can visualize how the image forms relative to changes in \(p\).

A common format is plotting \(q\) as a function of \(p\) on a graph, resulting in a hyperbolic curve, thanks to the relationship \(q = \frac{fp}{p - f}\). This curve highlights key phenomena, like the minimum distance configurations where \(p = 2f\) results in \(q = -2f\).

Such graphing helps in verifying theoretical calculations and understanding practical implications, like predicting where an image will form for a given object distance, or identifying scenarios where image and object distances result in equimagnification for precision tasks.

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Most popular questions from this chapter

An object is placed \(22.0 \mathrm{~cm}\) from a screen. (a) At what two points between object and screen may a converging lens with a \(3.00 \mathrm{~cm}\) focal length be placed to obtain an image on the screen? (b) What is the magnification of the image for each position of the lens?

A light bulb is \(3.00 \mathrm{~m}\) from a wall. You are to use a concave mirror to project an image of the bulb on the wall, with the image 3.50 times the size of the object. How far should the mirror be from the wall? What should its radius of curvature be?

Figure P34.99 shows a simple version of a zoom lens. The converging lens has focal length \(f_{1}\) and the diverging lens has focal length \(f_{2}=-\left|f_{2}\right| .\) The two lenses are separated by a variable distance \(d\) that is always less than \(f_{1}\). Also, the magnitude of the focal length of the diverging lens satisfies the inequality \(\left|f_{2}\right|>\left(f_{1}-d\right) .\) To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius \(r_{0}\) entering the converging lens. (a) Show that the radius of the ray bundle decreases to \(r_{0}^{\prime}=r_{0}\left(f_{1}-d\right) / f_{1}\) at the point that it enters the diverging lens. (b) Show that the final image \(I^{\prime}\) is formed a distance \(s_{2}^{\prime}=\left|f_{2}\right|\left(f_{1}-d\right) /\left(\left|f_{2}\right|-f_{1}+d\right)\) to the right of the diverging lens. (c) If the rays that emerge from the diverging lens and reach the final image point are extended backward to the left of the diverging lens, they will eventually expand to the original radius \(r_{0}\) at some point \(Q .\) The distance from the final image \(I^{\prime}\) to the point \(Q\) is the effective focal length \(f\) of the lens combination; if the combination were replaced by a single lens of focal length \(f\) placed at \(Q\), parallel rays would still be brought to a focus at \(I^{\prime} .\) Show that the effective focal length is given by \(f=f_{1}\left|f_{2}\right| /\left(\left|f_{2}\right|-f_{1}+d\right) .\) (d) If \(f_{1}=12.0 \mathrm{~cm}, f_{2}=-18.0 \mathrm{~cm},\) and the separation \(d\) is adjustable between 0 and \(4.0 \mathrm{~cm}\), find the maximum and minimum focal lengths of the combination. What value of \(d\) gives \(f=30.0 \mathrm{~cm} ?\)

You hold a spherical salad bowl \(60 \mathrm{~cm}\) in front of your face with the bottom of the bowl facing you. The bowl is made of polished metal with a \(35 \mathrm{~cm}\) radius of curvature. (a) Where is the image of your \(5.0-\mathrm{cm}\) -tall nose located? (b) What are the image's size, orientation, and nature (real or virtual)?

A concave mirror has a radius of curvature of \(34.0 \mathrm{~cm}\). (a) What is its focal length? (b) If the mirror is immersed in water (refractive index 1.33 ), what is its focal length?

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