/*! 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
The smallest distance is 4f, confirmed by the graph.

Step by step solution

01

Understand the Lens Formula

The lens formula is given by the equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length of the lens, \( d_o \) is the distance of the object from the lens, and \( d_i \) is the distance of the image from the lens. For real images formed by converging lenses, both \( d_o \) and \( d_i \) are positive.
02

Express Total Distance

The total distance between the object and the image is given by \( D = d_o + d_i \). Our goal is to find the minimum value of \( D \).
03

Combine Equations

Substituting \( d_i = \frac{d_o f}{d_o - f} \) from the lens formula into \( D = d_o + d_i \), we get \( D = d_o + \frac{d_o f}{d_o - f} \).
04

Differentiate and Solve for Minimum

To find the minimum value of \( D \), take the derivative of \( D \) with respect to \( d_o \), and set it to zero. This gives us \( D' = 1 - \frac{f^2}{(d_o - f)^2} = 0 \). Solving this, we find \( d_o = 2f \) and \( d_i = 2f \), so the smallest distance \( D = d_o + d_i = 4f \).
05

Graph the Function

Graph the function \( D = d_o + \frac{d_o f}{d_o - f} \) with respect to \( d_o \) and observe its behavior. The graph will show a minimum distance, confirming the analytical result found earlier.
06

Interpret the Graph

Upon graphing, the function \( D(d_o) \) has a vertex at \( d_o = 2f \), showing that the smallest distance occurs here, which agrees with the calculation from Step 4.

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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 crucial for understanding how lenses form images. In optics, it is expressed as: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]where:
  • \( f \) is the focal length of the lens.
  • \( d_o \) is the distance from the object to the lens.
  • \( d_i \) is the distance from the image to the lens.
This equation connects these three quantities, allowing you to calculate the position of the image if the object position and focal length are known. For converging lenses, which we use here, both \( d_o \) and \( d_i \) are positive values.
The lens formula is derived from the principles of geometric optics and is applicable for both real and virtual images. It is essential when determining image characteristics, such as position and nature. By substituting known values into this formula, we can calculate the unknowns for specific optical scenarios.
Focal Length
Focal length is a vital attribute of a lens that tells how strongly the lens converges or diverges light. It is the distance from the lens to the point where parallel rays of light converge to a point (focus).
An important concept to remember about focal length is:- Shorter focal lengths result in stronger convergence of light waves, producing a larger image.- Longer focal lengths mean a weaker convergence, producing a smaller image.
In our exercise, knowing the focal length \( f \) helps us determine the minimum distance between the object and its real image using the lens formula. When the object is placed at a distance of \( 2f \) from the lens, the image also forms at \( 2f \) on the opposite side.
Understanding focal length is essential for applications in photography, vision correction, and even in eyeglasses. It significantly influences the magnification and field of view of optical instruments.
Real Image
A real image is formed whenever light rays actually converge at a point. Unlike virtual images, which seem to be located at a point from which the rays appear to diverge, real images can be projected onto a screen.
Real images are typically formed by converging lenses or mirrors. In the context of the exercise, we were concerned with a converging lens creating a real image at \( d_i = 2f \). This means the light rays from the object get bent by the lens and meet at this point, creating an actual image.Key features of real images include:
  • They can be captured on a film or a sensor, like in cameras.
  • They are usually inverted, meaning they appear upside-down relative to the object.
  • They are formed when the object is placed beyond the focal point of the lens.
When working through our problem, identifying the conditions under which a lens can form a real image helps predict where this image will appear and how it can be used in practical applications.

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Most popular questions from this chapter

In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the bending occurs at the cornea, whose vertex is 2.60 cm from the retina. What should be the radius of curvature of the cornea such that the image of an object 40.0 cm from the cornea's vertex is focused on the retina?

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

A screen is placed a distance \(d\) to the right of an object. A converging lens with focal length \(f\) is placed between the object and the screen. In terms of \(f\), what is the smallest value \(d\) can have for an image to be in focus on the screen?

The eyepiece of a refracting telescope (see Fig. 34.53) has a focal length of 9.00 cm. The distance between objective and eyepiece is 1.20 m, and the final image is at infinity. What is the angular magnification of the telescope?

The cornea behaves as a thin lens of focal length approximately 1.8 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 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 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.)

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