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Chapter 18: Problem 4

Prove that for a converging lens, for the case \(d_{i}=d_{0}\), that \(d_{i}=d_{0}=2 f\).

Short Answer

Expert verified
For a converging lens where \( d_i = d_o \), both \( d_i \) and \( d_o \) equal \( 2f \).

Step by step solution

01

Recall the Lens Formula

The lens formula is given by \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. For a converging lens, this formula relates these three distances.
02

Set Object and Image Distances Equal

According to the problem, we have \( d_i = d_o \). Let's substitute \( d_i = d_o \) in the lens formula: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_o} = \frac{2}{d_o}. \]
03

Solve for Object Distance

To find \( d_o \), solve the equation from Step 2 for \( d_o \): \[ \frac{1}{f} = \frac{2}{d_o} \] which simplifies to \[ d_o = \frac{2}{f}. \] Hence, \( d_o = 2f \).
04

Verify the Solution for Image Distance

Since \( d_i = d_o \), it follows directly that \( d_i = 2f \). Therefore, \( d_i = d_o = 2f \), as required.

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Key Concepts

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

Understanding Converging Lenses
Converging lenses, often referred to as convex lenses, are thicker in the middle and thinner at the edges. They have a unique property that makes parallel rays of light converge to a single point called the focal point. This is due to the lens's shape, which bends the incoming light rays towards the principal axis of the lens. Converging lenses play a critical role in devices like cameras, glasses, and microscopes where they focus light to create clear images.

Converging lenses transform the way light travels:
  • Light passing through it bends towards the principal axis.
  • The focal point is where the light converges.
  • Images formed by converging lenses can be real or virtual, depending on the object's distance from the lens.
Focal Length and its Importance
The focal length of a lens is the distance from the center of the lens to the focal point. It determines the lens's power to converge or diverge light rays. For a converging lens, the focal length is positive. A shorter focal length indicates a stronger lens that bends light more sharply.

Calculating the focal length aids in lens design and functionality:
  • Short focal lengths lead to larger magnifications.
  • Longer focal lengths provide a wider field of view.
  • It is crucial for determining how the lens forms an image.
Determining Image Distance
Image distance refers to how far the image forms from the lens itself. According to the lens formula, \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), the image distance \(d_i\) changes depending on where the object is placed relative to the lens.

The image distance indicates how and where the image forms:
  • When the object is closer than one focal length, the image appears virtual and upright.
  • At distances greater than the focal length, a real, inverted image forms.
  • Solve for \(d_i\) using the lens equation to find precise image placement.
Understanding Object Distance
Object distance is the measure from an object to the lens itself. It significantly impacts the nature and characteristics of the image formed by the lens.

The concept of object distance is crucial:
  • When an object is placed exactly at twice the focal length, the image formed is real and the same size as the object.
  • Object placement changes the size, type (real or virtual), and orientation (inverted or upright) of the image.
  • Develops a deep understanding of how lenses manipulate light to create images.

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Most popular questions from this chapter

A plane mirror essentially has a radius of curvature of infinity. Using the mirror equation, show that (a) the image of a plane mirror is always virtual, (b) the image is "behind" the mirror the same distance as the object is in front of the mirror, and (c) the image is always upright.

Is our sense of touch a reliable measure of temperature? Explain.

Show that the magnification factor for a mirror or lens \(M=d_{i} / d_{0}\) (sign convention omitted) is the lateral magnification, or the ratio of the height (lateral size) of the image to that of the object. (Hint: Draw a ray diagram.)

Using the thin-lens equation and the magnification factor, show that for a spherical diverging lens the image of a real object is always virtual, upright, and reduced. Does the same apply for a spherical diverging mirror?

Explain what characteristics make convex spherical mirrors applicable for store monitoring and concave spherical mirrors applicable as flashlight reflectors.
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