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Chapter 24: Problem 31

\(\cdot\) You are standing in front of a lens that projects an image of you onto a wall 1.80 \(\mathrm{m}\) on the other side of the lens. This image is three times your height. (a) How far are you from the lens? (b) Is your image erect or inverted? (c) What is the focal length of the lens? Is the lens converging or diverging?

Short Answer

Expert verified
(a) 0.6 m behind the lens, (b) Image is inverted, (c) Focal length is -0.9 m; lens is diverging.

Step by step solution

01

Understanding the Problem

You're asked to find the distance from the object (you) to the lens, determine if the image is erect or inverted, and find the focal length of the lens, knowing that the image is three times your height and it's projected onto a wall 1.80 m from the lens.
02

Calculate Magnification

Magnification is calculated by the ratio of the height of the image to the height of the object. Here, magnification, \( m \), is 3 (since the image is three times your height). Magnification is also equal to the negative ratio of the image distance \( d_i \) to the object distance \( d_o \): \( m = -\frac{d_i}{d_o} \). Thus, \( 3 = -\frac{1.8 \, \text{m}}{d_o} \) which gives \( d_o = -0.6 \, \text{m} \).
03

Determine Image Orientation

Since the magnification is positive and greater than 1, the image is inverted. In optical systems, a positive magnification indicates an inverted image.
04

Use Lens Formula

The lens formula relates object distance \( d_o \), image distance \( d_i \), and focal length \( f \): \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Substituting \( d_o = -0.6 \, \text{m} \) and \( d_i = 1.8 \, \text{m} \), calculate: \( \frac{1}{f} = \frac{1}{-0.6} + \frac{1}{1.8} \). Solve this equation to find \( f \).
05

Calculate the Focal Length

Simplify the equation: \( \frac{1}{f} = \frac{1}{-0.6} + \frac{1}{1.8} = -\frac{1}{0.6} + \frac{1}{1.8} = -1.67 + 0.556 \approx -1.11 \). So \( f \approx -0.9 \, \text{m} \).
06

Determine Lens Type

The focal length is negative, indicating that the lens is diverging. Diverging lenses have negative focal lengths.

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Key Concepts

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

Image Magnification
When we talk about image magnification in optics, we are referring to how much larger or smaller the image of an object appears when viewed through a lens. Magnification is calculated as the ratio of the image height to the object height.
In this exercise, the image magnification is 3. This means the image is 3 times the height of the actual object, which is you in this case.
  • A magnification greater than 1 indicates the image is larger than the object.
  • A negative magnification suggests that the image is inverted compared to the object.
In practice, you also calculate magnification using the distances from the lens:
The formula is: \[ m = -\frac{d_i}{d_o} \] where:
  • \( m \) is the magnification,
  • \( d_i \) is the image distance from the lens,
  • \( d_o \) is the object distance from the lens.
A positive magnification here indicates the image is inverted.
Diverging Lens
A diverging lens, also known as a concave lens, spreads light rays apart. Unlike converging lenses, which bring light rays together, diverging lenses cause the light rays that pass through them to spread away from a common point.
The most telling characteristic of a diverging lens is its focal length, which is always negative.
Here are some general features:
  • Images formed by a single diverging lens are always virtual. This means they cannot be projected onto a screen.
  • The images are typically smaller than the object (minified) and upright.
  • The image found using the exercise was negative, indicating a diverging lens was used without explicitly visualizing it.
  • Diverging lenses are often used in situations where you want to correct vision, disperse light, or switch projection fields.
In the exercise, the lens was diverging, indicated by a negative focal length found in the solution.
Focal Length Calculation
Calculating the focal length of a lens is crucial in understanding how it will bend light. The focal length is the distance between the lens and the point where it focuses parallel rays of light. The lens formula guides us in finding it:
\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] This formula connects:
  • \( f \) - the focal length of the lens,
  • \( d_o \) - the object distance, calculated from the object to the lens,
  • \( d_i \) - the image distance, from the lens to where the image forms on the other side.
By rearranging and solving this equation, you determine the focal length.
In the example provided, you substitute the values: \( d_o = -0.6 \, \text{m} \) and \( d_i = 1.8 \, \text{m} \). The result, \( f \approx -0.9 \, \text{m} \), indicates a diverging lens because of its negative value.Knowing the focal length allows better understanding and prediction of how the lens will perform or be utilized.

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Most popular questions from this chapter

Focus of the eye. The cornea of the eye has a radius of curvature of approximately \(0.50 \mathrm{cm},\) and the aqueous humor bbehind it has an index of refraction of \(1.35 .\) The thickness of the cornea itself is small enough that we shall neglect it. The depth of a typical human eye is around 25 5 \(\mathrm{mm}\) . (a) What would have to be the radius of curvature of the cornea so that it alone would focus the image of a distant mountain on the retina, which is at the back of the eye opposite the cornea? (b) If the cornea focused the mountain correctly on the retina as described in part \((a),\) would it also focus the text from a computer screen on the retina if that screen were 25 \(\mathrm{cm}\) in front of the eye? If not, where would it focus that text, in front of or behind the retina? (c) Given that the cornea has a radius of curvature of about \(5.0 \mathrm{mm},\) where does it actually focus the mountain? Is this in front of or behind the retina? Does this help you see why the eye needs help from a lens to complete the task of focusing?

\(\cdot\) The front, convex, surface of a lens made for eyeglasses has a radius of curvature of \(11.8 \mathrm{cm},\) and the back, concave, surface has a radius of curvature of 6.80 \(\mathrm{cm} .\) The index of refraction of the plastic lens material is 1.67 . Calculate the local length of the lens.

Focus of the eye. The cornea of the eye has a radius of curvature of approximately \(0.50 \mathrm{~cm},\) and the aqueous humor behind it has an index of refraction of \(1.35 .\) The thickness of the cornea itself is small enough that we shall neglect it. The depth of a typical human eye is around \(25 \mathrm{~mm}\). (a) What would have to be the radius of curvature of the cornea so that it alone would focus the image of a distant mountain on the retina, which is at the back of the eye opposite the cornea? (b) If the cornea focused the mountain correctly on the retina as described in part (a), would it also focus the text from a computer screen on the retina if that screen were \(25 \mathrm{~cm}\) in front of the eye? If not, where would it focus that text, in front of or behind the retina? (c) Given that the cornea has a radius of curvature of about \(5.0 \mathrm{~mm},\) where does it actually focus the mountain? Is this in front of or behind the retina? Does this help you see why the eye needs help from a lens to complete the task of focusing?

\(\bullet\) Examining your image in a convex mirror whose radius of curvature is \(25.0 \mathrm{cm},\) you stand with the tip of your nose 10.0 \(\mathrm{cm}\) from the surface of the mirror. (a) Where is the image of your nose located? What is its magnification? (b) Your ear is 10.0 \(\mathrm{cm}\) behind the tip of your nose; where is the image of your ear located, and what is its magnification? Do your answers suggest reasons for your strange appearance in a convex mirror?

\(\bullet\) A converging lens with a focal length of 12.0 \(\mathrm{cm}\) forms virtual image 8.00 \(\mathrm{mm}\) tall, 17.0 \(\mathrm{cm}\) to the right of the lens Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side o opposite sides of the lens?
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