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Chapter 34: Problem 34

A converging lens with a focal length of 7.00 \(\mathrm{cm}\) forms an image of a \(4.00-\mathrm{mm}\)-tall real object that is to the left of the lens. The image is 1.30 \(\mathrm{cm}\) tall and erect. Where are the object and image located? Is the image real or virtual?

Short Answer

Expert verified
Object is at 5.35 cm and image at 17.39 cm; image is virtual.

Step by step solution

01

Understanding the Problem

We are dealing with a converging lens having a focal length of \( f = 7.00 \, \text{cm} \). An object of height \( h_o = 4.00 \, \text{mm} \) forms an image that is \( h_i = 1.30 \, \text{cm} \). We need to find the position of the object and image, and determine the nature (real or virtual) of the image.
02

Use Magnification Formula

The magnification \( m \) is given by the formula \( m = \frac{h_i}{h_o} = \frac{-s_i}{s_o} \), where \( h_i \) and \( h_o \) are image and object heights, \( s_i \) is the image distance, and \( s_o \) is the object distance. Calculate \( m \):\[m = \frac{1.30 \text{ cm}}{0.40 \text{ cm}} = 3.25\]Since the image is erect, the magnification is positive, so \( m = 3.25 \).
03

Relate Image and Object Distances

Since the magnification for an erect image is \( m = \frac{s_i}{s_o} \), we have \( \frac{s_i}{s_o} = 3.25 \), hence \( s_i = 3.25 s_o \).
04

Use Lens Formula

The lens formula is \( \frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i} \). Substitute \( s_i = 3.25 s_o \) into the lens formula to find \( s_o \):\[\frac{1}{7.00} = \frac{1}{s_o} + \frac{1}{3.25s_o}\]Combine the terms:\[\frac{1}{7.00} = \frac{4.25}{3.25s_o}\]Solving for \( s_o \) gives: \[s_o = \frac{3.25 \times 7.00}{4.25} = 5.35 \, \text{cm}\]
05

Calculate Image Distance

Using \( s_i = 3.25 s_o \), calculate \( s_i \):\[s_i = 3.25 \times 5.35 \, \text{cm} = 17.39 \, \text{cm}\]
06

Determine Nature of the Image

Since \( s_i \) is positive, the image is formed on the opposite side of the lens to the object, indicating a virtual image.

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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 plays a crucial role when dealing with lenses, especially when you want to determine the position of an image or object. It is expressed as:\[ \frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i} \]where:
  • \( f \) is the focal length of the lens.
  • \( s_o \) is the object distance from the lens.
  • \( s_i \) is the image distance from the lens.
To find the image or object position, you only need two of these three parameters.
Using the formula involves substitution and sometimes rearranging the terms to solve for the unknown. In the case of a converging lens, the focal length and distances are often positive. Hence, the equation allows us to practically locate where the image will form based on the object鈥檚 position.
Remember, using the lens formula correctly can help you determine detailed characteristics of how light travels through a lens and forms an image.
Magnification
Magnification tells us how much larger or smaller an image is compared to the real object. It is defined by the formula:\[ m = \frac{h_i}{h_o} = \frac{-s_i}{s_o} \]where:
  • \( h_i \) is the image height, and \( h_o \) is the object height.
  • \( s_i \) is the image distance, and \( s_o \) is the object distance.
The magnification value can be either positive or negative.
If magnification is positive, this means the image is upright; if it is negative, the image is inverted.
In our example, the image was found to be upright and larger than the object, indicated by a positive magnification of 3.25. Understanding magnification helps determine the nature and orientation of the image formed by the lens.
Real and Virtual Images
Understanding the difference between real and virtual images is key when studying lenses. A real image is formed when light rays converge and actually meet at a point after passing through the lens. This kind of image can be projected on a screen as it is on the side opposite the object.
Virtual images, on the other hand, appear to be located on the same side as the object itself.
  • With converging lenses, if the image distance \( s_i \) is positive, the image is real.
  • If the image distance is negative, it indicates that the image is virtual.
In our case, since the object distance and image distance led to a positive \( s_i \) value, the image was noted as virtual, even though it appeared erect. This distinction is crucial for understanding where the image forms relative to the lens.

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Most popular questions from this chapter

A Glass Rod. Both ends of a glass rod with index of refraction 1.60 are ground and polished to convex hemispherical surfaces. The radius of curvature at the left end is \(6.00 \mathrm{cm},\) and the radius of curvature at the right end is 12.0 \(\mathrm{cm} .\) The length of the rod between vertices is 40.0 \(\mathrm{cm} .\) The object for the surface at the left end is an arrow that lies 23.0 \(\mathrm{cm}\) to the left of the vertex of this surface. The arrow is 1.50 \(\mathrm{mm}\) tall and at right angles to the axis. (a) What constitutes the object for the surface at the right end of the rod? (b) What is the object distance for this surface? (c) Is the object for this surface real or virtual? (d) What is the position of the final image? (e) Is the final image real or virtual? Is it erect or inverted with respect to the original object? (f) What is the height of the final image?

A spherical, concave shaving mirror has a radius of curvature of 32.0 \(\mathrm{cm}\) . (a) What is the magnification of a person's face when it is 12.0 \(\mathrm{cm}\) to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virtual? (c) Draw a principal-ray diagram showing the formation of the image.

The left end of a long glass rod 8.00 \(\mathrm{cm}\) in diameter, with an index of refraction of \(1.60,\) is ground and polished to a convex hemispherical surface with a radius of 4.00 \(\mathrm{cm} .\) An object in the form of an arrow 1.50 \(\mathrm{mm}\) tall, at right angles to the axis of the rod, is located on the axis 24.0 \(\mathrm{cm}\) to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?

A glass rod with a refractive index of 1.55 is ground and polished at both ends to hemispherical surfaces with radii of 6.00 \(\mathrm{cm} .\) When an object is placed on the axis of the rod, 25.0 \(\mathrm{cm}\) to the left of the left-hand end, the final image is formed 65.0 \(\mathrm{cm}\) to the right of the right-hand end. What is the length of the rod measured between the vertices of the two hemispherical surfaces?

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?
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