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Chapter 2: Problem 11

Can an image be larger than the object even though its magnification is negative? Explain.

Short Answer

Expert verified
Yes, an image can be larger than the object even though its magnification is negative. The negativity of the magnification only indicates the orientation (inverted) of the image and does not limit the possibility of the image being larger than the object. The inequality \(I > -O\) verifies the situation where an image can be larger than the object with a negative magnification.

Step by step solution

01

Define magnification

Magnification (m) is defined as the ratio of the size of the image (I) to the size of the object (O), which can be written as: \( m = \frac{I}{O} \) A negative magnification indicates that the image is inverted or opposite in orientation as compared to the object.
02

Explore the possibilities of the magnification being negative

A negative magnification means the image is inverted in comparison to the object's orientation. The equation for magnification, \( m = \frac{I}{O} \), tells us that when magnification is negative: \( m = -\frac{I}{O} \) The sign of m only refers to the orientation of the image, but it doesn't say much about the size of the image. To find out if the image can be larger than the object, let's consider the following inequality: \( -\frac{I}{O} > 1 \)
03

Rearrange the inequality and isolate the image size

To determine if an image can be larger than the object, we need to check if the inequality above is true. If it is true, that means that the magnification is negative, but the image is still larger than the object. Solving the inequality, we isolate the image size (I) and make the object size (O) a constant: \( I > -O \) Since the object size is a positive value (O > 0), the negativity of the magnification doesn't automatically prohibit the image from being larger than the object.
04

Conclusion

Based on our analysis, an image can indeed be larger than the object even though its magnification is negative. The negativity of the magnification only indicates the orientation of the image and does not limit the possibility of the image being larger than the object.

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Key Concepts

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

Image Orientation
When we talk about image orientation in optics, it refers to how an image appears relative to the original object. An image can either be upright or inverted. In optics, a negative magnification signifies that the image is inverted compared to the object. This means that while the object might appear right-side-up, the image will be upside-down. The orientation is crucial in applications where the direction matters, such as in certain optical devices and experiments.
Negative magnification can be slightly confusing at first:
  • "Negative" doesn't mean that the size is less or "below zero."
  • It simply shows that the image flips in orientation compared to the original object.
Learning this helps us understand why optical devices like telescopes and microscopes sometimes show inverted images.
Image Size
Image size in optics is determined by magnification, which is the ratio of the image dimensions compared to the object dimensions. The equation for magnification can be represented as \( m = \frac{I}{O} \), where \(I\) is the image size and \(O\) is the object size.
Understanding this relation means:
  • If \( m = 1 \), the image size is equal to the object size.
  • If \( m > 1 \), the image is larger than the object.
  • If \( m < 1 \), the image is smaller than the object.
However, in the case of negative magnification, though it doesn't directly affect size, it tells us that the image is inverted (as discussed in Image Orientation).
Object Size
The object size is crucial as it serves as a reference point for determining magnification and, subsequently, image size. Generally, in optics problems, the object is considered as a real and physical item whose dimensions we use to measure and calculate image properties.
In the magnification formula, \( m = \frac{I}{O} \), the object size \( O \) can change the resulting magnification factor:
  • A larger object with a fixed image size will result in a smaller magnification value.
  • Conversely, a smaller object size with a similar image size will yield a larger magnification.
In most optical exercises, understanding the constant nature of an object鈥檚 size allows us to focus on other variables affecting the image produced.
Inversion of Image
Inversion in optics refers to situations where an image is flipped upside down relative to the object. This often occurs in lens and mirror systems where the image might be projected or formed on the opposite side of the optical axis.
The inversion results from light bending and the geometry of lenses or mirrors. Key points to remember about inversion include:
  • An inverted image is still valid and often used in projections, like in cameras.
  • The inversion indicator, a negative magnification, clarifies that the image orientation is different from that of the object.
Understanding inversion is essential because it helps interpret what we see in optical devices, ensuring we don't misinterpret the actual arrangement of elements just because the image is flipped.

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Most popular questions from this chapter

Find the focal length of a meniscus lens with \(R_{1}=20 \mathrm{cm}\) and \(R_{2}=15 \mathrm{cm} .\) Assume that the index of refraction of the lens is \(1.5 .\)

How far from the lens must the film in a camera be, if the lens has a 35.0 -mm focal length and is being used to photograph a flower \(75.0 \mathrm{cm}\) away? Explicitly show how you follow the steps in the Problem-Solving Strategy: Lenses.

(a) What magnification is produced by a \(0.150 \mathrm{cm}-\) focal length microscope objective that is \(0.155 \mathrm{cm}\) from the object being viewed? (b) What is the overall magnification if an \(8 \times\) eyepiece (one that produces an angular magnification of 8.00 ) is used?

. \(\mathrm{A} 3 \mathrm{cm}\) tall object is placed \(5 \mathrm{cm}\) in front of a convex mirror of radius of curvature \(20 \mathrm{cm} .\) Where is the image formed? How tall is the image? What is the orientation of the image?

An object is located in air \(30 \mathrm{cm}\) from the vertex of a convex surface made of glass with a radius of curvature 80 cm. Where does the image by refraction form and what is its magnification?
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