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An object is \(15 \mathrm{cm}\) in front of a diverging lens with a focal length of \(-10 \mathrm{cm} .\) Use ray tracing to determine the location of the image. Is the image upright or inverted? Is it real or virtual?

Short Answer

Expert verified
The image is located at a distance of -6 cm from the lens, on the same side as the object. This indicates that it is a virtual image. It is also upright and reduced in size.

Step by step solution

01

Understanding The Problem Statement

The object is placed 15 cm in front of a diverging lens that has a focal length of -10 cm. Diverging lenses always produce virtual, upright, and reduced images.
02

Applying The Lens Formula

The lens formula 1/f = 1/o + 1/i allows us to find the image distance (i). Substituting the given values:- focal length (f) = -10 cm and object distance (o) = 15 cm, we will find the image distance via the formula.
03

Calculation

After substituting the values into the lens formula, we get: 1/i = 1/-10 cm - 1/15 cm. Solving the above equation gives the image distance i = -6 cm.
04

Interpret the Results

Since the image distance is negative, it means that the image is formed on the same side as the object placed, which is a characteristic of virtual images in lens. As it's formed by a diverging lens, the image is upright. The absolute value of image distance being less than the object distance suggests that it's reduced in size.

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Key Concepts

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

Lens Formula
Diverging lenses are fascinating because they bend light rays outward. To understand how images are formed, we use the lens formula:\[\frac{1}{f} = \frac{1}{o} + \frac{1}{i}\]Here, \( f \) is the focal length of the lens, \( o \) is the object distance, and \( i \) is the image distance. In the case of a diverging lens, the focal length \( f \) is always negative.

To solve for the image distance \( i \), we rearrange the formula. Substituting the values provided in the exercise: the focal length \( f = -10 \text{ cm} \) and the object distance \( o = 15 \text{ cm} \), into the formula, we calculate as follows:\[\frac{1}{i} = \frac{1}{-10 \text{ cm}} - \frac{1}{15 \text{ cm}}\]Solving this equation results in \( i = -6 \text{ cm} \). The negative sign indicates that the image is virtual, meaning it forms on the same side as the object.

Remember, using the lens formula helps us to mathematically predict where and how an image will form. This foundational principle is crucial for optics.
Virtual Image
A virtual image is quite different from what we might imagine when thinking about images in general. Unlike real images, which can be projected onto a screen, virtual images can't be captured easily because they don't actually exist at a specific location in space. Instead, they appear to be coming from a place behind the lens.

In the context of diverging lenses, virtual images have a few key characteristics:
  • They are always formed on the same side as the object. This opposite placement from real images is a fundamental characteristic.
  • The image is upright compared to the object.
  • It is smaller and reduced in size relative to the original object.
For example, if you use a magnifying glass (a common diverging lens), and place it close to an object, you will notice that the image appears upright and smaller than the actual object. Understanding virtual images helps us comprehend how lenses alter our perception of reality.
Ray Tracing
Ray tracing is a powerful tool that allows us to visually predict where an image will form through a lens. This method involves drawing a few specific rays from the object through the lens to determine the path light travels and, therefore, the image's location.

Here is how ray tracing works for diverging lenses:
  • **First Ray:** Draw a ray parallel to the principal axis. Upon reaching the lens, this ray will refract and diverge away as if it came from the focal point on the same side as the object.
  • **Second Ray:** Aim this ray through the center of the lens. This ray will continue its path in a straight line, unchanged by the lens.
The intersection of the backward extensions of these divergent rays gives the position of the virtual image. In our exercise's scenario, these rays help confirm that the image is located at -6 cm, consistent with the lens formula calculation.

Ray tracing is critical as it offers a visual way to understand and validate the abstract calculations made with equations. This approach helps build a deeper understanding of how light and lenses interact.

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Most popular questions from this chapter

There is an interesting optical effect you have likely noticed while driving along a flat stretch of road on a sunny day. A small, distant dip in the road appears to be filled with water. You may even see the reflection of an oncoming car. But, as you get closer, you find no puddle of water after all; the shimmering surface vanishes, and you see nothing but empty road. It was only a mirage, the name for this phenomenon. The mirage is due to the different index of refraction of hot and cool air. The actual bending of the light rays that produces the mirage is subtle, but we can make a simple model as follows. When air is heated, its density decreases and so does its index of refraction. Consequently, a pocket of hot air in a dip in a road has a lower index of refraction than the cooler air above it. Incident light rays with large angles of incidence (that is, nearly parallel to the road, as shown in Figure P18.83) experience total internal reflection. The mirage that you see is due to this reflection. As you get nearer, the angle goes below the critical angle and there is no more total internal reflection; the 鈥渨ater鈥 disappears! Which of these changes would allow you to get closer to the mirage before it vanishes? A. Making the pocket of hot air nearer in temperature to the air above it B. Looking for the mirage on a windy day, which mixes the air layers C. Increasing the difference in temperature between the pocket of hot air and the air above it D. Looking at it from a greater height above the ground

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