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A 1.0 -cm-tall candle flame is \(60 \mathrm{cm}\) from a lens with a focal length of \(20 \mathrm{cm}\). What are the image distance and the height of the flame's image?

Short Answer

Expert verified
The image's distance is 30 cm from the lens and is formed on the opposite side of the lens, and the height of the flame's image is 0.5 cm and is inverted.

Step by step solution

01

Calculate image distance

We can calculate the image's distance using the lens formula: \(1/f = 1/v - 1/u\). We know that the focal length \(f=20\) cm and the object distance \(u=-60\) cm. Substituting these values into the formula, we get \(\frac{1}{20} = \frac{1}{v} - \frac{-1}{60}\). Solve for \(v\) to get the image distance.
02

Solve for \(v\)

First, simplify by performing addition: \(\frac{1}{20} = \frac{1}{v} + \frac{1}{60}\). You can then rewrite the equation to \(\frac{1}{v} = \frac{1}{20} - \frac{1}{60}\). Simplifying the right hand side we get \(\frac{1}{v} = \frac{2}{60}\). Therefore, \(v = 60/2 = 30cm\). This indicates that the image is formed 30cm on the opposite side of the lens.
03

Calculate height of the flame's image

We calculate the height of the flame's image (\(h'\)) using the magnification formula (\(h'/h = -v/u\)). We have \(h=1cm\), \(v=30cm\) and \(u=-60cm\). Substituting these values, we get \(h'/1 = -30/-60\). This simplifies to \(h' = 1/2 = 0.5cm\). This negative sign indicates that the image is inverted.

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Key Concepts

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

Lens Formula
In the realm of geometrical optics, the lens formula connects the focal length, object distance, and image distance in a simple equation. It is represented as \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \). Here, \( f \) stands for the focal length of the lens, \( v \) is the image distance from the lens, and \( u \) is the object distance from the lens. This equation is crucial for determining where an image will form when looking through a lens.
This formula applies to both concave and convex lenses but be sure to pay attention to the sign conventions, which dictate how distances are measured relative to the lens. For example, the object distance \( u \) is usually negative when the object is in front of the lens, which means the formula accounts for directions. Knowing this, you can use the lens formula to find unknown values when given the other two.
Image Distance
Image distance, represented as \( v \) in the lens formula, is the distance from the lens to where the image appears. Understanding how this value is calculated helps in predicting where the real or virtual image will form.
  • Real images, which can be projected onto a screen, are found on the opposite side of the lens from the object and have positive image distances.
  • Virtual images appear on the same side as the object and have negative image distances.

In the example problem, the image distance was calculated using the lens formula as \( v = 30 \) cm, showing that the image is a real one, as it formed on the opposite side of the object.
Magnification
Magnification relates to how much larger or smaller an image appears relative to the actual object. The formula for magnification \( m \) is given by \( m = \frac{h'}{h} = -\frac{v}{u} \), where \( h' \) is the height of the image and \( h \) is the height of the object.
This equation tells us two main things:
  • If the magnification is positive, the image is upright relative to the object.
  • If it is negative, the image is inverted.

In our exercise, the magnification was calculated as \(-\frac{30}{-60} = \frac{1}{2}\), meaning the image height is half that of the object, and it is inverted.
Focal Length
The focal length \( f \) of a lens is the distance from the lens to its focal point, where parallel rays of light meet. It is a fundamental property that influences how the lens bends light.
Depending on the type of lens, the focal length can be positive or negative:
  • For converging lenses, like convex lenses, the focal length is positive.
  • For diverging lenses, like concave lenses, it is negative.

The given problem used a focal length of \( 20 \) cm, indicating a converging lens. This focal length helped determine where the image would form when combined with the object distance in the lens formula.

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Most popular questions from this chapter

\(A 1.0\) -cm-tall object is \(20 \mathrm{cm}\) in front of a concave mirror that has a \(60 \mathrm{cm}\) focal length. Calculate the position and height of the image. State whether the image is in front of or behind the mirror, and whether the image is upright or inverted.

You're helping with an experiment in which a vertical cylinder will rotate about its axis by a very small angle. You need to devise a way to measure this angle. You decide to use what is called an optical lever. You begin by mounting a small mirror on top of the cylinder. A laser \(5.0 \mathrm{m}\) away shoots a laser beam at the mirror. Before the experiment starts, the mirror is adjusted to reflect the laser beam directly back to the laser. Later, you measure that the reflected laser beam, when it returns to the laser, has becn deflected sideways by 2.0 mm. Through how many degrees has the cylinder rotated?

An object is \(6 \mathrm{cm}\) in front of a converging 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?

a. Estimate the diameter of your eycball. b. Bring this page up to the closest distance at which the text is sharp - not the closest at which you can still read it, but the closest at which the letters remain sharp. If you wear glasses or contact lenses, leave them on. This distance is called the nearpoint of your (possibly corrected) eye. Measure it. c. Bstimate the effective focal length of your eye. The effective focal length includes the focusing due to the lens, the curvature of the cornea, and any corrections you wear. Ignore the effects of the fluid in your eye.

A 1.0 -cm-tall object is \(75 \mathrm{cm}\) in front of a converging lens that \(\quad 7=\) has a \(30 \mathrm{cm}\) focal length. a. Use ray tracing to find the position and height of the image. To do this accurately, use a ruler or paper with a grid. Determine the image distance and image height by making measurements on your diagram. b. Calculate the image position and height. Compare with your ray-tracing answers in part a.

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