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Chapter 33: Problem 6

(II) A stamp collector uses a converging lens with focal length \(28 \mathrm{~cm}\) to view a stamp \(18 \mathrm{~cm}\) in front of the lens. (a) Where is the image located? (b) What is the magnification?

Short Answer

Expert verified
(a) The image is located 50.4 cm in front of the lens. (b) The magnification is -2.8.

Step by step solution

01

Identify Known Values

We are given that the focal length \( f \) of the lens is \( 28 \text{ cm}\) and the object distance \( d_o \) is \( 18 \text{ cm}\). We need to find the image distance \( d_i \) and the magnification \( m \).
02

Use the Lens Formula

The lens formula relates the focal length \( f \), the object distance \( d_o \), and the image distance \( d_i \): \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Substituting the known values, we have:\[ \frac{1}{28} = \frac{1}{18} + \frac{1}{d_i} \].
03

Solve for Image Distance

Rearrange the equation from Step 2 to find \( d_i \):\[ \frac{1}{d_i} = \frac{1}{28} - \frac{1}{18} \].Calculate the right-hand side:\[ \frac{1}{d_i} = \frac{18 - 28}{504} = -\frac{10}{504} = -\frac{5}{252} \].Therefore, \( d_i = -252/5 \approx -50.4 \text{ cm} \).
04

Determine the Nature of the Image

Since \( d_i \) is negative, the image formed is a virtual image located \( 50.4 \text{ cm} \) on the same side of the lens as the object.
05

Calculate the Magnification

The magnification \( m \) of a lens is given by \( m = \frac{d_i}{d_o} \). Using the values found:\[ m = \frac{-50.4}{18} \approx -2.8 \].
06

Interpret the Magnification

The magnification is \(-2.8\), indicating that the image is inverted relative to the object and is 2.8 times larger than the object.

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Key Concepts

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

Converging Lens
A converging lens, also known as a convex lens, is a type of lens that bends incoming parallel light rays towards a common point. This point is known as the focal point. Converging lenses are thicker in the middle than at the edges. They are widely used in optical devices such as cameras, magnifying glasses, and eyeglasses.
The lens focuses light to form images. These images can be real or virtual, depending on the position of the object in relation to the focal length. A converging lens is ideal for tasks requiring magnification, such as stamp collecting or reading small print.
Focal Length
The focal length of a lens is a crucial parameter in determining how the lens bends light. It is the distance between the center of the lens and the focal point. In simple terms, it helps to indicate how strongly the lens can converge or diverge light.
In mathematical terms, the focal length is denoted by the symbol \( f \). When calculating image distance or magnification, knowing the focal length is essential. For instance, in our exercise, the focal length is given as 28 cm, which tells us how light passing through the lens will converge.
A shorter focal length indicates stronger converging power, and vice versa. It's instrumental in lens formulas and determining the characteristics of the formed image.
Image Distance
Image distance, represented by \( d_i \), is the distance from the lens to the location of the formed image. Determining the image distance is key to understanding where the image will appear relative to the lens. This is achievable by utilizing the lens formula: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]
In our exercise, calculating image distance, we found it to be approximately -50.4 cm. The negative sign indicates that the image is virtual, meaning it appears on the same side as the object.
Understanding image distance helps ascertain key properties of the image, like whether it's real or virtual, and its orientation regarding the object.
Magnification
Magnification, in the context of lenses, refers to how much larger or smaller an image appears compared to the object itself. It is represented by the symbol \( m \) and can be calculated using the formula: \[ m = \frac{d_i}{d_o} \]
In the exercise, the magnification is calculated to be approximately -2.8. This value means the image is 2.8 times larger than the object. The negative sign indicates that the image is inverted (upside down in relation to the object). Such detail is crucial when using lenses to magnify objects, like stamps, making tiny details easier to view and analyze.
  • A magnification greater than 1 denotes a larger image.
  • A magnification less than 1 indicates a smaller image.
  • A negative magnification signals an inverted image.

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Most popular questions from this chapter

(II) A lighted candle is placed 36 \(\mathrm{cm}\) in front of a converging lens of focal length \(f_{1}=13 \mathrm{cm},\) which in turn is 56 \(\mathrm{cm}\) in front of another converging lens of focal length \(f_{2}=16 \mathrm{cm}\) (see Fig. \(47 ) .(a)\) Draw a ray diagram and estimate the location and the relative size of the final image. (b) Calculate the position and relative size of the final image.

(II) An object is located \(1.50 \mathrm{~m}\) from an \(8.0-\mathrm{D}\) lens. By how much does the image move if the object is moved \((\) a) \(0.90 \mathrm{~m}\) closer to the lens, and ( \(b\) ) \(0.90 \mathrm{~m}\) farther from the lens?

(II) A planoconvex lens (Fig. 33-2a) has one flat surface and the other has \(R=15.3 \mathrm{~cm} .\) This lens is used to view a red and yellow object which is \(66.0 \mathrm{~cm}\) away from the lens. The index of refraction of the glass is 1.5106 for red light and 1.5226 for yellow light. What are the locations of the red and yellow images formed by the lens?

(III) In the "magnification" method, the focal length \(f\) of a converging lens is found by placing an object of known size at various locations in front of the lens and measuring the resulting real-image distances \(d_{i}\) and their associated magnifications \(m\) (minus sign indicates that image is inverted). The data taken in such an experiment are given here:\(\begin{array}{rrrrrr}{d_{\mathrm{i}}(\mathrm{cm})} & {20} & {25} & {30} & {35} & {40} \\ {m} & {-0.43} & {-0.79} & {-1.14} & {-1.50} & {-1.89}\end{array}\) (a) Show analytically that a graph of \(m\) vs. \(d_{\text { i should }}\) produce a straight line. What are the theoretically expected values for the slope and the \(y\) -intercept of this line? [Hint: \(d_{\mathrm{o}}\) is not constant.] \((b)\) Using the data above, graph \(m\) vs. \(d_{\mathrm{i}}\) and show that a straight line does indeed result. Use the slope of this line to determine the focal length of the lens. Does the \(y\) -intercept of your plot have the expected value? (c) In performing such an experiment, one has the practical problem of locating the exact center of the lens since \(d_{\mathrm{i}}\) must be measured from this point. Imagine, instead, that one measures the image distance \(d\) from the back surface of the lens, which is a distance \(\ell\) from the lens's center. Then, \(d_{i}=d_{1}^{\prime}+\ell .\) Show that, when implementing the magnification method in this fashion, a plot of \(m\) vs.di will still result in a straight line. How can \(f\) be determined from this straight line?

(II) (a) How far from a 50.0-mm-focal-length lens must an object be placed if its image is to be magnified \(2.50 \times\) and be real? \((b)\) What if the image is to be virtual and magnified \(2.50 \times ?\)
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