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Magazine Chapter 34: Problem 104
An object is placed 18.0 \(\mathrm{cm}\) from a screen. (a) At what two points between object and screen may a converging lens with a 3.00 -cm focal length be placed to obtain an image on the screen? (b) What is the magnification of the image for each position of the lens?
Short Answer
Expert verified
Place the lens at 3 cm or 15 cm from the object. Magnifications are -5 and -1/5, respectively.
Step by step solution
01
Identify the Given Information
We have an object, a screen, and a lens described by its focal length. The distance between the object and the screen is given as 18.0 cm, and the focal length of the lens is 3.00 cm.
02
Understand Lens Equation
The lens-maker's formula is: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] where \( f \) is the focal length, \( d_o \) is the object distance from the lens, and \( d_i \) is the image distance from the lens. To form an image on the screen, \( d_o + d_i = 18.0 \text{ cm} \). We need to solve for both \( d_o \) and \( d_i \).
03
Express Image Distance in Terms of Object Distance
Using \( d_o + d_i = 18 \), we can express \( d_i \) as \( 18 - d_o \). Substitute \( d_i = 18 - d_o \) into the lens equation to get: \[ \frac{1}{3} = \frac{1}{d_o} + \frac{1}{18 - d_o} \].
04
Solve for Object Distance
Multiply through by \( d_o(18 - d_o) \) to clear the fractions: \[ d_o (18 - d_o) = 3(18) \]. Simplify to form a quadratic equation: \[ 3d_o^2 - 18d_o + 54 = 0 \].
05
Simplify and Solve the Quadratic Equation
Divide by 3 to simplify: \[ d_o^2 - 6d_o + 18 = 0 \]. Use the quadratic formula \( d_o = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = -6 \), and \( c = 9 \): \[ d_o = \frac{6 \pm \sqrt{36 - 36}}{2} \]. Simplifying gives \( d_o = 3 \text{ cm} \) or \( d_o = 15 \text{ cm} \).
06
Calculate Image Distance for Each Lens Position
For \( d_o = 3 \text{ cm} \), \( d_i = 18 - 3 = 15 \text{ cm} \). For \( d_o = 15 \text{ cm} \), \( d_i = 18 - 15 = 3 \text{ cm} \). In both cases, the symmetric placement gives either distance as the other.
07
Calculating Magnification for Each Position
Magnification \( m \) is given by \( m = - \frac{d_i}{d_o} \). For \( d_o = 3 \text{ cm} \), \( m = -\frac{15}{3} = -5 \). For \( d_o = 15 \text{ cm} \), \( m = -\frac{3}{15} = -\frac{1}{5} \).
08
Interpretation of Results
The lens can be placed at \( 3 \text{ cm} \) or \( 15 \text{ cm} \) from the object. At these positions, the magnifications will be \(-5\) and \(-\frac{1}{5}\) respectively.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Lens Equation
The heart of understanding how a lens forms an image lies in the lens equation. The lens equation relates the object distance, image distance, and focal length of the lens. It is given by:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]where:
- \( f \) is the focal length of the lens. It indicates how strongly the lens converges or diverges light.
- \( d_o \) is the object distance, i.e., how far the object is from the lens.
- \( d_i \) is the image distance, measuring how far the image forms from the lens.
Object Distance
The object distance is an essential element in understanding lens behavior. It represents how far the object is from the lens. In our exercise, the lens was placed at two distinct distances from the object to form a clear image on a screen. We denote this distance as \( d_o \).
To find the positions where the lens can create an image, we used the sum of object and image distances, affirming:\[d_o + d_i = 18 \text{ cm}\]Substituting in the lens equation allows us to express \( d_i \) as a function of \( d_o \):\[d_i = 18 - d_o\]This transformation is critical in setting up a solvable equation to find out where exactly to place the lens to achieve focused images.
To find the positions where the lens can create an image, we used the sum of object and image distances, affirming:\[d_o + d_i = 18 \text{ cm}\]Substituting in the lens equation allows us to express \( d_i \) as a function of \( d_o \):\[d_i = 18 - d_o\]This transformation is critical in setting up a solvable equation to find out where exactly to place the lens to achieve focused images.
Image Distance
The image distance tells us how far the formed image is from the lens. Using the relationship between object and image distances, this value becomes understandable. In our case, we defined it as the difference between the total distance and the object distance:\[d_i = 18 - d_o\]This expression was substituted into the lens equation to form a quadratic equation:\[\frac{1}{3} = \frac{1}{d_o} + \frac{1}{18 - d_o}\]On solving, it was revealed that the lens could be positioned where:
- \( d_o = 3 \text{ cm} \)
- \( d_o = 15 \text{ cm} \)
Magnification Calculation
Magnification provides insight into how the size of the image changes compared to the object. It is calculated using the formula:\[m = -\frac{d_i}{d_o}\]This reveals the size relation of the image to the object, with the negative sign indicating image inversion by the lens.
In our setup:
In our setup:
- For \( d_o = 3 \text{ cm} \), magnification \( m = -\frac{15}{3} = -5 \). This means the image is five times larger than the object and inverted.
- For \( d_o = 15 \text{ cm} \), magnification \( m = -\frac{3}{15} = -\frac{1}{5} \). Thus, the image is one-fifth the size of the object, again inverted.
