QQuestionMathematics
QuestionMathematics
An object is placed 22.0 cm from a screen.
## Part A
At what two points between the object and screen, a converging lens with a 3.00 cm focal length will be placed to obtain an image on the screen?
Enter your answers numerically in centimeters separated by a comma.
## Part B
What is the magnification of the image for each position of the lens?
Enter your answers numerically separated by a comma.
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Answer
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Step 1:: To find the two points where the converging lens should be placed, we need to determine the image distance (v) for the given object distance (u) using the lens equation:
\frac{1}{f} = \frac{1}{u} + \frac{1}{v}
where f is the focal length of the lens.
Step 2:: First, let's find the image distance (v) for the object placed at 22.0 cm from the screen.
\frac{1}{3.00 \text{ cm}} = \frac{1}{22.0 \text{ cm}} + \frac{1}{v}
Substitute the values into the lens equation:
Step 3:: Solve for v (image distance):
v = \frac{1}{\frac{1}{3.00 \text{ cm}} - \frac{1}{22.0 \text{ cm}}}
Step 4:: Calculate the image distance (v):
v \approx 8.82 \text{ cm}
Step 5:: Since the image is formed on the screen, the first position of the lens should be between the object and the screen, such that the image distance is 8.82 cm from the lens.
Step 6:: To find the second position of the lens, we can use the thin lens equation again, but this time we will consider the image as a virtual object and the screen as the image location.
The new object distance (u) will be the negative of the image distance found in Step 4, and the new image distance (v) will be the distance between the lens and the screen, which is 22.0 cm - 8.82 cm = 13.18 cm.
Step 7:: Substitute the values into the lens equation:
\frac{1}{3.00 \text{ cm}} = \frac{1}{-8.82 \text{ cm}} + \frac{1}{13.18 \text{ cm}}
Step 8:: Solve for the new object distance (u):
u \approx -5.82 \text{ cm}
Step 9:: Therefore, the second position of the lens should be approximately 5.82 cm from the screen.
Step 10:: The two points where the converging lens should be placed are approximately 5.82 cm and 8.82 cm from the screen.
Final Answer
Step 11: To find the magnification of the image for each position of the lens, we can use the magnification formula: M = -\frac{v}{u} where M is the magnification, v is the image distance, and u is the object distance. Step 12: For the first position of the lens (u = - 5.82 cm, v = 8.82 cm): M_1 = -\frac{8.82 \text{ cm}}{- 5.82 \text{ cm}} M_1 \approx 1.515 Step 13: For the second position of the lens (u = 22.0 cm, v = 13.18 cm): M_2 = -\frac{13.18 \text{ cm}}{22.0 \text{ cm}} M_2 \approx 0.599
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