QQuestionAnatomy and Physiology
QuestionAnatomy and Physiology
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.
Attachments
6 months agoReport content
Answer
Full Solution Locked
Sign in to view the complete step-by-step solution and unlock all study resources.
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:
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
Need Help with Homework?
Stuck on a difficult problem? We've got you covered:
- Post your question or upload an image
- Get instant step-by-step solutions
- Learn from our AI and community of students