*PLEASE READ MY QUESTION FOLLOWING THIS PROBLEM* An object is placed 58.0 cm fro
ID: 2157573 • Letter: #
Question
*PLEASE READ MY QUESTION FOLLOWING THIS PROBLEM*
An object is placed 58.0 cm from a screen. (a) Where should a converging lens of focal length 6.5 cm be placed to form an image on the screen?
I've seen this problem worked out before and can't make sense of it. Please explain in terms of p=object distance and q=image distance. Also, please clarify why people are saying that q=58-p. I thought 58 was p. What variable is assigned to the didtance we're solving for (distance from the screen). I thought that was p, but it doesn't produce the right answer when I plug into M=-q/p equation.
PLEASE HELP!!! I've been stumped on this one for over an hour. Please don't just throw numbers, I need to know where they came from and why.
Explanation / Answer
a) di + do = 58cm
=>do = 58-di
1/f = 1/di + 1/do
=>1/6.5 = 1/di + 1/58-di = 58/(58-di)di
=>58di - di2 = 377
=>di2 - 58di + 377 = 0
=>Shortest distance = 7.46cm
Farther distance = 50.54 cm
b)M = -di/d0 = -di/(58-di)
If placed at shorter distance, Magnification = -0.1476
If placed at farther distance, Magnification = -6.7755
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