Lenses and optics: solving for focal point.

[ilovedeathcab]

Lenses and optics: solving for focal point. (SOLVED)

(2 points, penalty of 0.5)
When a flower pot is placed 1.2 meters on the left of a magnifying glass (converging lens), an image is formed on a screen 0.37 meters to the right of the lens. Find the focal length of the lens. Give your answer ...

* in terms of meters. Do not include units in your answer. They will be assumed to be meters.
* to a precision of hundreths of a meter. Use a format like #.## in your answer.
* to an accuracy of 0.02 m. Answers within 0.02 meters of the correct answer will be tolerated.

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I know that with a converging lens, the focal point value will be positive.
I started with the basic equation
1/f = 1/do + 1/di

1/f = 1/1.2m +1/.37m
f = .833333333m + 2.7027m

f = 3.536 m
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I entered this value and it counted it as incorrect. can somebody please tell me what I'm doing wrong? i would appreciate it. thanks ! http://moodle.oakwood.k12.oh.us/file.php/72/Unit_13_quiz_images/quiz13q3.jpg

 

[rl.bhat]

f = 3.536 m
This should be
1/f = 3.536

 

[nlightner]

Hello, it seems like you have made a mistake in your calculation. The correct equation to use is 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance. In this case, the object distance is 1.2 meters and the image distance is 0.37 meters. Plugging these values into the equation, we get:

1/f = 1/1.2 + 1/0.37
1/f = 0.833 + 2.703
1/f = 3.536
f = 1/3.536
f = 0.283 meters

Therefore, the focal length of the lens is 0.283 meters, or 28.3 centimeters. Please double check your calculations and make sure to use the correct equation when solving for focal point. Keep up the good work in your studies of lenses and optics!