Optics Problem: A real object and a converging lens

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Homework Help Overview

The problem involves a real object placed in front of a converging lens, with a specific focal length. Participants are discussing the characteristics of the resulting image, including its distance from the lens, whether it is real or virtual, and its size relative to the object.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are examining the application of the lens formula and discussing the implications of positive and negative image distances. There is also consideration of how image height relates to image distance, and some participants express confusion regarding the nature of the image based on the calculated values.

Discussion Status

There is an ongoing exploration of the problem, with various interpretations of the image characteristics being discussed. Some participants are questioning the correctness of the original poster's calculations and the assumptions made about the image type and size. Guidance has been offered regarding the relationship between image distance and height, but no consensus has been reached.

Contextual Notes

Participants note that the problem is part of a multiple-choice assignment, which may complicate the interpretation of the results. There is also mention of potential discrepancies in the provided answer choices compared to the original problem setup.

anomalocaris
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Homework Statement



A real object is placed 9.00 cm from a converging lens that has a focal length of 24.0 cm. The image is________.

My classmate answered "14.4 cm from the lens, virtual, and enlarged," but apparently this is incorrect. I solved it out and chose the option "14.4 cm from the lens, real, and enlarged," though I believe the image should be virtual since my "q" value was negative/<0. I do not have the correct answer since this assignment is still active, so I would really like to understand if I'm taking the wrong steps before my final next week. Any suggestions would be very much appreciated!

Homework Equations



\frac{1}{f} = \frac{1}{p} + \frac{1}{q}
where f=focal distance=24.0 cm
p=distance of real object from the mirror=9.0cm
q=distance of image from the mirror

I also know that f>0 for a converging lens and when q>0, the image will be real.

The Attempt at a Solution



\frac{1}{24.0 cm} = \frac{1}{9.0 cm} + \frac{1}{q}
and so moving things around: \frac{1}{24.0 cm} - \frac{1}{9.0 cm} = \frac{1}{q}

\frac{1}{q}=-\frac{5}{72}
q=(-\frac{5}{72} )^-1 =-14.4

My q is negative, so it should virtual, but irregardless, the computer finds this answer unfit! I'm still a little confused at knowing if it will be enlarged or not, especially because no heights were given for the object.
 
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I believe your answer is correct. Remember that image height is proportional to the image distance. The greater the distance, the greater the height (it will be upside down though if its real). Therefore, the image will be magnified larger if the image distance is greater than the object distance.
 
Have you tried inserting the answer as -14.4 cm from the lens? Also it would be upright, the image that is.
 
I think my classmate is right as well. The answers are multiple choice, so -14.4 cm is not an option unfortunately. I think my professor "borrows" his questions from other sources and sometimes he changes them slightly but forgets to change the choices of answers. Thank you all, I will have to ask about it in class or office hours.
 
your classmate is correct
 

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