Optics Problem: A real object and a converging lens

AI Thread Summary
A real object is placed 9.00 cm from a converging lens with a focal length of 24.0 cm, leading to confusion about the image characteristics. The calculations indicate that the image distance (q) is -14.4 cm, suggesting it is virtual, but this contradicts the expected outcome for a converging lens where q should be positive for real images. The discussion highlights the importance of understanding the sign conventions in optics, particularly that a negative q indicates a virtual image. Participants also note that the image height is proportional to the image distance, impacting whether the image appears enlarged or inverted. The thread concludes with a suggestion to seek clarification from the professor regarding the discrepancies in the assignment.
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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.
 
I agree with your classmate.
 
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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