Optics Problem: A real object and a converging lens

In summary: The image is virtual, upright, and enlarged.In summary, the conversation discussed a problem involving a converging lens and a real object placed 9.00 cm away from it with a focal length of 24.0 cm. The correct answer was determined to be -14.4 cm from the lens, virtual, and enlarged. There was some confusion over the sign of q and the resulting image being upright or inverted, but it was determined that the image would be upright and magnified due to the relationship between image distance and image height.
  • #1
anomalocaris
23
0

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



[itex]\frac{1}{f}[/itex] = [itex]\frac{1}{p}[/itex] + [itex]\frac{1}{q}[/itex]
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



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

[itex]\frac{1}{q}[/itex]=-[itex]\frac{5}{72}[/itex]
q=(-[itex]\frac{5}{72}[/itex] )^-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.
 
Last edited:
Physics news on Phys.org
  • #2
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.
 
  • #3
I agree with your classmate.
 
  • #4
Have you tried inserting the answer as -14.4 cm from the lens? Also it would be upright, the image that is.
 
  • #5
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.
 
  • #6
your classmate is correct
 

Related to Optics Problem: A real object and a converging lens

1. What is a real object in optics?

A real object in optics is an object that exists in the physical world and can be seen by the naked eye. It is not a virtual or imaginary object.

2. What is a converging lens?

A converging lens is a type of lens that causes parallel rays of light to converge or come together at a single point after passing through the lens. It is thicker in the middle and thinner at the edges.

3. How does a converging lens form an image?

A converging lens forms an image by bending the rays of light that pass through it, causing them to converge at a single point. The image can be either real or virtual, depending on the location of the object in relation to the lens.

4. What are the characteristics of an image formed by a converging lens?

The characteristics of an image formed by a converging lens include its size, orientation, and location. The image is usually smaller than the object, inverted, and located on the opposite side of the lens from the object.

5. How can I calculate the image distance and magnification of a converging lens?

The image distance and magnification of a converging lens can be calculated using the thin lens equation, which is 1/f = 1/do + 1/di, where f is the focal length of the lens, do is the distance of the object from the lens, and di is the distance of the image from the lens. The magnification is given by M = -di/do.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
399
  • Introductory Physics Homework Help
Replies
1
Views
840
  • Introductory Physics Homework Help
Replies
1
Views
975
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
488
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
812
  • Introductory Physics Homework Help
Replies
3
Views
1K
Replies
11
Views
415
  • Introductory Physics Homework Help
Replies
1
Views
4K
Back
Top