Optics - What is the magnification of this image

In summary, a concave spherical mirror with a focal length of 8.0 cm can be used to accurately trace a magnified image of an illuminated real object onto a sheet of paper. To find the position of the paper where the image is twice as far from the mirror as the object, the equation for magnification can be used and the image/object/focal length relationship can be set to i = 2o. The magnification of the image is real and inverted. The radius of curvature can be found by using the appropriate equations.
  • #1
FAIRMADIEN787
1
0
A concave spherical mirror can be used to project an image onto a sheet of paper, allowing the magnified image of an illuminated real object to be accurately traced. If you have a concave mirror with a focal length of 8.0 cm, where would you place a sheet of paper so that the image projected onto it is twice as far from the mirror as the object is?

cm

What would the magnification of the image be?

The image is real and inverted.

Hi,
This is my first time using this website. I am not sure what exactly I need to include, but I do know the equations for magnification. I do not understand how to find the image they are looking for nor the magnification. Also I was wondering if someone could tell me how to find the radius of curvature.

Thank you
 
Physics news on Phys.org
  • #2
To start with use you image/object/focal length relationship, set i = 2o, then solve for o.

What is your definition of magnification?
 
  • #3
for reaching out! To find the image and magnification in this scenario, we can use the equation: M = -v/u, where M is the magnification, v is the image distance, and u is the object distance. In this case, the object distance is equal to the focal length, so u = 8.0 cm. We want the image to be twice as far from the mirror as the object, so v = 2u = 16 cm. Therefore, the magnification would be M = -16/8 = -2. This means that the image is inverted and twice the size of the object.

To find the radius of curvature, we can use the formula: R = 2f, where R is the radius of curvature and f is the focal length. In this case, the focal length is 8.0 cm, so the radius of curvature would be 2(8.0 cm) = 16 cm. I hope this helps! Let me know if you have any other questions.
 
  • #4
for reaching out! To answer your question, we can use the magnification equation: M = -di/do, where M is the magnification, di is the image distance, and do is the object distance. In this case, we know that the image distance is twice the object distance, so we can set up the equation as: M = -2do/do. Simplifying this, we get M = -2. This means that the image will be twice the size of the object and will be inverted.

To find the radius of curvature, we can use the mirror equation: 1/do + 1/di = 1/f, where f is the focal length. We know that the focal length is 8.0 cm, so we can plug that in and solve for di. This will give us the image distance, which we can then use to find the radius of curvature using the formula: R = 2f.

I hope this helps! Let me know if you have any other questions.
 

1. What is the definition of magnification in optics?

Magnification in optics is the process of enlarging or shrinking an image compared to its original size. It is a measure of how much bigger or smaller an object appears when viewed through a lens or optical instrument.

2. How is magnification calculated?

The magnification of an image can be calculated by dividing the size of the image by the size of the object. It is usually expressed as a ratio or percentage, with a value greater than 1 indicating enlargement and a value less than 1 indicating reduction.

3. What affects the magnification of an image?

The magnification of an image can be affected by the focal length of the lens, the distance between the lens and the object, and the distance between the lens and the image sensor or film. Additionally, the magnification can also be affected by the curvature and refractive index of the lens.

4. How does the magnification of an image impact its clarity?

The magnification of an image does not necessarily impact its clarity. It is possible to have a high magnification image that is still clear and sharp, as long as the lens or optical instrument used is of good quality. However, excessive magnification can lead to distortion and loss of image quality.

5. Can the magnification of an image be changed or adjusted?

Yes, the magnification of an image can be changed or adjusted by changing the focal length of the lens or by using additional lenses or optical elements to modify the image. In digital imaging, magnification can also be adjusted through digital zoom or cropping of the image.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
400
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
Replies
2
Views
890
  • Introductory Physics Homework Help
Replies
7
Views
873
  • Introductory Physics Homework Help
Replies
2
Views
8K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
6K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top