Finding final image through a lens

In summary, the problem involves an object situated at various points relative to V, with light reflected by a mirror and the final image viewed by an observer. Equations are used to determine the location of the image, with the final result being R/4 to the right of V.
  • #1
coltin.walsh
7
0

Homework Statement



An object is situated at various points relative to V. The light is then reflected by the mirror, and the final image is viewed by the observer.

http://www.flickr.com/photos/55153239@N03/6196518731/

Homework Equations



1/s + 1/s' = 2/R
na/s + nb/s' = (nb-na)/R

The Attempt at a Solution



The final statement of the problem confused me. It says the final image is viewed from the left of the figure. Therefore, you the mirror create an image inside the lens, and the light would then be refracted to the viewer outside the lens?

for d) I used the second equation to refract the light into the lens, to get s' = -3/2 R. This then acts as the object for the mirror a distance 5/2 R. Using the first equation, i get s' = 5/8 R. This would act as the object for the second refraction outside the lens. the end result is R/4 to the right of V. Is that correct?
 
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  • #2
Hi coltin.walsh, welcome to PF.
"This would act as the object for the second refraction outside the lens. the end result is R/4 to the right of V."
Check this last step. Show your calculation.
 
  • #3
so i would have s = 3/8 R. and using the second equation:

1.5/(3R/8) + 1/s' = 0
s' = -R/4
image is located a distance R/4 to the right of V.
 
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1. How do you calculate the magnification of a lens?

The magnification of a lens can be calculated by dividing the height of the image by the height of the object. This can be represented by the equation M = hi/ho, where M is the magnification, hi is the height of the image, and ho is the height of the object.

2. What is the difference between a real image and a virtual image?

A real image is formed when light rays actually converge at a point and can be projected onto a screen, while a virtual image is formed when light rays appear to be coming from a point, but do not actually converge and cannot be projected onto a screen.

3. How does the distance between an object and a lens affect the size of the image?

The distance between an object and a lens, known as the object distance, affects the size of the image through the magnification equation. As the object distance increases, the image becomes smaller in size, and as the object distance decreases, the image becomes larger in size.

4. Can a lens produce an image that is both real and inverted?

Yes, a lens can produce an image that is both real and inverted, as long as the object is placed beyond the focal point of the lens. In this case, the image will be projected on the opposite side of the lens and will be inverted compared to the object.

5. What is the difference between a convex lens and a concave lens?

A convex lens is thicker in the center and thinner at the edges, causing light rays to converge and produce a real image. A concave lens is thinner in the center and thicker at the edges, causing light rays to diverge and produce a virtual image.

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