Diverging Spherical Lens Proof

In summary, using the thin-lens equation and the magnification factor, it can be shown that for a spherical diverging lens, the image of a real object will always be virtual, upright, and reduced with a magnitude of the magnification factor being less than 1. This is due to the fact that the focal point of the lens is virtual and the image distance is always negative, resulting in a positive magnification factor.
  • #1
emperrotta
11
0

Homework Statement


Using the thin-lens equation and the magnification factor, show that for a spherical diverging
lens the image of a real object is always virtual, upright, and reduced?

Homework Equations


Thin lens equation 1/p + 1/i = 1/f
p = distance of object from the lens
i = distance of image created by the lens from the lens
f = the focal length of the lens

magnification factor equation:
M = -(i/p)
M = magnification factor
p = distance of object from the lens
i = distance of image created by the lens from the lens

The Attempt at a Solution


I believe I have everything figured out except proving that the image will always be reduced. Here is what I have so far:
Starting with the thin lens equation:
1/p + 1/i = 1/f

Rearranging in order to solve for i:
i = [tex]\frac{pf}{p-f}[/tex]

With a diverging lens, as light rays pass through the lens they are refracted away from the central axis of the lens. Thus, there is no focal point that the light rays refract towards. Rather, the backwards extensions of the rays extend towards a virtual focal point. Since the focal point of the lens is virtual, f in the equation above is negative. Giving f a negative value, the above equation becomes:
i = [tex]\frac{-pf}{p+f}[/tex]

Since the object distance is considered to be always positive in these cases, then we know that the value for i will be negative. A negative value for i means the image is virtual.

Plugging a negative value for i into the magnification factor equation:
M = -(i/p)
we know M will be positive, which means the image is upright.

Now, here is where I get stuck. I don't know where to go from here in order to prove that the image will always be reduced. I know that a reduced image would mean that the magnitude of the magnification factor has is a fraction. In order to show that algebraically, I would have to show that i is always less than p. I just don't know how to do that, or maybe I am just looking at this the wrong way. Any help would be much appreciated.
 
Physics news on Phys.org
  • #2
M=-i/p = f/(f+p)<1
 
  • #3


Your approach so far is correct. To show that the image is always reduced, we can use the magnification factor equation M = -(i/p) and substitute in the equation for i that we derived earlier: i = -pf/(p+f). This gives us:

M = -(-(pf/(p+f))/p) = pf/(p+f)

Now, we know that f is always negative for a diverging lens, so we can substitute in -f for f in the equation above:

M = p(-f)/(p+(-f)) = -pf/(p-f)

Since p is always positive, we can conclude that the magnitude of the magnification factor, M, will always be less than 1. This means that the image will always be reduced in size compared to the object. Therefore, we have shown that for a spherical diverging lens, the image of a real object will always be virtual, upright, and reduced.
 

1. What is a diverging spherical lens?

A diverging spherical lens is a type of lens that causes light rays to spread out, or diverge, as they pass through it.

2. How does a diverging spherical lens work?

A diverging spherical lens works by refracting, or bending, light rays away from its center. This is due to the curved shape of the lens, which is thicker at the edges and thinner in the center.

3. What is the purpose of a diverging spherical lens?

The purpose of a diverging spherical lens is to correct for nearsightedness (myopia). By causing light rays to diverge, the lens helps to focus the image on the retina, rather than in front of it.

4. How is the proof of a diverging spherical lens performed?

The proof of a diverging spherical lens involves measuring the focal length of the lens, which is the distance from the lens to the point where the light rays converge. This can be done using a variety of methods, such as the lensmaker's formula or the thin lens equation.

5. What are some common applications of diverging spherical lenses?

Diverging spherical lenses are commonly used in eyeglasses and contact lenses to correct for nearsightedness. They are also used in optical instruments such as telescopes and microscopes to help focus light and produce clear images.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
339
  • Introductory Physics Homework Help
Replies
1
Views
917
  • Introductory Physics Homework Help
Replies
2
Views
615
  • Introductory Physics Homework Help
Replies
5
Views
977
  • Introductory Physics Homework Help
Replies
2
Views
948
  • Introductory Physics Homework Help
Replies
3
Views
808
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
Back
Top