- #1
emperrotta
- 10
- 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 = \frac{pf}{p-f}
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 = \frac{-pf}{p+f}
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.