1. Derive the thin lens equation (equation 1) for a divergent lens, using the di
ID: 2241831 • Letter: 1
Question
1. Derive the thin lens equation (equation 1) for a divergent lens, using the diagram in figure 1. (Hint: use similar triangles).
2. What is the difference between a real image and a virtual image? Is the image produced by a converging lens real or virtual?
Derive the thin lens equation (equation 1) for a divergent lens, using the diagram in figure 1. (Hint: use similar triangles). What is the difference between a real image and a virtual image? Is the image produced by a converging lens real or virtual? 1/o + 1/i + 1/fExplanation / Answer
A real image is formed when light rays originating from a point on one side of a lens (i.e.
the object) are refracted by the lens so that they focus (come together) to a point on
the other side of the lens at the image location. This happens when the object is father
away from the (converging) lens than the focal length.
A virtual image is formed when light rays originating from a point on one side of a lens
(i.e. the object) are refracted by the lens so that they diverge (move apart from each
other) on the other side of the lens. When these rays are traced back in a straight line
(ignoring that they were actually bent by the lens) then appear to diverge from a
point on the same side of the lens as the object (this is the location of the virtual
image). This happens when the object is closer to the (converging) lens than the focal
length.
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