Optics - Convex Lens with Virtual Image

Question 1

convex lens with virtual image

Answer 1

• We saw that if you place an object O at an image distance do that is beyond the focal point F (do > f), the lens will bring the rays back together to focus at a point. At that point, the rays add to give a bright image. Such an image, where the rays actually meet to make a bright point, is called a “Real Image.”
• As you move the object O in closer to the lens, the image I moves farther away. When the object reaches the Focal Point F, the rays that hit the lens come out parallel; that’s what the Focal Point means. However, parallel rays never come together; they never focus to give an image. So, no real image will form if the object is at the focal point (do = f).
• If the object is INSIDE the focal point (do < f) of a convex lens, the rays are diverging so strongly that the lens cannot bring them back together. It does bend them closer together, so they aren’t diverging as much as before, but they are still diverging when they leave the lens. That means they will never converge to form a real image: objects inside the focal point do NOT form real images.
• However… if you look at the rays on the right side, they are diverging, but if you trace them backward, it APPEARS as if they DID cross at a point on the left side:
• The rays do not actually cross at that point. But looking at the rays on the right, it sure looks as if they did. From those rays, you would think that the rays actually originated at the point on the left where they seemed to cross. That point is therefore an Image of the Object. However, the rays didn’t actually cross there; if you put a screen or film there, you would not see a bright image, because the rays weren’t actually there. The image is not a Real Image; it is what we call a “Virtual Image.”
• So, objects inside the focal point of a convex lens form Virtual Images. Their rays don’t come together to focus, but if you trace them back, they appear to have crossed at the Image point.
• (You might think, since they aren’t “real,” that virtual images aren’t important. As a near-sighted person, I beg to differ. I wear glasses that bend the light rays to form virtual images that happen to be in focus. Everything I see is a virtual image, so I’m very glad they exist.)
• You can calculate the image distance, height, and magnification for virtual images using the same equations as before. There are two main differences in the answers. First, since the image is on the “wrong” side of the lens (the left instead of the right), the image distance is Negative.
• +di = Real Image
• -di = Virtual Image
• Also, notice on the diagram that the virtual image is not flipped upside down, as the real image was. That means hi is +. And that means M is + as well.
• +M = Upright virtual image
• -M = Inverted real image
• So, when you are calculating things, watch for the + and – signs. They tell you if the image is real or virtual. Conversely, if the problem tells you the image is virtual, make sure to use –di and +M.

Question 2

real image

Answer 2

• where rays actually cross

Question 3

virtual image

Answer 3

• one where rays appear to cross, but don’t actually