Image distance and magnification: image formed by two lenses, virtual object case. скачать в хорошем качестве

Image distance and magnification: image formed by two lenses, virtual object case.

Another lens combination problem, but this time the first image becomes a virtual object for the second lens! 🧠 Access full flipped physics courses with video lectures and examples at https://www.zakslabphysics.com/ In the image formed by two lenses, virtual object case, we apply the lens equation twice and find the final image distance and magnification. We start with the ray diagram for two converging lenses. We take all three principal rays off the object and pass them through the first lens to form an inverted real image on the right side of the second lens. With the first image on opposite side of second lens, we have a virtual object for the second refraction. We can't simply use the first image as the second object, so the ray diagram is extra tricky. We have to find three rays to use as principal rays through the second lens, but these rays are not coming from the first image. We quickly see two of the rays that converged to form the first image that are also principal rays for the second refraction, but the third one requires some imagination. We find a ray travelling from the direction of the focal point of the second lens that converges to the first image. This ray is refracted parallel to the principal axis, and now we have three rays converging to our final image. This is where the inverted real image of the arrow head will form. We read off the grid to approximate the image distance using a ray diagram, and we count grid spaces to approximate the magnification using a ray diagram. To analytically find the image formed by two lenses, apply the lens equation twice. We index the first object with a 1 and find the first image distance using the lens equation. We calculate the second object distance based on the lens separation. Because this image is on the opposite side of the second lens, we have a virtual object so the object distance is negative. We apply the thin lens equation a second time to find the image distance relative to the second lens. This agrees very nearly with the approximation from the ray diagram. After repeated use of the magnification equation we find the magnification for two lenses by multiplying the magnifications at each step. We obtain a magnification very nearly equal to the magnification predicted by the ray diagram.