How far from the lens will the image be?Is the image real or virtual,and either reduced,enlarged,or same size as object?What is the height of the image?
2007-04-26 14:25:29 · 2 answers · asked by xxgodsgift2youxx 1 in Science & Mathematics ➔ Physics
See the ref. The applicable formula for distance is 1/S1 + 1/S2 = 1/f, where S1 and S2 are object and image distances and f is the focal length. With S1=10.9 and f= 7.4 this gives S2=23.045 cm. The magnification M = -S2/S1 = -2.114, so the image is enlarged (abs(M)>1), and with negative magnification, inverted. Knowing the magnification you can calculate the height of the image. Since S2 is positive the image is real.
What's important is keeping the signs of S1 and S2 straight. A positive S1 is on one side of the lens and a positive S2 on the other side. This of course means the image is projected (rather than seen by looking through the lens in the direction of the object) and therefore is real. Also f is negative for a diverging lens and positive for converging.
2007-04-27 17:21:31 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
you may desire to state no remember if the image projected on the demonstrate has a magnification ingredient of a million, under a million, or greater desirable than a million.the right ingredient of magnification would desire to be time-venerated to furnish a superbly precise answer.
2016-12-10 12:29:32 · answer #2 · answered by hinokawa 4 · 0⤊ 0⤋