What magnification will be produced by a lens of power –4.00 D (such as might be used to correct myopia) if an object is held 25.0 cm away?
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OpenStax College Physics for AP® Courses, Chapter 25, Problem 48 (Problems & Exercises)

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This is College Physics Answers with Shaun Dychko. What magnification will be produced by a lens with the power of negative 4.00 diopters given an object held 25.0 centimeters away and we'll change this object distance into meters— 25.0 times 10 to the minus 2 meters— and it's pretty important to change it into meters because this diopters is reciprocal meters and so these meters here will work well with the diopters, which are actually reciprocal meters. Okay! We have magnification is the negative of the image distance divided by the object distance, we need to figure out what this image distance is because that will be... that's the only thing we don't know in this equation, we know the object distance. So we use the thin lens equation to figure out the image distance: we know that 1 over object distance plus 1 over image distance is 1 over focal length and we can use this relation between power and focal length to substitute P in place of 1 over f so the power of the lens is the reciprocal of the focal length so 1 over f is P. And then subtract 1 over d o from both sides and we get 1 over d i then is the power minus 1 over d o; I like to write this 1 over something as the something to the power negative 1 this exponent negative 1 means take the reciprocal of the base of this power and so we have 1 over d o is d o to the power negative 1 and we have a convenient button on our calculator which says exponent negative 1. So we want to raise both sides of this to the exponent negative 1 to solve for d i on the left so d i on the left is P minus d o to the negative 1 all to the negative 1. Okay! So this can get replaced in place of d i in our magnification formula and so that's what I have done here. So magnification then is negative of negative 4 diopters—so that's the power— minus 25.0 times 10 to the minus 2 meters to the negative 1 and then raise that difference to the exponent negative 1, divide by 25.0 times 10 to the minus 2 meters—object distance— and the magnification is positive 0.500.