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(a) What magnification is produced by a 0.150 cm focal length microscope objective that is 0.155 cm from the object being viewed? (b) What is the overall magnification if an 8x eyepiece (one that produces a magnification of 8.00) is used?
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  1. -30
  2. -240
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OpenStax College Physics Solution, Chapter 26, Problem 27 (Problems & Exercises) (2:01)

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This is College Physics Answers with Shaun Dychko. We have a microscope that has a 0.150 centimeter focal length on the objective and the objective lens is 0.155 centimeters from the object. So, Do is that and F is that number and magnification is negative of image distance divided by object distance. Now, we don't know what the image distance is but we can figure it out knowing the focal length and the object distance. So, the thin lens equation says the reciprocal of focal length is the reciprocal of the image distance plus the reciprocal of the object distance and then we'll subtract one over Do from both sides and we get one over image distance is one over focal length minus one over object distance. We can write this as a single fraction by multiplying the top and bottom by Do here and top and bottom by focal length there and we end up with Do minus focal length divided by this common denominator, focal length times object distance. And then, when we want to solve for the image distance, we flip both sides and we get image distance is focal length times object distance divided by object distance minus focal length. So, now we know the image distance. And, we can substitute that in for  Di in our magnification formula. So, we have negative one over  Do from the formula and then we're going to multiply it by this fraction, which is Di. F Do over Do minus F. And, the Do's cancel there and the magnification then is negative focal length divided by object distance minus focal length. So, that's negative of 0.15 centimeters divided by 0.155 centimeters minus 0.15 centimeters, which is negative 30. The negative sign means the image is flipped compared to the object orientation. Now, if there was an eye piece with magnification eight placed in series with this objective lens, we would end up with a total magnification which is a product of their respective magnifications, which is negative 30 times eight, which is negative 240.