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$\bullet$ Resolution of a microscope. The image formed by a microscope objective with a focal length of 5.00 $\mathrm{mm}$ is 160 $\mathrm{mm}$ from its second focal point. The eyepiece has a focal length of 26.0 $\mathrm{mm}$ . (a) What is the angular magnification of the microscope? (b) The unaided eye can distinguish two points at its near point as separate if they are about 0.10 $\mathrm{mm}$ apart. What is the minimum separation that can be resolved with this microscope?
a) 317.3b) $3.15 \times 10^{-4} \mathrm{mm}$
Physics 102 Electricity and Magnetism
Physics 103
Chapter 25
Optical Instruments
Electromagnetic Waves
Reflection and Refraction of Light
Cornell University
Rutgers, The State University of New Jersey
University of Washington
McMaster University
Lectures
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Okay, so we have this formula that magnification is equal to 25 centimeters. Time's image distance over focal length of the objective Time spoke a wink of the eye piece. Ah, foreign enough to Reno on to find magnification. We just need to know as prime, which is where the images form. So the image is formed. Ah, Image has a focal length off 0.5 centimeters are five millimeters and it is 16 millimetre 16 centimeters of 1 16 millimeters from the the second focal point. So that adds up to 16.5 centimeters. Right, So you plug in values now 25. This is part here, by the way. So 25 times 16.5 over 0.5 for the objective and 2.6 for the eyepiece gives us 317 times magnification for part B. We want to find the minimum separation. And so ah, rule of thumb is that the magnification or the absolute magnification are sorry. Angular magnification is equal. Teo Ah, the the distance over the minimum separation calling it men. So men minimum separation will be a distance which has 0.1 millimeters Ah, over like actual separation 0.1 millimeters over magnification, which is three seventeen 17. So this gives us 3.2 times 10 to the negative for millimeters of minimum difference.
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