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A typical red blood cell subtends an angle of only$1.9 \times 10^{-5}$ rad when viewed at a person's near-point distance of25 $\mathrm{cm} .$ Suppose a red blood cell is examined with a compound microscope in which the objective and eyepiece are separated bya distance of 12.0 $\mathrm{cm} .$ Given that the focal length of the eyepieceis $2.7 \mathrm{cm},$ and the focal length of the objective is $0.49 \mathrm{cm},$ findthe magnitude of the angle subtended by the red blood cell whenviewed through this microscope.

$3.3 \times 10^{-3} \mathrm{rad}$

Physics 103

Chapter 27

Optical lnstruments

Wave Optics

Cornell University

Simon Fraser University

McMaster University

Lectures

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Okay, so we know that a compound microscope has objective focal lens sports here. Objective off zero point 49 ST Emitters. And we have a focal eyepiece lens off 2.7 saying team enters OK, the objective in the eyepiece that separated by a distance l that we're gonna call 12 same team enters, right? And this problem, we want to calculate what is the angular focal angle off the image. So we know that to calculate this to calculate the magnitude of the angle way can't say that a magnitude is the anger off the image divided by the angle off the objects. So if we want the angle angle off, the image is just the magnification multiplied by the young. Go off the object. Okay, we know that the angle of the object ISS 1.9 1.9 times 10 to the power off minus five, Reg. So how are we going to calculate this magnification in here? Because we do not have the magnification to calculate the angle. But we know that the magnification is defined as the distance of the image times the near point distance, divided by the focal objective and the focal eyepiece, and we just need to discover, then the distance off the image. So then we can calculate everything. So let's think we know that in the chapter we learned that the distance off the image is just did two blends subtract off the focal eyepiece? So this is just 12 minus 2.7. So this is equals to 9.3 ST Emitters. That's the distance of the image. Now let's calculate the magnification modification will be 9.3 times 25 of the near point zero point 49 of the objective times 2.7 off the eyepiece and all these gives us 176 off magnification. Now that we have the magnification, we can finally calculate the angle. Just gotta be 176 times 1.9 time stands to the power off minus five. So the angle and the final answer to this problem is just three point tree times 10 to the power off minor street Reg. And that's the answer to this problem. Thanks for

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