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# A microscope has an objective lens with a focal length of 16.22 mm and an eyepiece with a focal length of 9.50 mm. With the length of the barrel set at 29.0 cm, the diameter of a red blood cell’s image subtends an angle of 1.43 mrad with the eye. If the final image distance is 29.0 cm from the eyepiece, what is the actual diameter of the red blood cell? Hint: To solve this question, go back to basics and use the thin-lens equation.

## 8.823\times 10^{-7} m

#### Topics

Electromagnetic Waves

Wave Optics

### Discussion

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### Video Transcript

in this question we have to take close to 1.43 mil iridium. That is 10 to the power minus three brilliant. And distance at d equals to 29 centimeters. That is 10 to the power minus two m. So the M. S. Uh height will be a dish equals two D. 20 to So substituting the values, we get 29 multiplayer beta into the power -2 and off. Mhm 1.4, 3 muscular by 10 to the power -3 Radiant. Okay, so from here, after solving we get actually equals to 0.4147 multiplied by 10 to the power -3 m. Okay, so the magnification of the microscope will be small. M equals two minus L by F. Not or D by F. E. So substituting the values given we have al equals to 29 cm. They were by far is focal length of the objective lens, which is 16 point double to mm. Yeah, Manipulated by D is 25 cm distinctive. His understands divided by focal length of the eyepiece is 9.5 cm mm. Okay, 9.5 mm. So from here after solving, we get magnification. M equals 2 -4 70. So the real height or Diameter of the blood cell will be as equal to minus actually developed by M. So substituting the values we get minus of zero 4147 multiplied by 10 to the power minus three m, divided by magnification. M s minus 4 70. Okay, so from here we get As equals 8.8-3 muscular by 10 to the power -7 meter. So this is the answer for the question. Okay.

#### Topics

Electromagnetic Waves

Wave Optics