Skip to main content
Ch 34: Geometric Optics

Chapter 34, Problem 34

Resolution of a Microscope. The formed by a microscope objective with a focal length of 5.00 mm is 160 mm from its second focal point. The eyepiece has a focal length of 26.0 mm. (a) What is the angular magnification of the microscope?

Verified Solution
Video duration:
6m
This video solution was recommended by our tutors as helpful for the problem above.
294
views
Was this helpful?

Video transcript

Hi everyone. In this practice problem, we're being asked to calculate the total angular magnification of a microscope. We will have a microscope composed of two optical components, a converging lens. L one with a focal length of 0.7 centimeter is forming an image 15 centimeter to the right of the second focal point of L one. The image formed by L one is very close to the focal point of the second converging lens. L two and the focal length of L two is 2. centimeter. We're being asked to calculate the total angular magnification of the microscope. And the options given are a 15 B 100 and C 234 and D 534. So for this problem, for a microscope, the overall angular magnification or M of the compound microscope is the product of the two magnification or M one and M two. So in this case, total M will actually be just the multiplication of M one multiplied by M two. The first of these factors are M one, M one is, is going to be negative S one prime divided by S one where S one is the object distance and S one prime is the image distance respectively. All of this are for the first converging lens or L one, the real image produced by the first converging lens is going to be very close to the second converging lens as it is given in the problem statement. Therefore, we have to use M two where M two is going to equals to centimeter, which is uh just the eyesight distance divided by F two where F two is given to be 2.3 centimeter or the focal length of L two given in the problem statement. So the total angular magnification or M will be equals to M one multiplied by M two. And all of this is going to be in absolute value. So we are going to neglect the negative sign and um multiplying this, we will then get S one prime multiplied by a 25 centimeter divided by S one multiplied by F two just like so awesome. So now we can actually um start by calculating S one prime. So the image distance S one prime is going to equal to 0.7 centimeter which is the focal length of the uh first lens or the first converging lens L one plus 15 centimeter, which is the actual image distance. And that will give us one prime value of 15. centimeter. Now we want to use the object distance lens equation to get S one or the actual object distance. So one divided by S one plus one, divided by S one prime will be equals to one divided by F one. And this is the object distance lens equation. Uh Now we wanna actually substitute all of our values. So one divided by SS one plus one divided by S one prime is 15.7 centimeter. Will it be equals to one divided by F one? Which is given to be 0.7 centimeter and then rear through rearrangement, we will then get S one to then be equals to 15.7 centimeter multiplied by 0.7 centimeter divided by 15.7 centimeter minus 0. centimeter. And this is just through um some rearrangement where one divided by S one will equals to one divided by 0. centimeter minus one divided by 15.7 centimeter. And then you wanna uh take the common denominator where one divided by S one will be equals two, 15.7 centimeter minus 0.7 centimeter divided by 15.7 centimeter multiplied by 0.7 centimeter. And then we will take the end first of that to get our S one. And in calculating S one, we will then get S one to be equals to 0. centimeter. So known that we have S one S one prime and we also know what our F two is from the problem statement, we can actually calculate the overall or total angular magnification of our microscope. Where then our M will then equals to I'm just going to rewrite the equation that we have S one prime multiplied by 25 centimeter. All of that going to be divided by S one multiplied by F two. And then now we can substitute our information. So S one prime is 15.7 centimeter multiplied by 25 centimeter divided by S one S one S 0.73 centimeter and F two is given to be 2.3 centimeter. Calculating this, we will get our total angular magnification to then be equals to 234. So our M values or total angular magnification is going to be 234 which will correspond to option C in our answer choices. So option C will be the answer to this particular practice problem and that'll be it for this video. If you guys still have any sort of confusion, please make sure to check out our other videos on similar topics on our website. But other than that, that will be it for this one. Thank you.
Related Practice
Textbook Question
BIO Contact Lenses. Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or ) is the same as the distance from the lens to that object (or ). A certain person can see distant objects well, but his near point is 45.0 cm from his eyes instead of the usual 25.0 cm. (a) Is this person nearsighted or farsighted?
310
views
Textbook Question
The focal length of a simple magnifier is 8.00 cm. Assume the magnifier is a thin lens placed very close to the eye. (b) If the object is 1.00 mm high, what is the height of its formed by the magnifier?
493
views
Textbook Question
The focal length of the eyepiece of a certain microscope is 18.0 mm. The focal length of the objective is 8.00 mm. The distance between objective and eyepiece is 19.7 cm. The final formed by the eyepiece is at infinity. Treat all lenses as thin. (b) What is the magnitude of the linear magnification produced by the objective?
291
views
Textbook Question
A telescope is constructed from two lenses with focal lengths of 95.0 cm and 15.0 cm, the 95.0-cm lens being used as the objective. Both the object being viewed and the final are at infinity. (b) Find the height of the formed by the objective of a building 60.0 m tall, 3.00 km away.
466
views
Textbook Question
A concave mirror has a radius of curvature of 34.0 cm. (b) If the mirror is immersed in water (refractive index 1.33), what is its focal length?
409
views
Textbook Question
BIO (a) Where is the near point of an eye for which a contact lens with a power of +2.75 diopters is prescribed?
317
views