Problem

. The objective lens and the eyepiece of a micros…

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Problem 34 Hard Difficulty

An insect 1.2 $\mathrm{mm}$ tall is placed 1.0 $\mathrm{mm}$ beyond the focal
point of the objective lens of a compound microscope. The objective lens has a focal length of $12 \mathrm{mm},$ the eyepiece a focal length of 25 $\mathrm{mm}$ . (a) Where is the image formed by the objective lens and how tall is it? (b) If you want to place the eye-piece so that the image it produces is at infinity, how far should this lens be from the image produced by the objective lens?
(c) Under the conditions of part (b), find the overall magnification of the microscope.

Answer

a) Hence, the height of the image is $[14.4 \mathrm{mm}]$
b) 25 $\mathrm{mm}$
c) $m=\frac{h^{\prime \prime}}{h}$

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Video Transcript

So we have ah, objective lens for ogling 12 millimetres. We have, ah, an insect that's placed one millimeter away from it. So that's 12 plus warning was 13 millimeters as your object distance. Therefore ah, image distance is just one over F minus one over us, and that's 12 millimeters and that's 13 millimeters. So we get image distance off 1 56 millimeters or 15.6 centimeters on. Then we want magnification. Samantha Kishan is just negative image distance over object distance on DSO that's needed 1 56 over 12.8 What I meant 13. And so that's, um, 12 times on DH. So we want the app Milanese. Absolute value of this magnet of this magnification. Ah, in our ah calculation of image height, so object. And so that's just 12 times object Hide. Um, Mrs One playing two millimeters and that gives us 14.4 millimeters for for the height of the image of the insect. Okay, in part B, we're looking at an object that with an image that's very far west, the images that infinity so one over infinity goes to zero. No. So we're left for its one over F is one over s, which, of course, means that focal length is equal toe object Distance on vice versa. So the focal length is simply the smuggling of the hi piece Rich is 25 millimeters, or 2.5 centimeters, and that turned out to do the object distance. Um, and finally, in part C, we want the magnification. We want that overall magnification, so there are well, so there's this one. There's there's him patch objective, the linear magnification of the objective lens times the angular magnification on. So this we found to be 12 and someone is just 25 over the focal length, which is 2.5 right on these airboats and centimetres. And so that's 1 20 has a magnification, Um, and we'll put a negative sign. They're just too, just to be consistent. But that's 1 20 times a few using this, this exact formula. But if you're using this formula, which is also in the book, Ammi calls 25 centimeters times as prime over F one times have to if two different focal, Lexie remembers. So this becomes 25 times as prima's 15.6 centimeters and F one is 12 millimeters, so 1.2 centimeters enough two is 2.5 on DSO. Ah, this gives us a magnification of 1 30 and and so there's a slight difference there. But I will note that both both these answers are acceptable.

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