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Question 6.1 Homework UnansweredIn observing and examining Bacillus bacteria with the IOOx objective, you estimate that 40 bacteria would fit across the diameter of...

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Question 6.1 Homework UnansweredIn observing and examining Bacillus bacteria with the IOOx objective, you estimate that 40 bacteria would fit across the diameter of the FOV: What is the estimated size of the bacillus bacteria?Select an answerand submit For keyboard navigation, use the up/down arrow keys -lect an answer0.04 micrometers (um)micrometers (um)0.4 micrometers (um)40 micrometers (um)

Question 6.1 Homework Unanswered In observing and examining Bacillus bacteria with the IOOx objective, you estimate that 40 bacteria would fit across the diameter of the FOV: What is the estimated size of the bacillus bacteria? Select an answerand submit For keyboard navigation, use the up/down arrow keys - lect an answer 0.04 micrometers (um) micrometers (um) 0.4 micrometers (um) 40 micrometers (um)



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A dissecting microscope is designed to have a large distance between the object and the objective lens. Suppose the focal length of the objective of a dissecting microscope is $5.0 \mathrm{cm},$ the focal length of the eyepiece is $4.0 \mathrm{cm},$ and the distance between the lenses is $32.0 \mathrm{cm} .$ (a) What is the distance between the object and the objective lens? (b) What is the angular magnification?

So we have how many bacterium or bacteria or Located in the intestinal tract of a human where 100th of the volume of the intestinal tract is occupied by bacteria. And so we have the volume of the intestinal track. This would be equal to the area for the cross sectional area multiplied by the length. This would be equaling then pi r squared times L. Are being. Of course the radius of the intestinal tract. Now we can say that the volume of the bacteria would be equaling to the length of the bacteria, Cute modeling the bacteria as a Cube essentially. So we also have. Then the number of bacteria multiplied by the volume of a single bacteria would be equal to the volume of the intestinal tract divided by 100. Because of 10 1/100 of the intestinal track is going to be bacteria. So the number of bacteria would then be equal to the volume V. Divided by 100 multiplied by the volume of a single bacterium. And so we can then substitute. This would be equal to pi r squared l divided by 100 multiplied by the length of the bacterium quantity cubed. And so we can Settelf and would be equaling two pi multiplied by our in this case two centimeters or 20.2 m quantity squared, Multiplied by the length of seven m. This would then be divided by 100 and multiplied then by 10 to the negative six m or one microphone Raised to the 3rd power. And so we find that the end is going to be equal to approximately nine times 10 to the 13th. Uh So bacteria the microbes and microorganisms nine times 10 to the 13th. Microorganisms uh contained in the average intestinal tract. So uh and then for part B. Are these bacteria dangerous? No they are actually. Well this would be your answer. Pro A. And then B. We can say no not dangerous. In fact not dangerous. They are beneficial. This is kind of the more biology but it is beneficial to have these microorganisms in your body simply because they are allow you to help digest the food and actually get the nutrients from the food and efficiently process the food in order for it to provide energy for your body. So no the not dangerous their beneficial to one's intestinal tract. And it would actually be a massive problem if you didn't have this bacteria in your in your intestinal tract that is the end of the solution. Thank you for watching.

Hi there. So for this problem we have a microscope of the type that is shown in this figure, we are given the focal length of the objective, we are going to call this F. O. B. And that has a value of four cm. And we are also given the focal length for the I. P. S. And that is equal to eight cm. Now the distance between the lenses it is a given volume of 25 cm and that of course is the sum of all of that. We are going to call this the distance Eld and what we need to obtain for part A of this problem is the tooth length adds. Mhm. Yeah. Mhm. Now, from the figure above we can see that this distance from the length of the objective and the length of the I PS. That value is given that is the distance. ELT. Now um from here we will have that that distance elf is the sum of the distance X plus the focal distance of the objective plus the focal distance of the eye piece. So in here what we need to do is to solve for the distance eggs. So we will have that, that distance adds is equal to the length Eld minus the focal distance of the objective minus the focal distance of the I PS. And we just need to substitute all of these values. So we will have for the distance l we are given that that is a value of 25 centimeters minus the focal distance of the objective. That is also given that is four centimeters And minus the focal distance of the eye piece which is given and that is eight cm. So from here we obtain that the um tooth length is equal to 13 center matures no for part B of this problem we are told that the if the image I that is in, in in this figure that is shown in this figure, the image I This one right here. The problem is that is that is to be just inside the focal point at Prime one. How far from the objective should should the object B. No. To calculate that, we use the following equation that is um that is that one over the object distance plus one over the image distance is equal to one over the focal distance of the objective. Now and here since we want the object distance, what we need to do is to solve for the the object distance be so doing that, we will find that that is equal to the Well, we'll have the here we'll have that, that is one over the focal distance of the objective -1 over the Image Distance and this elevated to the -1. No. And the image distance in this case, as you can see from the figure, is this some of the, is the sum of the focal distance of the objective and the distance adds. So we will have the thirties focal distance of the objective plus the and length of the tooth that is ads. So in here with some those two values that we know it, Their values and we will obtain that that is 17 cent immature. Now what we need to do is to introduce that value into the equation for the image distance be so we will have that, that is um won over difficult um Yeah The focal distance of the objective which is four cm -1 over the image distance that is 17 cm and all of these elevated to the -1. Now solving for this, we obtain about you of 5.23 cm. Now for parsi of this problem we are asked about the lateral magnification m. Of the objective. Now we know that the lateral magnification is defined as minus the ratio of the image distance over the object distance speed. Now in this case the image is the one that we have in here that we calculate in part a this one right here, 17 so we have minus 17 centimeters. And the object distance, the one that we just have obtained in part B of this problem, that is 5.23 centimeters. So from this we obtain a negative value of minus 3.25. Now, for part D of this problem we are asked about the angular magnification of of the eyepiece. No, we know that the angular magnification of the I ps is going to be the length ELT that is given over the difficult distance of the eye piece. Now from here, we just substitute those values 25 centimeters over the focal distance of the eyepiece. That is also given and that is eight centimeters. And from here we obtain a value of 3.13. And in the final question for this problem is the overall magnification of the microscope. Now we know that the overall magnification of the microscope is the product between the magnification of the objective and the magnification of the eyepiece. So we will have magnification of the objective time, the angular magnification for the eye piece. So we saw, we substitute those two values, so we will have minus three point 25 times 3.13. So from here we obtain a value of -10.2. So this is a solution for this problem.

Hi there. So for this problem we have a microscope of the type that is shown in this figure. Now the focal length of the objective is given and we call that F. O. P. And that is a body of force into matures. Yeah. And that of the eyepiece that we call F E Y. Is equal to a sentiment yours and the distance between the lenses. The distance adds studies in here. Well, sorry the list then the total distance L'd. This total distance is equal to held and that distance is equal to 25 centimeters now for part A of this problem we need to determine the distance adds which is this distance right here in the figure now, which corresponds to the distance of the tube Now, um as you can see from the picture, the distance ELT is equal to the sum of the focal distance of the objective plus the focal distance of the eyepiece plus the distance acts. So what we need to do is to solve for the distance adds the line of the tooth. So we just passed these two values to the left side of this equation. So we will have l minus the focal distance of the objective, miners, the focal distance of the pipes and we substitute those values in here. So we have For L 25 cm minus difficult distance of the objective which which is for centimeters minus eight centimeters. So from here we obtain a value off 13 centimeters now for part B of this problem, what we need to calculate is, and we are told that if the image I as a shown in this figure, the image I okay, it tells us that is to be inside the focal point F one prime. So what we need to determine is how far from the objective should the object B. Now, what we need to calculate in this case is what we are going to call the distance P of the object. No, to obtain this, we saw from these following equation that is one over the object distance plus one over the image distance, it's equal to one over the focal distance of the objective. So and we know also that the image distance is equal to the phone call distance of the objective. Plus the distance adds since we already know these two values, we just simply substitute that in here. So it is forcing temperatures Plus 13 cm. So this will give us a value of 17cm now substituting into this equation and solving for P for the object distance we will find that this is equal to the product between the image distance times the object distance Over the image distance minus the focal distance of the objective. So we substitute all of these values we have for the image distance 17 cm. The object and the faculty since of the objective is for centimeters and in the denominator we have 17 cm -4 cm. So from here we obtain a positive value of 5.23 cm now for part C of this problem we are asked about the lateral magnification of the objective. Now to obtain that we know that the lateral magnification is the finest minus the ratio between the image distance and the object distance. So we use simply substitute those values in here. We have for the image distance, 17th and amateurs and for the object distance we have 5.23 Centim. So from here we obtain about you for the magnification of -3.25. Now for party of this problem we are asked about the angular magnification of the I PS. Now we know that the angular magnification for that I PS. Is the distance. ELT over the um difficult distance of the I PS. So and here we substitute those values. We have 25 centimeters for the distance L'd and for the um focal distance of the pipes, we have aids and damages. So solving for this we will find that This magnification is 3.13. And now for party of this problem we are asked about the overall magnification of the microscope And that is the product between the magnification of the objective tiny magnification of the eyepiece. So we just substitute those two values in here. So we have minus 3.25 times 3.13 And this is equal to -10.2. So this is a solution for this problem

Part A of our question wants us to calculate the length of the tube. We're gonna call that length s that's equal to the distance between the eyepieces l minus the focal length of each So minus F. Savo the focal length of the objective and the minus. FCB the focal into the eyepiece plugging those values. And we find that this is equal to 13 centimeters. Weakened box said then is their solution for part a part B wants us to find the distance to the to the object that produces a kn image here. So this would be p using the lens equation. Here we have part B. We have one over, uh f sub zero. The focal length of the objective is equal to won over I plus one over p. Here I is equal to the focal length of the object or the objective. Plus the length of the tube s so rearranging this equation to solve for P, we find that P is equal to won over uh, the focal length, the objective one over at zero, minus one over I, which we just indicated waas at zero plus s and this is all to the minus one. So that's how we got p out of the denominator. We find that this is equal to 5.23 centimeters making box. I didn't. Is our solution for B. Carsey wants us to find the lateral magnification. We can call this imp sub l and this is equal to minus I over key. Well, as we said, I is equal to, uh, with the focal length of the, uh, objective. So this would be minus F zero plus the two blank s all divided by p plugging those values. And we find the lateral magnification is equal to negative 3.25 Part D wants us to find the angular magnification. We'll call this, uh, in fada. This would be equal to, uh, l divided by fcv. So plucking those values and we find that this is equal to 3.13 African boxing is their solution for D. Then lastly, Part E wants us to find the total magnification. We'll just call this end, and that would be equal to himself. L well supplied by and Fada. So to find this overall magnification, we just simply plug in the values for a minute and melon and data multiplied together and we find this is equal to negative 10.2 weaken box. That is the total magnification of the microscope.


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