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.