Put this question. We're told to look at figure p 23.5 to to see that the object is midway between a lens with a focal focal length. F of negative 16.7 centimeters in a mirror with the radius of curvature are 20 centimeters and they're separated by a distance D equal to 25 centimeters. And for part, A wants us to consider only the light that leaves the object and travels first towards the mirror and locate the final image formed by this system. Okay, so to do this, we're gonna use the lens equation which says that one over F which is equal to two over the radius of curvature since the mirror is the first thing it interacts with here, this is going to be equal to one over p one. The distance to the object from the mere plus one over kyu won the distance to the image formed by this. Well, P one is 12.5 centimeters, So we can rearrange this to solve for Kyu won. We see that q one is equal to ah p one times are divided by two times p one minus R so Q one comes out to equal that d centimeters. It's positive. Therefore, it's to the left of the mirror so we can use Kyu won the image produced by this as the object for the second part here. So for part B to find P to the object for the lens, P two is equal to the distance between the two minus Q one. So this comes out to equal negative 25 centimeters. Now, if you go back to this page, we can use this equation here, which I'm gonna underline. One over f is equal to one over p one plus one over kyu won. We can rearrange that then to solve for Q except for this time instead of one. Were using some script too. Right? So que tu is equal to p two times the focal length of the lens divided by P two minus the focal length of the lens playing those values and we find that que tu is equally negative. 50 0.3 centimeters. Weekend box said in his air solution for part A Oh, this wasn't part. Be sorry about that. This was all part of a here. So to race this okay, part be asked if this image that is produced is riel or virtual well, To do that, we just have to look at the sign of Q two. And since que Tu is negative, it's virtual. So if you two would have been positive, it would have been real making box in it. And this is their solution for part B. Parsi asked. If, uh, if the image that is produced is upright or if it's inverted To do that, we need to know the sign of the magnification. So magnification of thing Entire system is equal to the magnification produced by the mirror, which is Kyu won over P one, multiplied by the magnification produced by the lips so minus Q two over Key, too. And we know all those values, so we complex them in. We find that this is equal to positive 8.5 So since the magnification is positive, that means it's upright, so we can say, uh, in is positive. So it's up right? Making box. It is their solution for C in the part D asked for the value of this magnification. We already found that in part, see, so the value of the magnification M is equal to 8.5 making box set in as our solution for D in the final part of our question.