1. The problem statement, all variables and given/known data A compound microscope consists of two converging lenses. Lens 1, called the objective, is a distance 61.0 mm from lens 2, called the eyepiece. The objective is placed in between the object and the eyepiece. The objective has a focal length of 15.0 mm, and the eyepiece has a focal length of 25.5 mm. If the microscope is used to examine an object placed 24.1 mm in front of the objective, a) what is the distance of the final image with respect to the eyepiece? b) Is the image real or virtual? c) Is the image inverted or non-inverted? d) What is the overall magnification of the microscope? 2. Relevant equations 1/do+1/di = 1/f M=-di/do M=m1m2 3. The attempt at a solution This is what I did for part a and b. 1/24.1+1/di = 1/15 which gives me 39.725. Then 61-39 because they are both between lens 1 and 2. which gives 21. then 1/21 + 1/di = 1/25.5 which gives -128.405 mm. The answer is 130mm to the left of the eyepiece so I got that part right. The because the image distance is negative the image is virtual. Thus I thought this would mean the image would be upright sice the di value is negative and the lens are converging. However it is not can anyone help me with this?