1. The problem statement, all variables and given/known data A compound microscope has an objective of focal length 12 mm and an eyepiece of focal length 50 mm. The lenses have a separation of 90 mm and an object of height 0.3 mm is placed 15 mm from the objective. Calculate: (a) the distance of the intermediate image from the objective; (b) the distance of the final image from (i) the eyepiece, (ii) the objective; (c) the size of the intermediate image; (d) the angular magnification. Answers: (a) 60 mm; (b) (i) 75 mm, (ii) 15 mm; (c) 1.2 mm; (d) 33.3 2. The attempt at a solution (a) (1 / u) + (1 / v) = (1 / f) u = 15 mm, f = 12 mm. Plug in and v = 60 mm (c) (h / ho) = (v / u) ho = 0.3 mm, v = 60 mm, u = 15 mm. Plug in and h = 1.2 mm (d) M = β / α α = ho / D = 0.3 / 250 = 1.2 * 10-3 β = h / (90 mm - 30 mm) = 0.04 M = 33.3 But (b)? I used this graph from the book for my question. And according to this graph the distance of the final image (final virtual image) from the eyepiece is D and is 250 mm. And the distance from FVI to the objective should be more than 15 mm. Maybe I don't quite understand what shall I find. Really stuck on this one. Any help please?