1. The problem statement, all variables and given/known data A man who is farsighted and standing in front of a flat mirror without glasses at some distance to see his image in focus. When he puts on his glasses he comes closer for seeing himself in focus to a distance that is 6 times shorter of the previous distance. a. what is the shortest distance he can see without glasses? b. what is his Diopters c. when he puts on his wife's glasses over his he can see from 20cm distance, what kind of a sight does she have? 2. Relevant equations guessing 1/f=D 1/f = 1/u + 1/v 3. The attempt at a solution a. with glasses he should see from a correct 25cm exactly, because of the mirror the distance is cut by half so 12.5cm distance from mirror because the distance is 6 times more without glasses - his distance is 75 and viewing distance of 150cm b. to put an object in 25cm into virtual image at (-)150cm results in D=3.993 c. object in 20cm needs to get to (-)25cm so with a second lens should put 25 into (-)150cm with that in mind : 1/20 + 1/(-25) = 0.01 so her diopter is 0.01 [m^-1] so I'm guessing she is farsighted... (??) I will be happy for you to review my answer and feedback either on method and results. Thanks.