- #1
Rawr
- 15
- 0
A 1.5-cm tall object is placed 0.50 m to the left of a diverging lens with a focal length of 0.20 m. A converging lens with a focal length of 0.17 m is located 0.08 m to the right of the diverging lens. What is the height and orientation with respect to the original object of the final image.
Okay, first, I find the distance of the image with respect to the diverging lens first.
-1/.2 = 1/di + 1/.5
di = -0.1428...
Then, I add that distance image to the converging lens distance to get the object distance for the converging lens.
0.08 + -.1428.. = .2228 m
Then I use the lens equation again except with that object distance and the focal length of the converging lens.
1/.17 = 1/di + 1/.2228
di = 1.395...
Then I take the magnification equation and use hi/h0 = -di/d0
hi / (1.5) = -(1.395)/(.5)
Solve for hi.. and get -4.185.
Apparently, that is not the answer. The answer key says that it is 1.4 cm with an inverted image. I don't really know how they get 1.4. Where did I go wrong in my work? Or am I wwwaaayy off with what I'm doing?
Okay, first, I find the distance of the image with respect to the diverging lens first.
-1/.2 = 1/di + 1/.5
di = -0.1428...
Then, I add that distance image to the converging lens distance to get the object distance for the converging lens.
0.08 + -.1428.. = .2228 m
Then I use the lens equation again except with that object distance and the focal length of the converging lens.
1/.17 = 1/di + 1/.2228
di = 1.395...
Then I take the magnification equation and use hi/h0 = -di/d0
hi / (1.5) = -(1.395)/(.5)
Solve for hi.. and get -4.185.
Apparently, that is not the answer. The answer key says that it is 1.4 cm with an inverted image. I don't really know how they get 1.4. Where did I go wrong in my work? Or am I wwwaaayy off with what I'm doing?