- #1
7Kings
- 9
- 0
I recently took a test on lenses. One of the questions was "The image formed by a converging lens is real and is n times the size of the object. If the focal length of the lens is f, the distance from the lens to the image is?
Now, the answer is f(n+1) but I can't seem to figure out why. Heres my work so far and what i get. (di being distance of the image, do being distance of the object, f being focal length, and n being magnification
1/di + 1/do = 1/f so therefore 1/di = 1/f- 1/do
n = -di/do so therefore do=-si/n
from combining those two i come up with 1/di = 1/f - -n/di which equals 1/di = 1/f + n/di
then i got a common denominator and cmae up with 1/di = (di + fn) / (f*di)
then i cross multiplied and came up with f*di = di(di+f*n) so i crossed out the two di's and got f = di+fn. and then simply solved for di (what we're looking for) and got di=f-fn and from that di=f(1-n).
Now, the answer says its supposed to be f(n+1)...where did i go wrong? I assume i messed up with a sign somewhere but i can't seem to figure out where.
Now, the answer is f(n+1) but I can't seem to figure out why. Heres my work so far and what i get. (di being distance of the image, do being distance of the object, f being focal length, and n being magnification
1/di + 1/do = 1/f so therefore 1/di = 1/f- 1/do
n = -di/do so therefore do=-si/n
from combining those two i come up with 1/di = 1/f - -n/di which equals 1/di = 1/f + n/di
then i got a common denominator and cmae up with 1/di = (di + fn) / (f*di)
then i cross multiplied and came up with f*di = di(di+f*n) so i crossed out the two di's and got f = di+fn. and then simply solved for di (what we're looking for) and got di=f-fn and from that di=f(1-n).
Now, the answer says its supposed to be f(n+1)...where did i go wrong? I assume i messed up with a sign somewhere but i can't seem to figure out where.