Answer :
Answer:
it can be 15 or 75
Explanation:
using this equation
[tex] \frac{1}{f} = \frac{1}{so} + \frac{1}{si} [/tex]
f = focal length
so = object distance
si = image distance
since we do not now what lens used, I assume from the problem the lens are positive (convex) but I don't know the image position because lens have two side.
1. the positif lens with negative image distance (it means object and image are in the same side of lens)
1/25 = 1/so - 1/37.5
1/so = 1/25 + 1/37.5 = 1/15
so = 15 cm in front of the lens (same side with the image)
2. the positif lens with positive image distance (it means object and image are not in the same side of lens)
1/25 = 1/so + 1/37.5
1/so = 1/25 - 1/37.5 = 1/75
so = 75 cm in front of the lens (not in the same side with the image)
Answer:
Its 15 this is the same as my question except the distance is given
Explanation: