To solve this problem, we can use the magnification formula for a converging lens:
Magnification (m) = -v/u
Where: m = Magnification v = Distance of the image from the lens (in this case, the height of the image) u = Distance of the object from the lens (in this case, the height of the object)
Given that the image formed by the lens is 3 times as tall as the object, we can write:
m = -v/u = -3
Now, let’s substitute the given focal length of the lens into the lens formula:
1/f = 1/v – 1/u
Substituting the given focal length of 12 cm:
1/12 = 1/v – 1/u
Now, let’s rearrange the equation to solve for v:
1/v = 1/12 + 1/u
Substituting the value of m = -3 into the magnification formula, we get:
-3 = -v/u
Rearranging the equation to solve for v:
v = 3u
Now, let’s substitute this expression for v into the equation we derived from the lens formula:
1/3u = 1/12 + 1/u
To simplify the equation, we can find a common denominator:
1/3u = (u + 12)/12u
Now, let’s cross-multiply and solve for u:
12u = 3u(u + 12)
12u = 3u^2 + 36u
3u^2 + 36u – 12u = 0
3u^2 + 24u = 0
Dividing by 3:
u^2 + 8u = 0
Factoring out u:
u(u + 8) = 0
So, either u = 0 or u = -8
Since the distance cannot be negative, we discard u = -8.
Therefore, the distance of the image from the object is u = 0.
This means that the image is formed at infinity, which implies that it is a virtual image