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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

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