A lens of focal length 12cm forms an image three times the size of the real object, what is the distance between the object and the image

Answer 1

To solve this problem, we will use the lens formula:

1/f = 1/v - 1/u

Where:
f = focal length
v = image distance
u = object distance

Given:
f = 12 cm (focal length)
The image is three times the size of the real object.

Let's assume the size of the real object is h, and the size of the image is 3h.

Using the magnification formula:

magnification (m) = -v/u = height of image/height of object

We can substitute the given values:

- v/u = 3h/h

Simplifying the equation, we get:

- v/u = 3

Since h cancels out, we know that magnification is only dependent on the ratio of v/u.

Now, let's substitute the values into the lens formula:

1/12 = 1/v - 1/u

Since magnification (m) = -v/u = 3, we can replace -v/u with -3:

1/12 = 1/v + 1/(3u)

To simplify the equation, let's find a common denominator:

1/12 = (1/(3u) + 1/v)((3u)(v)/(3u)(v))

1/12 = (v + 3u)/(3uv)

Cross-multiplying, we get:

1 = 4uv + 36u

Rearranging the equation:

4uv + 36u - 1 = 0

Solving this quadratic equation for u, we find two roots. However, we are only interested in the positive root, as the distances must be positive values.

u ≈ 0.004 cm

Now, substituting this value into the lens formula:

1/12 = 1/v - 1/0.004

Solving for v, we find:

v ≈ 0.00412 cm

Therefore, the approximate distance between the object and the image is 0.00412 cm.