An object 1.5cm high 12.0cm from a diverting lens of focal length 3.0cm. Find the length and position of the image

Answer 1

To solve this problem, we can use the lens formula which states:

1/f = 1/v - 1/u

where:
f = focal length of the lens (3.0 cm)
v = distance of the image from the lens (unknown)
u = distance of the object from the lens (12.0 cm)

We can rearrange the formula to solve for v:

1/v = 1/f + 1/u

Now, let's substitute the given values:

1/v = 1/3.0 + 1/12.0

Simplifying the equation:

1/v = (4 + 1) / 12.0 = 5/12.0

Now, find the reciprocal of both sides:

v = 12.0/5 = 2.4 cm

So, the length of the image is 2.4 cm.

To find the position of the image, we can use the magnification formula which states:

m = -v/u

where:
m = magnification
v = distance of the image from the lens (2.4 cm)
u = distance of the object from the lens (12.0 cm)

Now, let's substitute the given values:

m = -2.4 / 12.0 = -0.2

Since the magnification is negative, it tells us that the image is inverted.

Therefore, the length of the image is 2.4 cm and its position is 0.2 times the distance of the object from the lens, in this case, -2.4 cm.