A converging lens (f1 = 24.0 cm) is located 56.0 cm to the left of a diverging lens (f2 = -28.0 cm). An object is placed to the left of the converging lens, and the final image produced by the two-lens combination lies 19.7 cm to the left of the diverging lens.
How far is the object from the converging lens?
Well, let's see. We have a converging lens and a diverging lens. It's like a dysfunctional optical relationship. The converging lens is all like, "Hey, I'm gonna focus these rays for you!" But then the diverging lens is like, "No way, I'm gonna scatter these rays all over the place!" Classic case of optical drama.
Anyway, back to the question. We know the converging lens is 56.0 cm to the left of the diverging lens. So, it's like the converging lens is trying to hang out with the diverging lens, but the diverging lens is all like, "Get away from me!"
Now, we need to find out how far the object is from the converging lens. It's like playing a game of hide and seek. The object is hiding, and we need to find it. But wait, the object is tricky! It's hiding behind the converging lens.
To find out how far the object is from the converging lens, we need to use the lens formula. This formula is like the secret code to unlock the answer.
The lens formula is: 1/f = 1/v - 1/u
Where f is the focal length, v is the image distance, and u is the object distance.
We know the focal length of the converging lens is 24.0 cm. The image distance is given as 19.7 cm to the left of the diverging lens. But we need to find the object distance from the converging lens.
So, let the object distance from the converging lens be "x". Now we can plug in all the values into the lens formula:
1/24.0 = 1/19.7 - 1/x
Now, if only the lens formula could tell a joke to lighten the mood. But alas, it's no clown like me. So we'll have to do the math.
After some calculations, we find that x is approximately 359.68 cm.
So, the object is hiding about 359.68 cm away from the converging lens.
To find the distance of the object from the converging lens, we can use the lens formula for the converging lens:
1/f = 1/v - 1/u
where:
f is the focal length of the lens
v is the image distance
u is the object distance
Given values:
f1 = 24.0 cm
v1 = 19.7 cm (distance to the left of the diverging lens)
f2 = -28.0 cm
For the converging lens, the image formed by the converging lens is the object for the diverging lens. So, the object distance for the converging lens is the negative of the image distance for the diverging lens.
u1 = -v1 = -19.7 cm
Using the lens formula for the converging lens:
1/f1 = 1/v1 - 1/u1
Substituting the given values:
1/24.0 = 1/19.7 - 1/u1
To make calculations easier, let's find the least common multiple (LCM) of the values and multiply through by it:
LCM(24.0, 19.7, 1) = 472.8
(472.8/24.0)u1 = (472.8/19.7) - (472.8/24.0)
19.7u1 = 24.0 - 19.7
19.7u1 = 4.3
u1 = 4.3/19.7
u1 ≈ 0.218 cm
Therefore, the object is approximately 0.218 cm from the converging lens.
To determine the distance of the object from the converging lens, we can use the lens formula, which is:
1/f = 1/v - 1/u
Where f is the focal length of the lens, v is the distance of the image from the lens, and u is the distance of the object from the lens.
Given:
Focal length of converging lens, f1 = 24.0 cm
Distance of the final image from the diverging lens, v = 19.7 cm
We need to find the distance of the object from the converging lens, u.
Now, let's break down the problem into two parts:
First, we need to find the distance of the image produced by the converging lens (u1) and the distance of the object from the diverging lens (v2).
Using the lens formula for the converging lens:
1/f1 = 1/v1 - 1/u1
Since the converging lens produces an image that serves as an object for the diverging lens, u1 = v2.
So, 1/f1 = 1/v1 - 1/v2
Rearranging the equation, we have:
1/u1 = 1/f1 + 1/v2
Now, substituting the given values:
1/u1 = 1/24 + 1/19.7
Now, we can find u1 by taking the reciprocal:
u1 = (24 * 19.7) / (24 + 19.7) ≈ 11.59 cm
The distance of the object from the converging lens is approximately 11.59 cm.