Solving the Problem of Converging Lenses' Focal Point

[raiieanna]

I am having trouble with my homework. I know the answer but I can't figure out how to get there. The question is "An image of a flower is produced on a screen by a lens 1.60m from the screen. The image of the flower has a linear magnification of -2.5. What is the focal point of the lens?" I know the 1/o + 1/i = 1/f. I also know that
m = -i/o. Also, the image distance 1.6m. So, shouldn't -2.5=-1.6/o? But when I do this I get the wrong answer! Please help me!

 

[Integral]

You have
$$\frac 1 o + \frac 1 i = \frac 1 f$$
and
$$-2.5 = -\frac i o$$
and
$$i= 1.6m$$
so
$$i = 2.5 o$$
$$1.6m = 2.5o$$
$$o = \frac {1.6} {2.5}m$$

Can you finish it now?

 

[M98Ranger]

Hi there,

I can understand how this problem can be confusing, but let's break it down step by step to solve it.

First, we know that the distance between the lens and the screen is 1.60m. This is the image distance (i) in the formula 1/o + 1/i = 1/f.

Next, we are given the linear magnification (m) of -2.5. This means that the image is 2.5 times smaller than the object. Since the formula for linear magnification is m = -i/o, we can rearrange it to solve for the object distance (o). This gives us o = -i/m.

Now, we can substitute the values we have into the formula 1/o + 1/i = 1/f. This gives us 1/(-i/m) + 1/i = 1/f. Simplifying this equation, we get -1/i + 1/i = 1/f. This cancels out, leaving us with 1/f = 0. Therefore, f = infinity.

This might seem like the wrong answer, but remember that we are dealing with a converging lens. This means that the focal point is actually on the opposite side of the lens from the object. In other words, the focal point is behind the lens.

To find the actual focal point, we need to use the sign convention for lenses. In this case, since the object is on the same side as the focal point, the object distance (o) is positive. And since the image is on the opposite side of the focal point, the image distance (i) is negative. This means that the formula becomes 1/o + 1/i = 1/f. Substituting in the values we have, we get 1/1.60 + 1/-1.60 = 1/f. Solving for f, we get f = -0.64m.

So, the focal point of the lens is -0.64m, or 64cm, behind the lens.

I hope this helps you understand how to solve this problem. Just remember to pay attention to the sign convention and the direction of light rays when dealing with converging lenses. Best of luck with your homework!