# Homework Help: Focal length of a lens

1. Feb 23, 2013

### JoeN

1. The problem statement, all variables and given/known data

My physics coursework hypothesis: "As the thickness of the lens increases, the focal length decreases". I tested this using a ray box, a stand to hold lenses and a screen to see where the light is most focused, which is then measured with a metre ruler. The distance 'u' (light to lens) was kept constant. A concave lens had to be placed directly in front of the ray box to make the rays parallel.

However, I am having difficulty differentiating between the equation 1/u +1/v = 1/f and the focal length. The focal length is shown on diagrams as being the distance from the middle of the lens to where the rays intercept? What does the equation, then, state? (I know what u and v are).

Also, would the thickness of the lens essentially be the radius of curvature in this scenario, as both sides of the lens have an equal radius?

Finally, due to a lack of data online, I'm unsure whether my results are accurate. The data is displayed below:

lens (mm) u/focal length (m) power (D) f (equation)

4 0.474 2.11 0.184

5 0.259 3.86 0.139

7 0.166 6.02 0.107

8 0.105 9.52 0.078

15 0.046 21.74 0.040

A thing I should note is that while the order of thicknesses is correct, I'm not highly confident that the value for the thicknesses are correct... We had to measure them ourselves, and the equipment wasn't sophisticated at all so the outcome varied among most people.

Note that the distance 'u' is set at a constant 0.3m.

The graph for the focal length shows a curve: is this correct, or should the relationship generally be linear? On the other hand, the graph for power is virtually completely linear, which I would assume is correct?

2. Relevant equations

1/u + 1/v = 1/f
Power (D) = 1/f

3. The attempt at a solution

1) Is the f value in the equation referring to the point where the rays converge, whereas in our experiment using the screen we were measuring the image? (Apparently they are different?)

2) Or would they be two different values, with the radius of curvature being just 2/thickness?

3) Apart from one bit of data lying a tad to the side of the line of best fit, the points all fit the curve pretty well, so I guess that increases my confidence. They're still not as neat as I would like, though.

2. Feb 23, 2013

### PeterO

The formula involves 3 variables. Given any two of them, you can calculate the third.

if u = 50 and v = 20, then the left hand side = 1/50 + 1/20 = 7/100

That tells us 1/f = 7/100 or f = 100/7 so the focal length ≈ 14.2.

Note: if you have used parallel light as your "input", then u = ∞ .

3. Feb 23, 2013

### JoeN

I did use parallel light as the input, but how could that make u infinity?..

And thanks, but I think you missed the question. I know how to use the formula, and know that the output is focal length, but why is this a different value to v? In diagrams the focal length is the middle of the lens to the point where the rays converge.

4. Feb 23, 2013

### haruspex

Light from an object at infinity, and only from such an object, would consist of parallel rays. Your use of the concave lens to make the rays parallel created an image of the light source at infinity. That image is the effective light source from there on.
Yes, the focal length is to the point where (originally) parallel rays converge, but that's not the same as the point where the rays (or back extrapolations of the rays) originating at a given point converge. Draw a convex lens and a line through its (horizontal) axis. Draw a vertical stick standing up from the axis at some distance from the lens. A ray originating at the top of the stick and running parallel to the axis is deflected through the focal point on the far side of the lens. A second ray from the same point but passing through the middle of the lens is undeflected. Extrapolating both rays back through the lens they converge at the location of the image of the point.
No. For a start, that would be inverted - for a given diameter, the thicker the lens the smaller the radius of curvature. The precise relationship you can calculate easily enough by Pythagoras. Draw a lens as two segments of a circle back-to-back and mark the centre of curvature to one side. Draw a right-angled triangle joining the centre of curvature, the centre of the lens, and one edge of the lens.
If you're comparing lenses of different diameters only by thickness then you've no idea how the focal lengths will vary.

5. Feb 24, 2013

### JoeN

I see. So would this affect the value for u? I was measuring from the filament of the lamp to the middle of the lens, but would it be accurate to measure from the concave lens to the middle of the convex lens in that case?

Not entirely sure I follow, sorry. If the focal length is the middle of the lens to the point where the rays converge, and this is a different value to 1/u+1/v, then what does the equation measure? Lens power is calculated using 1/f, where v = f right? I know you've given an explanation but I'm too muddled to make sense of it.

On a final note, is this equation preferable to use in comparing focal lengths in my case compared to lens to image?

So the radius of curvature is just the radius of the circle... or the hypotenuse? But not sure how I would do this as only known value is diameter..

If you're comparing lenses of different diameters only by thickness then you've no idea how the focal lengths will vary.[/QUOTE]

All the same diameter.

Is it the policy not to comment on data? A simple confirmation of whether the relationship between thickness of the lens and focal length is linear or curved would be useful!!

6. Feb 24, 2013

### PeterO

All the same diameter.

Is it the policy not to comment on data? A simple confirmation of whether the relationship between thickness of the lens and focal length is linear or curved would be useful!![/QUOTE]

When the value of u = ∞, the value of 1/u = 0

That rather simplifies your formula to 1/v = 1/f

Of course the chances if putting your convex lens in exactly the correct position so that the rays are exactly parallel is probably low, so you would be far better to repeat your experiment with a bare globe, a screen. The lens you are testing is then placed between the two in the positions that give a sharp image on the screen. The two positions for which that works lets you double check your measurements.

7. Feb 24, 2013

### haruspex

Yes. As far as the convex lens is concerned, the distance u is to the image created by the concave lens.
to the point where parallel rays are made to converge.
Look at it this way: Suppose we have an object at distance u and its image (the point where the rays from that object are made to converge) at distance v. For a given lens, the value 1/u + 1/v is a constant. When u=∞, 1/u = 0, so 1/v equals that constant. We call the value of v when u=∞ infinity the 'focal length', f, and the equation becomes 1/u+1/v=1/f.
Both.
Suppose the centre of curvature is at O and the centre of the lens at A. Mark one edge of the lens as B. OB = R, the radius of curvature. Let C be the point on the surface of the lens that lies on the projection of OA beyond A. OC=R. The diameter is 2*AB. Apply Pythagoras to find OA, then subtract from OC and double to get the thickness.
If you determine the relationship between thickness and radius of curvature, as described above, you'll see it's unlikely to be a straight line. There's also the question of how focal length varies as a function of radius, which may also be nonlinear.

8. Feb 25, 2013

### Basic_Physics

Your setup gave the focal length of the lenses directly. You do not need to calculate them with the formula. The image distances, v, will be focal lengths of the lenses directly. This is how the focal length is defined - it is the image distance for an object located far away. Because the lenses received parallel rays they focussed the rays at the focal point of the lens.

9. Mar 7, 2013

### JoeN

Thanks for the responses by the way, but really need a quick answer to this.

Does the distance from the lens to to the light sources affect the focal length? I would have assumed not, but in a preliminary test the lens was increased in distance from the light source and the focal length decreased. Is this correct?

And how does diameter affect focal length?.. I can never find any information, been looking for hours

Last edited: Mar 7, 2013
10. Mar 7, 2013

### haruspex

Focal length is a property of the lens, so cannot be affected by the position of the light source. But I suspect you are misrepresenting the test info. I can think of two possibilities:
1. The distance to the image changed, not the focal length.
2. There were two different lenses.
It doesn't. The focal length is determined by the curvature. However, a spherical surface on a lens does not give true focus. (Similarly, the ideal shape of a mirror is parabolic, not spherical.) As you increase the diameter of the lens, the outer portions will have a slightly different focal length, leading to "spherical aberration".

11. Mar 9, 2013

### JoeN

I thought as much for the distance. Thanks very much for everyone's help