What is the focal length of the lens?

[AN630078]

Homework Statement

Hello, I have a question below which I am rather uncertain how to approach concerning an investigation into the lens formula, I have attached the table of results from the question also. I am especially stumbling on sections of the question concerning uncertainty and would be very grateful of any help. I have attempted to answer all of the questions, and in places where there are an absence of solutions I have explained where I am stuck.
Sorry this is rather a lengthy post, clearly the questions lead on from another which is why I have posted them collectively in case amending one section may advance another area which I am having difficulty with. Thank you to anyone who replies.

In an experiment, students varied the distance (u) between a point source of light and a lens, and measured the corresponding distance (v) from the lens to a screen on which a real image was formed. The students found a range of positions on the screen for each distance u for which the image looked quite sharp. The students chose to record the the smallest and largest possible values of v for each value of u.

Question 1.
ii Suggest a reason why any uncertainty in the values of u can be neglected.
ii. Describe the variation in the uncertainty shown by v

Question 2
i. Draw a table of the values of 1/u and the largest and smallest values of 1/v.
ii. Plot the data from your table in a graph of 1/v against 1/u. Include error bars and add a straight line of best fit.
iii. Use the graph to find the focal length of the lens.

Relevant Equations

1/u+1/v=1/f
y=mx+c

Question 1.
i Suggest a reason why any uncertainty in the values of u can be neglected.

I am really rather unsure how to answer this question but I have produced my opinion nonetheless.
One could assume that the uncertainty in object distance u, i.e., the distance from the lens to the light source is very small compared to the uncertainty of the image distance, i.e., the distance from the lens to the image formed on the screen.
This yields an object uncertainty of such a small magnitude that it becomes negligible and can be omitted.

Question 1 ii. Describe the variation in the uncertainty shown by v

I truly do not know how to answer, is this referring to the difference in the largest and smallest image distances?
If this is correct, then the variation in the uncertainty shown by the image distances greatly decreases as the object distance increases, for example this difference decreases from 0.2m at an object distance of 0.22m to 0.14m when u=0.28 to 0.03 m when u=0.33 etc.

Question 2
i. Draw a table of the values of 1/u and the largest and smallest values of 1/v.
I have calculated the reciprocal of the original values and presented these in a table which I have attached to 3.s.f.

ii. Plot the data from your table in a graph of 1/v against 1/u. Include error bars and add a straight line of best fit.

I do not have trouble in plotting this graph, however, I am uncertain of the the values to plot for the reciprocal of the image distance (1/v), would this be the average of the smallest and largest image distances?

i.e. when 1/u=4.54 m, would 1/v=(1.12+0.917)/2=1.019 m to 3.s.f ?

Also, I do not know how to include error bars in my graph, I have not been taught to do so. I know that percentage uncertainty= absolute error/measurement * 100%

The absolute uncertainty would be ±0.5 mm if a ruler was used with a resolution of 1mm.
How would this be applicable, if at all, to producing error bars. I have read my textbook but it does not explicate this at all, only the difference between random and systematic errors. Although I have also looked online I have not found a great deal of resources on how to produce error bars, and admittedly what I have found has been a little confounding and not especially applicable to this scenario.

iii. Use the graph to find the focal length of the lens.

Once I have produced my graph, I can compare the lens formula 1/u+1/v=1/f with that for a straight line graph; y=mx+c.
Clearly, the y-intercept of the graph will be the reciprocal of the focal length.

 

[Tom.G]

Following site found with:
https://www.google.com/search?&q=error+bars+on+graph

https://datavizcatalogue.com/methods/error_bars.html
The ends of the Error Bars would be at "...the smallest and largest possible values of v..."

If you wish to include the "most likely" (or Data Point) value of "v" on the error bars, in this case it would typically be the geometric mean of the two "v" values. Considering this is probably an Algebra 1 course, you could probably get away with using the average, the results will be very close to each other.

Geometric mean here would be:
SQRT( V_SMALLEST x V_LARGEST ).

Cheers,
Tom

 

[AN630078]

Following site found with:
https://www.google.com/search?&q=error+bars+on+graph

https://datavizcatalogue.com/methods/error_bars.html
The ends of the Error Bars would be at "...the smallest and largest possible values of v..."

If you wish to include the "most likely" (or Data Point) value of "v" on the error bars, in this case it would typically be the geometric mean of the two "v" values. Considering this is probably an Algebra 1 course, you could probably get away with using the average, the results will be very close to each other.

Geometric mean here would be:
SQRT( V_SMALLEST x V_LARGEST ).

Cheers,
Tom

Thank you for your reply.

Question 2 i. So the error bars I am required to plot would here represent the maximum and minimum values of v?
Also, when drawing the error bars would this be extending from the mean data point of "v" to to the maximum and minimum values of v to exhibit the uncertainty of a data point.
The shorter the error bar the more concentrated the values of data are to the average whereas the longer the error bar the more dispersed the values and thus the less reliable the average point plotted is.
Thank you for showing me how to find the average value of v using the geometric mean that is very helpful. Is it better suited to this data than the arithmetic mean as it better describes the central tendency? I have included a column for the geometric mean in my original table, which I have attached.

I have drawn and attached what I think the graph of 1/u against 1/v should look like. Would this be correct? Sorry I know that it is a little rough I have just jotted it down hastily.
I notice that the error bars extend further the greater the value of 1/u, as the values of 1/v exhibit a greater dispersion and spread of values. In comparison, the smaller values of 1/u, the spread of values of 1/v is more closely aligned to the geometric mean.

I have also plotted the data on desmos, just to try to be a little more precise perhaps than my graph, which I have also attached, and attempted to draw a line of best fit on it also, though I am not sure how to add error bars in desmos so these have been omitted.

ii. Comparing the lens formula 1/u+1/v=1/f with that for a straight line graph; y=mx+c the y-intercept of the graph will be the reciprocal of the focal length.
The y-intercept is ~ 1/5.45 m
Therefore, f~5.45 m ?

Question 1 ii. I think from observing the data of the image distance in the graphical form, I would more firmly comment that the uncertainty in v is greater as the object distance,u, increases. This is evident in the more pronounced error bars which are less concentrated at greater values if 1/u, compared to the concentrated data values of 1/v at lesser values of 1/u. Would this be correct. Also would my earlier response to question 1 i. be suitable?
Thank you very much again for your help

 

[Tom.G]

I agree with your tables and graph. Good job!

re. Question 2 i
Error bars: represent the Max & Min of 1/v, not of v as you asked. but they are shown correctly on graph.
Shorter & longer Error Bar description: Yes

Geometric Mean vs Arithmetic Average: The geometric mean gives a value that has the same multiplicative factor to the two limits (measurements). For instance two readings of 1 and 10.
The geometric mean is 3.317,
3.317 x 3.014 = 9.997
3.317 /3.014 = 1.1
The geometric mean is 3.16,
3.16 x 3.16 = 9.99
3.16 /3.16 = 1

re. Question 2 ii:
1/5.45m You missed taking the reciprocal on the next line.

re. Question 1 ii:
Since you are plotting reciprocal of the readings, you are also seeing the reciprocal. Please re-think that.

re. Question 1 i:
I don't know if your answer is applicable or not, but it doesn't seem quite right.

Anyone else here willing to contribute on this?

Cheers,
Tom

 

[AN630078]

I agree with your tables and graph. Good job!

re. Question 2 i
Error bars: represent the Max & Min of 1/v, not of v as you asked. but they are shown correctly on graph.
Shorter & longer Error Bar description: Yes

Geometric Mean vs Arithmetic Average: The geometric mean gives a value that has the same multiplicative factor to the two limits (measurements). For instance two readings of 1 and 10.
The geometric mean is 3.317,
3.317 x 3.014 = 9.997
3.317 /3.014 = 1.1

re. Question 2 ii:
1/5.45m You missed taking the reciprocal on the next line.

re. Question 1 ii:
Since you are plotting reciprocal of the readings, you are also seeing the reciprocal. Please re-think that.

re. Question 1 i:
I don't know if your answer is applicable or not, but it doesn't seem quite right.

Anyone else here willing to contribute on this?

Cheers,
Tom

Thank you for your reply. Oh so have I calculated the geometric mean incorrectly in my table? I used the formula you stated to be SQRT( VSMALLEST x VLARGEST ), but how is the geometric mean of 1 and 10, found by SQRT(1*10), equal to 3.317 in the example you have given? I calculate SQRT(10)=3.16 to 3.s.f?

I thought the reciprocal of 1/5.45 was 5.45/1 which is 5.45m?
Also, I realize that I am plotting the reciprocals of the object and image distances, however, I read that using a graph of 1/v against 1/u one can find the reciprocal of the focal length to be the value of the y-intercept.

I am still uncertain on question 1 i. also, why do you think that it does not appear quite right?

Thank you again for your reply and help

 

[Tom.G]

Thank you for your reply. Oh so have I calculated the geometric mean incorrectly in my table? I used the formula you stated to be SQRT( VSMALLEST x VLARGEST ), but how is the geometric mean of 1 and 10, found by SQRT(1*10), equal to 3.317 in the example you have given? I calculate SQRT(10)=3.16 to 3.s.f?

You are right! My mistake on that. (Oops, corrected in post 4, above) [I incorrectly computed SQRT( sum ) instead of SQRT( product )]

I thought the reciprocal of 1/5.45 was 5.45/1 which is 5.45m?

When doing symbolic math (re-arranging an equation), if you have a fraction in the denominator of an expression you can invert that fraction and move it to the numerator of the expression. You are taking the reciprocal of a fraction. Conceptually this is a 'double inversion' which yields the same final result. In your current problem you need the reciprocal of a number, 5.45, not the reciprocal of a fraction.

To find the numeric value of a reciprocal you have to do the division, for instance the reciprocal of 25 is 0.04.

I am still uncertain on question 1 i. also, why do you think that it does not appear quite right?

• The measurements were taken for u and v for magnifications both greater and less than unity.
• I assumed that the readings for both the u and v were taken by the same people using the same measuring device.

Therefore, I would expect the errors for both u and v would be similar when their absolute values were similiar.

Cheers,
Tom

 

[AN630078]

You are right! My mistake on that. (Oops, corrected in post 4, above) [I incorrectly computed SQRT( sum ) instead of SQRT( product )]

When doing symbolic math (re-arranging an equation), if you have a fraction in the denominator of an expression you can invert that fraction and move it to the numerator of the expression. You are taking the reciprocal of a fraction. Conceptually this is a 'double inversion' which yields the same final result. In your current problem you need the reciprocal of a number, 5.45, not the reciprocal of a fraction.

To find the numeric value of a reciprocal you have to do the division, for instance the reciprocal of 25 is 0.04.

• The measurements were taken for u and v for magnifications both greater and less than unity.
• I assumed that the readings for both the u and v were taken by the same people using the same measuring device.

Therefore, I would expect the errors for both u and v would be similar when their absolute values were similiar.

Cheers,
Tom

Thank you very much for your reply. Oh so in order to find the numeric value of the reciprocal would this be 1/5.45=0.183486...~0.18 ? Would this mean the focal length is 0.18m?

What do you mean in your statement by unity? Yes, I agree with the rest of your statement though, that you would expect similar error for v and u when their absolute values are similar.
Thank you again for your help

 

[Tom.G]

What do you mean in your statement by unity?

Unity magnification is when the Image size is the same size as the Object. (the ratio of their sizes is 1:1) In this experiment that definition is difficult to detect because the light source is essentially a point. The other way to find it is when the distance from Object to Lens is the same as the Lens to Image distance, with the Image in focus. Or stated yet differently, when the lens focuses the Image with the lens half way between the Object and Image.

Would this mean the focal length is 0.18m?

That is the number I get.

A sanity to help confirm your calcs.:
Take advantage of entries in your 1:u 1:v table. If you look at table lines 3 & 4 you will see on line 3 that 1/u is larger than 1/v. But line 4 shows 1/v larger.

. . . . .1/u . 1/v
ln3: 3.03 2.56
ln4: 2.50 3.08

This shows that the unity magnification point is between those two distances.
A closely related shortcut is:
If the 1/u and 1/v are the same then the focal length is the distance between the Object and its Image. This is a direct result of the lens formula of 1/f = 1/u + 1/v

Cheers,
Tom

 

[AN630078]

Unity magnification is when the Image size is the same size as the Object. (the ratio of their sizes is 1:1) In this experiment that definition is difficult to detect because the light source is essentially a point. The other way to find it is when the distance from Object to Lens is the same as the Lens to Image distance, with the Image in focus. Or stated yet differently, when the lens focuses the Image with the lens half way between the Object and Image.

That is the number I get.

A sanity to help confirm your calcs.:
Take advantage of entries in your 1:u 1:v table. If you look at table lines 3 & 4 you will see on line 3 that 1/u is larger than 1/v. But line 4 shows 1/v larger.

. . . . .1/u . 1/v
ln3: 3.03 2.56
ln4: 2.50 3.08

This shows that the unity magnification point is between those two distances.
A closely related shortcut is:
If the 1/u and 1/v are the same then the focal length is the distance between the Object and its Image. This is a direct result of the lens formula of 1/f = 1/u + 1/v

Cheers,
Tom

Thank you for your reply. So you are saying that to find the focal length this is when 1/u and 1/v and the same, and the focal length corresponds to the distance between object and image. I cannot see anywhere when 1/u and 1/v are the same though?

Also, I had been taught that the 1/f was equal to the y-intercept of the graph, therfore, would it not be better to find the focal length from this intercept? Was f=1/5.45=0.18 m incorrect? Sorry that I still sound a little confused.

 

[Tom.G]

Either the 0.18m is correct or we both made the same mistake! That is the number I got... and it is compatible with the 'sanity check' with lines 3 & 4 in the 1:u 1:v table.

Yes, the graph y-intercept is at 1/f. I supplied the additional information on other methods more as comments and alternative ways to find or approximate the focal length. They can sometimes be useful as 'sanity checks' for a calculation or estimation. Sorry if they were confusing.

Cheers,
Tom

 

[AN630078]

Either the 0.18m is correct or we both made the same mistake! That is the number I got... and it is compatible with the 'sanity check' with lines 3 & 4 in the 1:u 1:v table.

Yes, the graph y-intercept is at 1/f. I supplied the additional information on other methods more as comments and alternative ways to find or approximate the focal length. They can sometimes be useful as 'sanity checks' for a calculation or estimation. Sorry if they were confusing.

Cheers,
Tom

Thank you for your reply. No I truly appreciate your help and welcome new methods, it is brilliant to have an alternative. So the difference between 3&4 in the tables does correspond to 0.18 m then? Could you show how this is true, if it is?

e.g. in line 3 Would it be that the difference between 1/u and 1/v be the difference is 3.03-2.56=0.47
so f=1/0.47=2.12

I think this is wrong I just cannot see how you have calculated 0.18m in this way. I am sorry to continue misunderstanding and do earnestly appreciate your help

 

[Tom.G]

I seem to be having trouble expressing the concept. I have asked others here, Science Advisors and Mentors, to come in and contribute. Please be patient, it sometimes takes two days or so for the right people to spot a request and respond.

Thanks,
Tom

 

[AN630078]

I seem to be having trouble expressing the concept. I have asked others here, Science Advisors and Mentors, to come in and contribute. Please be patient, it sometimes takes two days or so for the right people to spot a request and respond.

Thanks,
Tom

Oh ok, thank you very much for your help. I think you have been expressing the concept well, I am afraid it is just my own miscomprehension.

 

[Charles Link]

Hello.
The graph of $\frac{1}{v}=-\frac{1}{u}+\frac{1}{f}$ is a somewhat clumsy way to find the focal length. It works, but it is a little lengthy. One way that is easier is to just look at one pair of $\frac{1}{u}$ and $\frac{1}{v}$. You can add them together and get $\frac{1}{f}$, which you can see is approximately $5.5$ for every pair of inverse points. This means $f \approx \frac{1}{5.5}$.

 

[BvU]

a somewhat clumsy way to find the focal length

In the sense that it takes some processing.
NOT in the sense that a reduction of the uncertainty in the result is attempted by doing six observations !

 

[AN630078]

Hello.
The graph of $\frac{1}{v}=-\frac{1}{u}+\frac{1}{f}$ is a somewhat clumsy way to find the focal length. It works, but it is a little lengthy. One way that is easier is to just look at one pair of $\frac{1}{u}$ and $\frac{1}{v}$. You can add them together and get $\frac{1}{f}$, which you can see is approximately $5.5$ for every pair of inverse points. This means $f \approx \frac{1}{5.5}$.

Thank you for your reply and for suggesting the approach of adding 1/u to 1/v to find 1/f. Does this mean that the focal length is ~ equal to 5.5 m here?
Was I wrong in earlier stating it to be 1/5.45 ~ 0.18m?

 

[AN630078]

In the sense that it takes some processing.
NOT in the sense that a reduction of the uncertainty in the result is attempted by doing six observations !

Thank you for your reply also

 

[Charles Link]

In the sense that it takes some processing.
NOT in the sense that a reduction of the uncertainty in the result is attempted by doing six observations !

It would be far easier to compute $f$ individually for the 6 cases, and take the average. I think part of the exercise here is to teach the student to do a graphical analysis, with some algebra to make the data points into something that gives a linear function.
To add to my post 14, I think @Tom.G was trying to locate the point where $u=v$, as a simple check on the data. The result there is that $u=v=2f$. Using his approach, $u=v$ at approximately $u=.36$, making $f \approx .18$.

 

[Charles Link]

Thank you for your reply and for suggesting the approach of adding 1/u to 1/v to find 1/f. Does this mean that the focal length is ~ equal to 5.5 m here?
Was I wrong in earlier stating it to be 1/5.45 ~ 0.18m?

The $f \approx .18$ m is correct.

 

[AN630078]

It would be far easier to compute $f$ individually for the 5 cases, and take the average. I think part of the exercise here is to teach the student to do a graphical analysis, with some algebra to make the data points into something that gives a linear function.
To add to my post 14, I think @Tom.G was trying to locate the point where $u=v$, as a simple check on the data. The result there is that $u=v=2f$. Using his approach, $u=v$ at approximately $u=.36$, making $f \approx .18$.

Thank you again for your reply. Yes, I agree that I think this exercise is testing one's graphical analysis, however, I think I would perhaps take the average as you have suggested in future. Oh right I see, so f would not equal 5.45m?

I better understand what @Tom.G was suggesting by his 'sanity check'. Thank you again

 

[Charles Link]

$f =5.45 m$ is incorrect. Just from a practical standpoint, this number is very impractical. Typically a magnifying glass may have a focal length of about $8" \approx 20 \, cm=.2 m$, while focal lengths greater than one meter, e.g. $f=5.45$ m are something you might find for the objective lens or mirror on a very large telescope. In this case $\frac{1}{f}=5.45/m=\frac{1}{u}+\frac{1}{v}$, (basically for any pair of $u$ and $v$), making $f \approx .18$ m.