Finding the Focal Length: A Homework Challenge

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Homework Help Overview

The discussion revolves around a problem related to optics, specifically focusing on the calculation of focal length using the lens formula. Participants are exploring the relationship between object distance, image distance, and focal length, while also considering the implications of speed and velocity in this context.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss various methods for calculating image distances and velocities, questioning the accuracy of average versus instantaneous speed. There are attempts to differentiate the lens formula to derive relationships between distances and their rates of change.

Discussion Status

There is an ongoing exploration of different approaches to the problem, with some participants suggesting the use of derivatives for greater accuracy. Guidance has been offered regarding the differentiation of the lens equation, but no consensus has been reached on a definitive method or solution.

Contextual Notes

Some participants express confusion regarding the application of derivatives in this context, and there are mentions of significant digits and precision in calculations. The discussion also touches on the impact of time intervals on accuracy and potential roundoff errors.

Dan453234
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Homework Statement


Screen Shot 2016-05-05 at 11.40.31 PM.png


Homework Equations


1/do+1/di=1/f

The Attempt at a Solution


I tried finding the distance of image at 27m and then at 30.5m and taking the difference but that didn't work.
 
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Does that mean you tried to find the average speed of the image? I think their talking about "initial speed" indicates that they want instantaneous speed.
 
Your method gives the average velocity in the first second. It is a good approximation, if you calculate the image distances with enough significant digits (five at least). Show your work.
 
jtbell said:
Does that mean you tried to find the average speed of the image? I think their talking about "initial speed" indicates that they want instantaneous speed.
How would I perform the calculation with instantaneous speed?
 
ehild said:
Your method gives the average velocity in the first second. It is a good approximation, if you calculate the image distances with enough significant digits (five at least). Show your work.
I tried doing this but unfortunately this didn't give me the correct answer. Is there a more accurate way of doing this?
 
Dan453234 said:
How would I perform the calculation with instantaneous speed?
How do you get instantaneous speed (or more precisely, velocity) from position?
 
jtbell said:
How do you get instantaneous speed (or more precisely, velocity) from position?
derivative?
 
Yup.
 
jtbell said:
Yup.
Ok cool, I'm still a little confused how i would apply it to this problem however.
 
  • #10
Dan453234 said:
I tried doing this but unfortunately this didn't give me the correct answer. Is there a more accurate way of doing this?
What did you get? Yes, taking the derivative of di would be more accurate, but not much different.
 
  • #11
Dan453234 said:
I'm still a little confused how i would apply it to this problem

You know the relationship between ##d_0## and ##d_i##: $$\frac 1 {d_o} + \frac 1 {d_i} = \frac 1 f.$$ Take the derivative with respect to t, of both sides of this equation, and you'll have a relationship between ##\frac {dd_o}{dt}## and ##\frac {dd_i}{dt}##.
 
  • #12
ehild said:
taking the derivative of di would be more accurate, but not much different.
With Δt = 1 s I get about a 13% difference in the final answer, using the full precision of my calculator.

I agree, if Dan shows us his working, we can tell him if he at least calculated that approximation correctly.
 
  • #13
jtbell said:
With Δt = 1 s I get about a 13% difference in the final answer.
Yes, it is true.
 
  • #14
Good luck... if I want to get up for work in the morning I need to go to bed now. :oldwink:
 
  • #15
jtbell said:
With Δt = 1 s I get about a 13% difference in the final answer, using the full precision of my calculator.

I agree, if Dan shows us his working, we can tell him if he at least calculated that approximation correctly.
I am getting .00011 if i do it using the approximation with average velocity. It marks this as incorrect.
 
  • #16
OK, is the image moving towards the lens, or away from it?

The method with differentiation must be more accurate. Differentiate the equation ##\frac {1} {d_o} + \frac {1} {d_i} = \frac {1} {f} ## with respect to time and solve it for di'.
 
  • #17
Dan453234 said:
I am getting .00011 if i do it using the approximation with average velocity.
Yes, that's what I got. Another way to get more accuracy would be to use a smaller time interval. In principle, if you make it small enough, the answer will get close enough to the exact answer to make your software happy. But you have to be very careful to avoid roundoff errors.
 
  • #18
jtbell said:
Yes, that's what I got. Another way to get more accuracy would be to use a smaller time interval. In principle, if you make it small enough, the answer will get close enough to the exact answer to make your software happy. But you have to be very careful to avoid roundoff errors.
Great i ended up doing this and got the right answer. Thanks!
 
  • #19
If you differentiate the equation ##\frac {1} {d_o} + \frac {1} {d_i} = \frac {1} {f}## you get ##-\frac {1} {d_o^2} d_o' -\frac {1} {d_i^2} d_i'=0##. Solve for di'.
 

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