Finding the Focal Length: A Homework Challenge

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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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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?
 
jtbell said:
How do you get instantaneous speed (or more precisely, velocity) from position?
derivative?
 
jtbell said:
Yup.
Ok cool, I'm still a little confused how i would apply it to this problem however.
 
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.
 
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}##.
 
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.
 
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.
 
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'.
 
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.
 
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!