# Lens question

1. May 5, 2016

### Dan453234

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

2. Relevant equations
1/do+1/di=1/f

3. 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.

2. May 5, 2016

### Staff: Mentor

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.

3. May 6, 2016

### ehild

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.

4. May 6, 2016

### Dan453234

How would I perform the calculation with instantaneous speed?

5. May 6, 2016

### Dan453234

I tried doing this but unfortunately this didn't give me the correct answer. Is there a more accurate way of doing this?

6. May 6, 2016

### Staff: Mentor

How do you get instantaneous speed (or more precisely, velocity) from position?

7. May 6, 2016

derivative?

8. May 6, 2016

Yup.

9. May 6, 2016

### Dan453234

Ok cool, I'm still a little confused how i would apply it to this problem however.

10. May 6, 2016

### ehild

What did you get? Yes, taking the derivative of di would be more accurate, but not much different.

11. May 6, 2016

### Staff: Mentor

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. May 6, 2016

### Staff: Mentor

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. May 6, 2016

### ehild

Yes, it is true.

14. May 6, 2016

### Staff: Mentor

Good luck... if I want to get up for work in the morning I need to go to bed now.

15. May 6, 2016

### Dan453234

I am getting .00011 if i do it using the approximation with average velocity. It marks this as incorrect.

16. May 6, 2016

### ehild

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. May 6, 2016

### Staff: Mentor

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. May 6, 2016

### Dan453234

Great i ended up doing this and got the right answer. Thanks!

19. May 6, 2016

### ehild

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'.