How Fast is the Image Moving in Optics?

[PhizKid]

Homework Statement

An object 20 cm. to the left of a converging lens with focal point 0.3 cm is moved to the left at a constant 5 cm/s. How fast is the image moving?

Homework Equations

v = vo + at
1/f = 1/u + 1/v

The Attempt at a Solution

Since there is no acceleration, I got 5 m/s as the velocity but I don't know how to treat a non-constant value of u. The units don't match up in the lens equation, so what should I do?

 

[NascentOxygen]

Can you express u in terms of v and f? Then differentiate both sides wrt time.

 

[PhizKid]

Can I just differentiate:
$\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ 0 = -\frac{1}{u^2} \cdot \frac{du}{dt} - \frac{1}{v^2} \cdot \frac{dv}{dt} \\ 0 = -\frac{1}{20^2} \cdot 5 - \frac{1}{0.305^2} \cdot \frac{dv}{dt} \\$
Having solved for v using the regular 1/f = 1/u + 1/v = 1/0.3 = 1/20 + 1/v, then solve for $\frac{dv}{dt}$?

I get dv/dt = -0.001. If this is correct, does this mean the image is moving away from the lens to the right, since the velocity towards the left is a positive value?

 

[NascentOxygen]

That's what I had in mind.

To check your answer, use this piece of knowledge: as the object moves nearer to infinity, its image moves ever closer to f. (Remember, a converging lens focuses the sun's rays right at f.)

 

[PhizKid]

My solution is incorrect then because the sign has opposite direction to the object which means it's moving away. What did I do wrong?

 

[Basic_Physics]

The object distance is a function of time:

u(t) = 0.05 t + 0.02

so I don't think one can just substitute values for

u² and v²

which means you are evaluating the differential equation at t = 0.

It might be easier to use Newton's formula

x₁ x₂ = f²

 

[NascentOxygen]

Having solved for v using the regular 1/f = 1/u + 1/v = 1/0.3 = 1/20 + 1/v, then solve for $\frac{dv}{dt}$?

I get dv/dt = -0.001. If this is correct, does this mean the image is moving away from the lens to the right, since the velocity towards the left is a positive value?

You should check this in a physics book: but the formula 1/f = 1/u + 1/v suggests to me that for u and v, the positive direction is measured outwards from the lens. So your negative speed indicates motion towards the lens. That is what I expected, it's moving closer to the focus.

 

[PhizKid]

My professor says the answer is wrong because it doesn't include focal length which affects the velocity.

 

[Basic_Physics]

1/f = 1/u + 1/v

= (v + u)/uv

uv = fv + fu

uv - fv = fu

v(u - f) = fu

v = fu/(u - f)

 

[PhizKid]

Solving for v and then differentiating the equation is mathematically the exact same thing as differentiating the equation then solving for dv/dt..

 

[haruspex]

My professor says the answer is wrong because it doesn't include focal length which affects the velocity.

That's a puzzling statement. You used the focal length to compute v, so it is 'included'. Symbolically, you have deduced $\frac{dv}{dt} = -\left(\frac vu\right)^2\frac{du}{dt} = -\left(\frac f{u-f}\right)^2\frac{du}{dt}$

 

[PhizKid]

Right, and he insists that the physics is incorrect even though I showed him that (v/u) = (f/(u-f)) he says that it's physically impossible.

 

[NascentOxygen]

You could check your answer by approximating the calculus. For example, determine v and u when the object is in a particular position, then take a tiny time interval, say 0.001 secs, and determine the new v and u after that elapsed time (using plenty of significant figures on your calculator). Knowing ∆u and ∆t, you can calculate a good approximation to its velocity. No one could argue with that approach!

BTW, I'm confident the analytical method is correct. But if your professor insists that it be answered differently (even if he is wrong) you should do it the way he demands. He is the one who determines your grade.

 

[rude man]

The formula haruspex came up with is correct except I would substitute u = (-20cm - 5cm/s*t) to make dv/dt a function of constants and time only.

This assumes by "focal point" was intended "focal length". Is that what it's called in the UK or Australia or ?

 

[haruspex]

This assumes by "focal point" was intended "focal length". Is that what it's called in the UK or Australia or ?

No, you're right, it should be focal length. Focal point is a place, not a distance.

 

[rude man]

No, you're right, it should be focal length. Focal point is a place, not a distance.

So I suppose they could have meant v = 0.3cm instead of f = 0.3cm? Wouldn't make much of a difference I suppose.

PS does everyone use u for object and v for image distance nowadays? I always had p and q. u and v should be reserved for velocity ... in fact the OP used v for both image distance and velocity, just to cite one example ...