Optics question - Image formation

[Oshada]

Homework Statement

Homework Equations

Thin lens equation: 1/f = 1/s + 1/s'

The Attempt at a Solution

I tried to define image and object distances for both instances and equate them. Didn't work out :(

Any help is welcome!

 

[ehild]

The object and image distances are different in those cases. What is the image distance if the object distance is s? How is related s to f and D?


ehild

 

[Oshada]

I meant equating like this: apply thin lens equation twice, equate first and second instances (1/s + 1/(D-s) = 1/t + 1/(D-t), where s and t are the distances from the object to the lens). Am I on the right track?

 

[ehild]

Well, it is true but you just canceled the known f.

Express both s and t with f and find their difference.

ehild

 

[Oshada]

I get t - s = sqrt(D(D-4f)) + D for some reason. If you could help me with the working out that'd be great!

 

[ehild]

How did you get it? Show in detail, please!

ehild

 

[Oshada]

Sure thing:

Thin lens equation for s: 1/f = 1/s + 1/(D-s); making s the subject; s = +/- (sqrt(D(D-4f)) + D)/2. Similarly for t, t =+/- (sqrt(D(D-4f)) + D)/2. So if I take the positive (pr negative) answers for both of them t - s = 0. If I take one positive and one negative I get the answer I mentioned above.

 

[ehild]

s = +/- (sqrt(D(D-4f)) + D)/2. Similarly for t, t =+/- (sqrt(D(D-4f)) + D)/2. So if I take the positive (pr negative) answers for both of them t - s = 0. If I take one positive and one negative I get the answer I mentioned above.

Take care with the parentheses. The +/- sign is in front of the square root. The solution of a quadratic equation ax²+bx+c =0 is

x₁₂=[-b±sqrt(b²-4ac)]/(2a),

so

s=(+/- sqrt(D(D-4f))+D)/2, and the same for t.

s + t = D. So you have to choose one root for s and the other one for t.

ehild

 

[Oshada]

How can we get that s + t = D? Also, I thought t - s = d? I'm a bit confused right now. Please enlighten me

 

[ehild]

The way of light rays is reversible. What is object in the firs case, it is image in the other case. An the object distance +image distance = D.

Anyway, you get two possible values for s, the object distance.
So you can move the lens from one to the other to get a sharp picture again. The distance between these positions is the difference between the object distances.

ehild

 

[Oshada]

The image is formed on the screen on both occasions (there's a hint after the question that I didn't put up). But it certainly does look like s = t'. I'm not sure where the d is to be honest.

 

[ehild]

See the attached picture.


ehild