How Does Lens Separation Affect Image Distance in a Dual Lens System?

[lacrosse1440]

Blanking quite a bit on this one

Homework Statement

Assuming that paraxial light from a distant object enters lens 1, find the image distance s′2 (due to the refraction at lens 2) as a function of the lens separation d. Your answer will also contain f, the unspecified focal length of the converging lens.

Homework Equations

f1=f
f2=-2f

The Attempt at a Solution

s'1 = s1f1/(s1-f1)
s2 = d-s'1
s'2 = s2f2/(s2-f2)

But what is s1 since it has to be in terms of f and d? Is s'1 = f since it's a distant object?

But then that would make s1 = infinity, and we are asked to find

"Suppose the camera is trained on a distant object of angular size α radians (α << 1) when viewed by the unaided eye. What is the lateral size y′ on the CCD array of the image of that distant object produced by the zoom lens? Your result will contain d, f and also α."

and

"Effective focal length of the zoom lens: If the camera is fitted with a simple converging lens of focal length F, that distant object would produce an image on the CCD array of lateral size equal to αF. Set the (magnitude of) your result of (d) equal to αF to determine the ”effective focal length” F of the zoom lens as a function of d and f. Work out expressions for F at d = 0 and d = D in terms of f"

Which wouldn't make sense if the magnification is 0. (-s'2/s = 0)

 

[anathema]

Hello, I understand that you may be feeling overwhelmed by this problem. Don't worry, I'm here to help! Let's break down the problem into smaller steps and see if we can figure out the solution together.

First, let's consider the image distance s'1. As you correctly stated, for a distant object, s'1 will be equal to the focal length f of the lens. This is because for a distant object, the object distance s1 is essentially infinity, so the equation s'1 = s1f1/(s1-f1) simplifies to s'1 = f.

Now, let's move on to finding s2. We know that s2 = d-s'1, so we can substitute in our previous result for s'1 to get s2 = d-f.

Next, we can use this value for s2 to find s'2. You have the correct equation for s'2, but we need to substitute in our value for s2. This gives us:

s'2 = (d-f)(-2f)/((d-f)-(-2f))

= (-2f(d-f))/(d+f)

= (-2fd+2f^2)/(d+f)

Now, we can simplify this expression further by using the fact that f1 = f and f2 = -2f. This means that d = f1+f2. Substituting this into our equation for s'2, we get:

s'2 = (-2fd+2f^2)/(f1+f2+f)

= (-2fd+2f^2)/(2f+f)

= (-2fd+2f^2)/3f

= (-2d+2f)/3

Finally, we can use this expression for s'2 to find the lateral size y' on the CCD array. The lateral size y' is given by the equation y' = s'2 tan(α), where α is the angular size of the object. Substituting in our expression for s'2, we get:

y' = (-2d+2f)/3 * tan(α)

= (-2d+2f)/3 * α

= (-2dα+2fα)/3

So, the lateral size y' on the CCD array is equal to (-2dα+2fα)/3. This expression contains d, f, and α,