Astronomical Lens Telescope Problem - Angular Magnification

[AXidenT]

Homework Statement

A convex lens forms an image 8.96 mm on a screen. Keeping the screenand the object fixed, the lens is moved through a distance of 15.4 cm and a sharp image of the object is again observed on the screen. If the length of the second image is 3.54 cm, calculate the focal length of the lens.

Homework Equations

1/f = 1/u + 1/v

M = v/u = Hi/Ho

The Attempt at a Solution

Hi1 = 0.896, Hi2 = 3.54

u2 = u1 + 15.4, v2 = v1 - 15.4 (Could be plus or minus for both).

1/u1 = 0.896/Ho

=> v2/u2 = 3.54/Ho = (v1-15.4)/(u1+15.4)

=> 1/(u1+15.4) + 1/(v1-15.4) = 1/u1 + 1/u2

I've tried solving from there, but always end up with negatives or weird answers. The book's answer was 10.3. Any hints or help please? :P

 

[ehild]

Homework Equations

1/f = 1/u + 1/v

M = v/u = Hi/Ho

The Attempt at a Solution

Hi1 = 0.896, Hi2 = 3.54

u2 = u1 + 15.4, v2 = v1 - 15.4

Think: the object and the screen are fixed at a distance, say D. What do you know about the sum of object distance and image distance, u+v?
The image is real in both cases. Where is the object with respect to the focal point?
The first image is much smaller than the second one. When is the object farther from the lens?

ehild

 

[AXidenT]

Well from that you could say uv/D = f

Both u and v > f.

The greater v is compared to u, the greater the magnification. As a result you could say v2 = v1+15.4 and u2 = u1-15.4

u1v1/D = (v1+15.4)(u1-15.4)/D => u1v1 = (v1+15.4)(u1-15.4) = u1v1 - 15.4v1 + 15.4v1 - 237.16

=> (237.16 + 15.4v1)/15.4 = u1

v1/u1 = 0.896/Ho = v1/((237.16 + 15.4v1)/15.4) => Ho = 0.896/v1/((237.16 + 15.4v1)/15.4)

(v1+15.4)/(u1-15.4) = 3.54/Ho = (v1+15.4)/(((237.16 + 15.4v1)/15.4)-15.4)

3.54/0.896/v1/((237.16 + 15.4v1)/15.4) = (v1+15.4)/(((237.16 + 15.4v1)/15.4)-15.4)

I think when solved that gets me a -ve value for v... :S

Is that even on the right track at all? It's down to one equation with one variable but woah... I'm starting to confuse myself I think. :P

I'll have another try after some electrical questions I think...

Thank you.

 

[ehild]

Well from that you could say uv/D = f

NO! The object and the screen are fixed and the lens is in between. So what can you say about the sum of v and u?

ehild