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

 

[rwisz]

I would approach this problem by first understanding the concept of angular magnification in telescopes. Angular magnification is the ratio of the angular size of an image to the angular size of the object. In this problem, the angular size of the object remains constant, but the angular size of the image changes as the lens is moved.

To solve this problem, we can use the equation M = v/u = Hi/Ho, where M is the angular magnification, v is the image distance, u is the object distance, Hi is the height of the image, and Ho is the height of the object. We know that the angular magnification remains constant, so we can set up the equation M1 = M2, where M1 is the angular magnification for the first image and M2 is the angular magnification for the second image.

We also know that the object distance remains the same for both images, so we can set u1 = u2. Using the given values, we can write the equations:

M1 = v1/u1 = Hi1/Ho

M2 = v2/u2 = Hi2/Ho

Since M1 = M2 and u1 = u2, we can set v1/u1 = v2/u2. Therefore, we can write the equation:

v1/u1 = v2/u2 = Hi1/Ho = Hi2/Ho

Substituting the given values, we get:

v1/u1 = 3.54/0.896 = 3.95

Since v1/u1 = 3.95, we can write the equation 1/u1 + 1/v1 = 3.95. We also know that the lens is moved through a distance of 15.4 cm, so we can write the equation u1 + v1 = 15.4.

Solving these two equations simultaneously, we get u1 = 7.71 cm and v1 = 7.69 cm. Now, using the equation 1/f = 1/u + 1/v, we can calculate the focal length of the lens:

1/f = 1/7.71 + 1/7.69

1/f = 0.258 + 0.260

1/f = 0.518

f = 1/0.518 = 1.93 cm

Therefore, the focal length of the