Help with focal point questions - OptiBoard [PDF]

Jun 10, 2008 - Where is the image positioned when light from a small source passes through a lens with a focal length of

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Help with focal point questions

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Thread: Help with focal point questions Thread Tools

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#1

06-10-2008, 01:37 AM

Whitwoo

Help with focal point questions

OptiBoard Apprentice Join Date: Location: Zealand Occupation: Posts:

Hi there,

May 2008 Auckland, New

I need some help on a few questions pleaaase :)

Dispensing Optician 20

4. An unknown lens has a focal length of +540mm, and is plano-convex in form. The radii are measured to be infinity and -305.95mm respectively. What is the refractive index? 5. If an object is placed 50mm in front of a lens and the image is found to be 160mm behind the lens, what power is the lens? 6. A projection sight-testing chart is set to produce an image at a distance of 6m. If the lens in the projector has a power of +10.00D, how far from the len would the slide have to be placed? (Light shining through the transparent slide acts as a source). 7. Where is the image positioned when light from a small source passes through a lens with a focal length of +285.71mm which is positioned 120mm away? 8. Calculate the magnification and type of image produced when an object is placed 145mm in front of a lens of -7.25D 9. An inverted image half the size of an object is required. If the object must be placed 120mm from the lens, what is the power of the lens, and where would the image form? 10. Calculate the power of the lens required to produce an upright virtual image which is 0.8x the size of the object, if the object distance is to be 100mm. 11. What is the size and position of the image produced when an object 36mm high is places 25mm outside the focal length of a +8.00D lens? NB: These are all assignment questions. I would love it if someone could give me a nudge in the right direction. I'm not here to cheat, I just don't have anyone to bounce ideas off over this side of the world :) Thanks to all :cheers: Reply With Quote

#2

06-10-2008, 11:31 AM

The majority of these questions pertain to the basic image construction principles of thin lenses, particularly conjugate object and image distances and linear magnification. You can read about these principles at Ray Tracing of Thin Lenses. This online course also includes a calculator that can be used to answer several of these questions. In the meantime:

Darryl Meister Carl Zeiss Vision

4. The focal length (f) is related to the radii (r1 and r2) and refractive index (n) of this lens by: Join Date: Location: United States Occupation: Posts:

May 2000 Kansas City, Kansas,

1 / f = (n - 1) (1 / r1 - 1 / r2) Rearrange to solve for the refractive index (n). And note that 1 / r1 is equal to zero, since r1 is infinitely long.

Lens Manufacturer 3,700

5. This problem relies on the basic conjugate foci formula: 1 / l' = F + 1 / l where you need the object distance (l) in meters, the focal power (F), and the image distance (l') in meters. You already have the object (-0.050 m) and image distances (+0.160 m), so rearrange to solve for the focal power (F). Note that the sign convention commonly used with these formulas requires that real objects have negative (-) object distances (l) and that real images have positive (+) image distances (l'). 6. See problem 5. 7. See problem 5. 8. This problem relies on both the conjugate foci formula (described after problem 5) and the lateral magnification (m) formula: m = l' / l where l' and l are again the image and object distances, respectively. First, calculate the image distance (l') for the object distance (-0.145 m) and focal power (-7.25 D) using the conjugate foci formula described after problem 5. Then, solve for the lateral magnification (m). If the magnification (m) is positive, the image is erect; if it is negative, the image is inverted. 9. This is similar to problem 8; in this case, however, you know the lateral magnification (m = -0.5, since the image is half the size and inverted) and the object distance (-0.120 m). Rearrange the lateral magnification formula to solve for the image distance (l'). 10. See problem 9. 11. You could solve this using Newton's relationship, but this would be introducing another concept, so I would stick to the two formulas that you have already been using with these problems (namely, the conjugate foci and lateral magnification formulas). First, calculate the primary focal length (f') of the lens using f' = -1 / F. Since the object is 0.025 m farther, you'll want to subtract 0.025 m from the primary focal length in order to determine the final object distance (l). Next, calculate the final image distance (l') from the object distance (l) and the focal power (+8.00 D) using the conjugate foci formula. Then calculate the magnification (m) from the image and object distances (l' and l) using the lateral magnification formula. Finally, multiply the actual object height (36 mm) by the lateral magnification (m) in order to determine the final image size and orientation. Feel free to post your answers for verification. Good luck!

Darryl J. Meister, ABOM Reply With Quote

#3

07-01-2008, 01:26 AM

The Zeiss Meister Master to the rescue again!!

rolandclaur OptiBoard Professional Join Date: Location: Occupation: Posts:

Apr 2005 San Diego Dispensing Optician 230

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