Optics - working with multiple lenses.

[Jazz House]

I have never used this template, but I will do my best. THe main problem involves assumption. My teacher isn't sure how to help with this. I thought that there might be some Optics experts here that could point me in the correct direction.

Homework Statement

I have been asked to do some calculations for my year 11 Optics Telescopes Assignment.

1- Calculate the height of the image of the moon when viewed through

(a)- maksutov-cassegrain telescope of focal length 1540mm
(b)- refractor telescope of focal length 600mm
(c)- Newtonian telescope of focal length 1000mm

where the moon is approximately 3471940000mm tall and 3.76289 x 10¹¹mm away from the viewer. Assume that light is not refracted upon entry to the atmosphere. These values I have researched and obtained myself.

Homework Equations

Derived from the lens/mirror formula and the magnification formula.

height_image = (height _object) (v) [ f^-1 - v^-1 ]

and

M=$$\frac{v}{u}$$

v- distance of object from mirror/lens
u- distance of image from mirror/lens

The Attempt at a Solution

I have substituted the values into the above equations and obtained the following results:

(a)

Maksutov-Cassegrain
height_image = (3471940000) (3.76289 x 10¹¹) [ 1540^-1 - (3.76289 x 10¹¹)^-1 ]
= 8.48 x 10⁷mm

I have followed this process for (b) and (c) and have made the same calculations for Uranus and a 1.8m tall person only 5km away for some different comparison.

4. Assumptions

THere is no problem with my ability to put numbers in formulas.

I just don't think the values I am getting are quite right. They suggest a very large image and really big magnification. Conceptually, to my teacher and me they don't seem quite right. I am just not sure if I am using the correct formulas.

I know the value of the focal length in these telescopes is derived from a combination of all the lenses involved. Is it right to assume that the mirror formula works for combinations of lenses as well as just single lenses/mirrors?

If not, is there a formula I can use for refractors or Newtonians or Maksutov-Cassegrains that works better?

Any help would be greatly appreciated.

Thanks a lot!

 

[Jazz House]

I did a bit of research on telescopes in my 40-year-old set of encyclopedias.

I have learned that the height of an image is equal to the product of the focal length of the telescope and the angle subtended by the object.

Does is equation have a name?? Can anyone point me towards a more up to date source?

 

[ehild]

Are you sure that the formula you applied is correct? You certainly have learned about lenses. Then you have to be familiar with the equations

1/v+1/u =1/f,

and that

height of image/height of object = u/v

( considering all quantities without sign).


The object Moon is very far with respect to the focal length. So 1/v is about zero, and the object is in the focus. Try to derive the equation for the object height by yourself. And check if the distance of Moon is correct.

ehild

 

[Jazz House]

Thanks! It's great to have a reply. The formula I derived is correct and that is confirmed by my teacher. I just manipulated a couple of formulas with some substitution and the like. Simple algebra. THe problem is that the formula i derived doesn't follow for combinations of lenses.

I tried the equation I found in the encyclopedias and the values are more reasonable. Much more reasonable. I like this equation much more. :)

 

[ehild]

The formula I derived is correct and that is confirmed by my teacher.

The formula

height image = (height object) (v) [ f-1 - v-1 ] is wrong , as the magnification is M=u/v, the same as the ratio (image height / object height).
ehild

 

[Jazz House]

Apologies for the poor quality picture.

 

[ehild]

You took v as distance of object, and u as distance of image. The magnification is defined as M=u/v. It is very much smaller than 1 in this case. Let the sun-rays fall to a piece of paper through a converging lens (magnifying glass) How big is the bright spot?

ehild

 

[Jazz House]

Ok. So now with the correctly defined magnification.

M=v/u

v~image distance
u~object distance

Therefore (similar derivation as before)

H_i=H_ou^-1($$\frac{1}{f}$$+$$\frac{1}{u}$$)^-1

 

[Jazz House]

I have redone the calculation. You are completely correct. Thanks for your help.

 

[ehild]

You are welcome.

ehild