望遠鏡可以增加目視亮度?

[WFPC2]

Mauro Da Lio 於2007年的回應挑起一番深入討論. 有列出一些相關文獻及其摘要. 絕非大鏡聚得夠多光讓cone cell能工作那麼簡單.
http://www.cloudynights.com/ubbthreads/ ... part/4/vc/

忽然令我想起一個問題,為甚麼不同天文書所標示天體的星等有很大的出入?

而且很多天文書都沒列明是藍色星等Mb或是目視星等Mv,就目視而言,應以Mv表示天體的星等,為甚麼把兩者的數值混在一起?

[kylileo]

Sorry for the mis-information.

If I keep focal length fixed and increase diameter,
soon the exit pupil will >7mm.
So faster optics will not get brighter eye view,
after reaching the 7mm limit.

So here's another don't know whether correct idea:
There are so binoviewer in the martket like-
http://www.telescopes.com/telescope-acc ... viewer.cfm

If we reverse it, and put 2 identicle telescope light
to 1 eye, 'perhaps' eye view birghtness can
increase , while keeping exit pupil the same.
But perhaps the 2 light ray will interfere each other...

Sorry for incorrect post, better read mor optics

[Wah!!]

But.... it seems that there is no way to put 2 light beams together.

[Skyobs]

But.... it seems that there is no way to put 2 light beams together.

I guess the binoviewer splits converging light cone into two paths and form left and right focuses. In the same way, it should be possible for the diverging cones from left and right focuses go through the reverse path and form one beam.

However, there may be a catch. If the two beam are from ordinary circular objective (instead of matched halves), will they merge as described without lossing part of the circular cone?

[鄧登凳]

望遠鏡讓我們看到暗的星雲星系. 直覺上和經驗上都是這樣. 但有點不對勁. 望大家給些意見.

這裡不考慮生理和心理因素,只考慮物理因素. 而看到的物體是有面積的(如星雲,星系,彗星,及一般風景). 設望遠鏡物鏡直徑為D,眼睛瞳孔直徑為P,放大倍數為m. 若要物鏡收集到的所有光都進入眼睛,則m的最小值為D/P. 以下就取m=D/P這個臨界值.

望遠鏡能收集到的光是眼睛的(D/P)^2倍 (集光面積的比例)
在望遠鏡中看到物體的面積是眼睛看到的m^2倍,也等於(D/P)^2倍.
一除之下,在望遠鏡中看到物體的表面亮度,應該跟沒有用望遠鏡看到的沒有分別. (m>D/P就只會更暗.)

若這分析是正確,那是否我們觀星經歷到的都是生理和心理因素? (以後有機會才討論這個.) 我也知道生理和心理對人類視覺是有很大影響,但想先搞清楚這個物理問題.

這個分析有問題。

無限遠的星光到眼是平行光束, 瞳孔約5mm限光後再由眼內的晶體聚焦於一點上, 那點包括了5mm圓面積的平行光線。

天文望遠鏡的像也是無限遠的距離, 目鏡出來的星光也是平行線, 同樣由眼的晶體聚焦於一點上, 那今次的平行光束的限制是物鏡的大小, 而不是瞳孔了(因望遠鏡將物鏡所有光束都集中到瞳孔大小的範圍內)。由於集光多了(D/P)^2倍, 那點的光度也是多了約(D/P)^2倍。

那放大又如何, 觀星者都知用大鏡 300X 觀星 (如天狼星吧), 那目視時星是一點, 300X 星都是一點, 不會變成300點闊的小圓碟。所以放大並不會將星光按比例分散的。

[Skyobs]

這問題是針對有面積的景像哦!

(題外話: 我用150mm鏡200X看天狼星,的確見到小圓碟,外面再加幾圈. 否則我應該已經看到它的伴星. )

[鄧登凳]

這問題是針對有面積的景像哦!

(題外話: 我用150mm鏡200X看天狼星,的確見到小圓碟,外面再加幾圈. 否則我應該已經看到它的伴星. )

那是衍射環(diffraction rings), 因沒有鏡是完美的, 如對焦及光軸正確, 中心還是一個光亮的小點。那起碼我想那個"碟"不是200點闊度吧。

在近乎無限遠的一個光點, 射出來的光束, 經眼內晶體聚焦後, 會是一點(基本凸鏡光學), 而經望遠鏡和眼的晶體的光徑(light path), 也是一點出, 聚焦於一點。只不過到達該點的途徑較多(也可用QED的probability function), 所以光點更亮。
(用不同口徑的放大鏡聚焦陽光的原理一樣)

那放大了又是怎樣, 疏散星團最好說明, 放大了每星還是一點(今次衍射環太弱, 不會有小碟!), 不過亮了, 星和星間的距離大了。因此由一點光源聚焦於一點的光學原理沒有變。

那月球如何, 肉眼看滿月, 大概你不會要太陽眼鏡吧。中口徑望遠鏡看滿月, 時間一長就受不了, 要加減光鏡。那是由於在月球上每一點反射出來的光子, 透過望遠鏡會有更大的機會率射到]視網膜上那個像素的點上。由月球上一點出來的光, 必須要歸於視網膜的一點上, 那才能成像。如果由月上一點反射出來的光, 經望遠鏡就變成了一個小碟(闊度等如倍數), 而月上每一點都是這樣, 那視網膜上是無數互相重疊的小碟, 那是不會成像的, 只會有朦朧的白光(如沒有調好焦的一樣)。

基本法, 一點光源(point source), 經聚焦光學系統歸於成像上的一點, 那一定是口徑愈大, 就有更多的光束會由光源去到像素的點上, 那光度一定增加。

[PTS]

My interpretation & summary after readind these post and thinking myself:

****
d: diameter
d^2: d times d
Note:
1. neglect non-linear respond of human eye
2.I am thinking refractors only in my though experiment
****

Case 1: Looking a single star
All light collected by telescope go to one single cell in your eye,
so brightness scale with d^2

This assumption may not be correct. The size of the "Airy Disc" must be calculated to match the mean size of the "cell" of your eye before you can assume this.
This is the same as if you are calculating "resolving power" for your CCD pixels.

PTS

[Chanlunlun]

My interpretation & summary after readind these post and thinking myself:

****
d: diameter
d^2: d times d
Note:
1. neglect non-linear respond of human eye
2.I am thinking refractors only in my though experiment
****

Case 1: Looking a single star
All light collected by telescope go to one single cell in your eye,
so brightness scale with d^2

This assumption may not be correct. The size of the "Airy Disc" must be calculated to match the mean size of the "cell" of your eye before you can assume this.
This is the same as if you are calculating "resolving power" for your CCD pixels.

PTS

Come on, Mr. PTS. ! The size of the Airy disc has nothing to do with this. the output from the exit pupil is always a parallel beam of light, except than you are using absurdedly high power in which case, the Airy disc spreds into a big disc seen through the eyepiece. In generel case of deep sky observing we usually employ low enough power ( 5 X per inch to 10X per inch ) to make sure the effect of diffraction does not comes into acccount.

When taking picture we must measure the actual size of the image formed on the CCD. When viewing visually the situation becomes more complex - though real image is still formed on our retina, but the light has to pass through the eyepiece first.

That's why we use ANGULAR magnification - in this which case the f/no. of the system fails to give a correct perception of how bright he image is seen by our eyes.

Mr. Tang's analysis is excellent - think about light as particles rather than continuouos form energy and you will understand why bigger aperture gives brighter image. Brighter image seen gives you perception of color.

Best regards
Chan Yuk Lun

[Wah!!]

I think projecting to the retina is similar to projecting to CCD.
Both system should have an equivalent focal length and focal ratio.
Focal ratio is the key that affect light density.

[Chanlunlun]

My interpretation & summary after readind these post and thinking myself:

****
d: diameter
d^2: d times d
Note:
1. neglect non-linear respond of human eye
2.I am thinking refractors only in my though experiment
****

Case 1: Looking a single star
All light collected by telescope go to one single cell in your eye,
so brightness scale with d^2

This assumption may not be correct. The size of the "Airy Disc" must be calculated to match the mean size of the "cell" of your eye before you can assume this.
This is the same as if you are calculating "resolving power" for your CCD pixels.

PTS

Come on, Mr. PTS. ! The size of the Airy disc has nothing to do with this. the output from the exit pupil is always a parallel beam of light, except than you are using absurdedly high power in which case, the Airy disc spreds into a big disc seen through the eyepiece. In generel case of deep sky observing we usually employ low enough power ( 5 X per inch to 10X per inch ) to make sure the effect of diffraction does not comes into acccount.

When taking picture we must measure the actual size of the image formed on the CCD. When viewing visually the situation becomes more complex - though real image is still formed on our retina, but the light has to pass through the eyepiece first.

That's why we use ANGULAR magnification - in which case the f/no. of the system fails to give a correct perception of how bright he image is seen by our eyes.

Mr. Tang's analysis is excellent - think about light as particles rather than continuouos form of energy and you will understand why bigger aperture gives brighter image. Brighter image seen gives you perception of color.

Best regards
Chan Yuk Lun

[Wah!!]

I think projecting to the retina is similar to projecting to CCD.
Both system should have an equivalent focal length and focal ratio.
Focal ratio is the key that affect light density.

Why do I think that eye is similar to CCD?
It is because when we see images through a telescope, the optical system is exactly the same as afocal astro-photography.

[鄧登凳]

I do not quite follow your argument.

In telescopic observations, a eye-piece is needed. The objective focus the parallel rays onto the focal point of the eye-piece and light comes off at the eye-piece as parallel rays again.

In astrophotography other than eye-piece projection, no eye piece is involved. The objective lens/mirror/optic system focus the parallel beams onto spots on the CCD/CMOS/film.

To me, the two are clearly different optic systems. So the photographic system cannot be used to consider the parallel-in-parallel-out observational set-up.

That said, most of us know that in photography, f-ratio is king rather than simple diameter of the objective. Why so? The answer is in the light metering systems of our cameras. The light metering system is a total over a large number of pixels system, it is not a number of photon on individual pixel system. So if you are talking about a total amount of light received by a photosensor of a certain area, it will be increased by diameter of objective and will be decreased by the focal length. The focal length operates by the fact the increasing the focal length, the angle of scenery (FOV) covered will be less. In other words, the photosensor is receiveing light from a smaller area of light source(s). Yet this analysis can neither be applied to a visual observation system with the eye-piece in place nor be applied to the intensity of light from a point source falling onto an individual pixel.

[Wah!!]

Parallel rays are only before telescope and after eyepiece, but light becomes a cone before projecting to the CCD/retina.

Also, we have to consider both light intensity and image area rather than light intensity only.

Using a telescope with lowest effective power, both total light intensity and image area on CCD/retina increase, light density (cd/m^2) remains unchanged.

[鄧登凳]

Parallel rays are only before telescope and after eyepiece, but light becomes a cone before projecting to the CCD/retina.

Also, we have to consider both light intensity and image area rather than light intensity only.

Using a telescope with lowest effective power, both total light intensity and image area on CCD/retina increase, light density (cd/m^2) remains unchanged.

Well, once again, an eye piece is not part of the normal astrophotography light path. If your "cone" on the retina is refering to the airy disc. The reply is in another post: http://www.hkastroforum.net/viewtopic.p ... 580#160580

Again, I am sorry that I cannot follow your argument, especially that I cannot see how the image area on CCD/retina increases. Moreover, you are using many terms where I have difficulties in discerning the meanings: "light intensity only", "total light intensity" and "light density". Could you clarify?