I think compound microscope will also work if focal length of eyepiece is kept smaller than that of objective lens. I am just a beginner in physical optics. I understand ray diagram of compound microscope. Kindly explain in an easy way.
[Physics] Why we take objective of short focal length and eye piece of long focal length in a compound microscope
geometric-opticsoptics
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It's like leverage. The longer the distance from the objective lens to the virtual image, the larger the virtual image.
Imagine there's a piece of frosted glass at the focal point. It will show the virtual image.
Now the eyepiece looks at that virtual image with a magnifying glass. That also makes it look bigger.
The Magnification is a combination of all of the focal lengths of the picture you have shown above. A real image is created by the objective and tube lens. This creates an image of what you have at the object plane that is magnified by:
$M = \frac {f_{tube lens}}{f_{objective}}$
So, if you were to measure the size of the image, it would be M times larger than the object that is placed to the left of the objective lens.
One common confusion is that many microscope objectives actually create an image all by themselves without a tube lens. There are several standards including objectives that create images 160 mm and 170 mm away from the microscope objective. In your diagram, it implies an infinity corrected objective lens. This means that the image created by the microscope objective is infinitely far from the objective lens. This might lead you to believe that since the light is collimated from the microscope objective, you can place the tube lens anywhere you want. That is not technically correct because of two factors: vignetting and the optical design of the tube lens.
Vignetting means that the light escapes the size of the lens. In your diagram, this would happen if the tube lens is too small. Many infinity corrected objectives are designed for tube lenses that are 180 mm from the objective lens. If the tube lens is not placed at a distance close to 180 mm, you can have vignetting or performance from optical aberrations may cause the image to degrade.
Now, take the final step to the eye of the observer. This is the eyepiece. Your eye prefers (is relaxed) when looking at infinity. Therefore, the eyepiece is typically designed to project the image created by the objective-tube lens pair to infinity. Your diagram actually shows the image at 25 cm instead of at infinity. For this case, the eyepiece is placed at nearly one focal length away from real image (image plane 3 in your diagram).
The final magnification is $ M_{total} = M \times \frac{25 cm}{f_{eyepiece}} $
There are additional considerations including:
- Working distance of the objective (distance between objective and object)
- Eye relief (distance between eyepiece and observer's eye)
- Pupil or eyebox (sometimes you look inside a microscope and it is black until you line up your eye with the microscope)
- Illumination (most objects require some external light source to illuminate them so you can see them!)
One last consideration. If you just want to put the image onto a camera. In which case, you don't need the eyepiece!
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[ Image from http://www.schoolphysics.co.uk/age16-19/Optics/Optical%20instruments/text/Microscope_/index.html]
Theoretically speaking, there is no problem at all which of the focal lengths is greater as the formula for magnification is given by:- $$M = \Bigg[\frac{D}{f_e}+1\Bigg]\left[\frac{v_o}{f_o}+1\right]$$ Therefore whether $f_O > f_e$ or $f_O < f_e$, you can still acheive magnification.
However, usually when you are usually observing with a compound microscope, conventionally your sample is extremely close to the objective and hence we use a small focal length for the objective and in most cases, the eye-piece ends up having a greater focal length. The reverse case will also work theoretically, but as I pointed out there are practical limitations.