# Microscope Optics: questions and calculations

• ChiralSuperfields
In summary, the conversation discussed a problem involving a microscope experiment with given distances between the objective lens and object, and between the objective lens and eyepiece. Relevant equations were provided, including the thin lens formula and magnification formula. The conversation also touched upon the accuracy of calculations and the performance of the microscope in terms of magnification.
ChiralSuperfields
Homework Statement
Relevant Equations
For this problem,

Distance from objective to object ##šš = 10.6cm##
Distance between the objective and eyepiece ##š· = 34cm##

For (b) I got ##d_I = 26 cm## and ##M_1 = -2.4## which means that firsts image is inverted and real

For (c) I got ##dI' = 35 cm## and ##M_2 = -1.3##. However, I thought ##dI' < 0## since the second image is virtual and inverted from a ray diagram.

For (d)

I got ##M = M_1M_2 = 3.2## which is interesting since the finial image is inverted

Can someone please tell me whether I am correct and how to tell the second image is virtual without drawing a ray diagram?

Many thanks!

Last edited:
Hi, seems to me there is a lot missing:
Given information?
Relevant equations?

ChiralSuperfields
BvU said:
Hi, seems to me there is a lot missing:
Given information?
Relevant equations?

Distance from objective to object ## d_O = 10.6cm##
Distance between the objective and eyepiece ##D = 34cm##

Is the given information (the data collected from the experiment)

Relevant equation:
- thin lens formula ##\frac{1}{d_O} + \frac{1}{d_I} = \frac{1}{f}##
- Magnification formula ##M = -\frac{d_{O}}{d_I}##

Many thanks!

Never saw a microscope do anything sensible at 10.6 cm from the object....

ChiralSuperfields and berkeman
BvU said:
Never saw a microscope do anything sensible at 10.6 cm from the object....

Well it did in the labs! We used a 150 mm and 75 mm convex lens to make a microscope and adjusted the distances accordingly to get a clear image.

Many thanks!

ChiralSuperfields said:
Well it did in the labs! We used a 150 mm and 75 mm convex lens to make a microscope and adjusted the distances accordingly to get a clear image
Do you realize there is some very useful information there?
BvU said:
Never saw a microscope do anything sensible at 10.6 cm from the object....
So this isn't some biologist's microscope, but a lab exercise to work out the principles. OK!

And now we have some input for the relevant equations !

Now we can gamble safely that the objective lens is the 75 mm one ( but it would have been much better if you had offered that voluntarily in the problem description ).

But not what you do for (c). What is your ##d_{O'}## ?
(Or better: post your work, clearly, step by step )

##\ ##

SammyS and ChiralSuperfields
BvU said:
Do you realize there is some very useful information there?

So this isn't some biologist's microscope, but a lab exercise to work out the principles. OK!

And now we have some input for the relevant equations !

Now we can gamble safely that the objective lens is the 75 mm one ( but it would have been much better if you had offered that voluntarily in the problem description ).

But not what you do for (c). What is your ##d_{O'}## ?
(Or better: post your work, clearly, step by step )

##\ ##

My ##d_{O'}## for is the real image formed by objective lens. I use that virtual object for the eyepiece.

##\frac{1}{26} + \frac{1}{dI'} = \frac{1}{15}##
##d_{I'} = 35 cm##

Many thanks!

ChiralSuperfields said:
My ##d_{O'}## for is the real image formed by objective lens. I use that virtual object for the eyepiece.

##\frac{1}{26} + \frac{1}{dI'} = \frac{1}{15}##
##d_{I'} = 35 cm##
How come you think this image is formed at a distance of 26 cm from the eyepiece ?

##\ ##

ChiralSuperfields
BvU said:
How come you think this image is formed at a distance of 26 cm from the eyepiece ?

##\ ##

True, that might be a mistake. I guess 26 cm is the distance from the real image from the objective lens not the eyepiece so I guess the distance to the real image from the eyepiece should be ##34 - 26 = 8 cm## this gives
the finial image to be - 17 cm from the objective lens.

Many thanks!

We are getting there, step by step

A remark about accuracy: you have input with 2 to 2Ā½ digit accuracy. You want to do your calculations with at least that and only round off properly at the end. So NOT ##d_{O'} = 8 ## cm but 8.355 and then ##d_{I'} = -18.86 ## cm, so you get -18.9 cm, not -17 !

So: what's the performance of your 'microscope' ?

##\ ##

Last edited:
ChiralSuperfields and SammyS
BvU said:
We are getting there, step by step

A remark about accuracy: you have input with 2 to 2Ā½ digit accuracy. You want to do your calculations with at least that and only round off properly at the end. So NOT ##d_{O'} = 8 ## cm but 8.355 and then ##d_{I'} = -18.86 ## cm, so you get -18.9 cm, not -17 !

So: what's the performance of your 'microscope' ?

##\ ##
Thank you for your help @BvU!

Sorry, what do you mean by performance of the microscope?

Many thanks!

Magnification

ChiralSuperfields
BvU said:
Magnification
Thank you for your help @BvU!
For the total magnification I get ##-2.4 \times 2.4 = -5.8## which means that it must be an enlarged and inverted image

Many thanks!

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