Microscope Optics: questions and calculations

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

Distance from objective to object $𝑑𝑂 = 10.6cm$
Distance between the objective and eyepiece $𝐷 = 34cm$

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!

 

[BvU]

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

 

[ChiralSuperfields]

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

Thank you for your reply @BvU!

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!

 

[BvU]

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

 

[ChiralSuperfields]

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

Thank you for your reply @BvU!

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!

 

[BvU]

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?

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 ).

Then I can understand your answer to the (b) (?) part.

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

$\$

 

[ChiralSuperfields]

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 ).

Then I can understand your answer to the (b) (?) part.

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

$\$

Thank you for your reply @BvU!

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!

 

[BvU]

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]

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

$\$

Thank you for your reply @BvU!

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!

 

[BvU]

We are getting there, step by step

A remark about accuracy: you have input with 2 to 2½ 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' ?

$\$

 

[ChiralSuperfields]

We are getting there, step by step

A remark about accuracy: you have input with 2 to 2½ 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!

 

[BvU]

Magnification

 

[ChiralSuperfields]

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!