Compound Microscope Homework: Angular Magnification & Object Distance

[dvolpe]

Homework Statement

A microscope has an eyepiece that gives an angular magnification of 5.25 cm for a final image at infinity and an objective lens of focal length 1.45 cm. The tube length is 16.1 cm.
a - What is the transverse magnification due to the objective lens alone?
b - what is the angular magnification due to the microscope?
c - how far from the objective should the object be placed?

Homework Equations

M transverse = -I/f obj.
M eyepiece = N/f eyepiece
d of image = tube length - f eyepiece

The Attempt at a Solution

I correctly got -11.5 for a and -60.5 for b. I am having trouble with c!
I used d image = 16.7 - 25/5.25 = 11.94
then (1/d image) + (1/d object) = 1/f obj
1/11.94 + 1/d obj = 1/1.45
I get d obj = 1.65 THIS IS WRONG - answer is 1.58 Please help.

 

[I like Serena]

I'm having trouble with your wordings and symbols.

How can "angular magnification" have a unit of "cm"? It should be dimensionless.

What do your symbols I and N mean?

Can I assume that your tube length corresponds to the distance between the back focal plane of the objective and the back focal plane of the eyepiece?

 

[dvolpe]

I is the tube length
the angular magnification is not in cm
N is the normal point for the eye - 25 cm
d is the distance to the lens (d image is distance of image to lens and d object is distance of object to lens)

 

[I like Serena]

I is the tube length
the angular magnification is not in cm
N is the normal point for the eye - 25 cm
d is the distance to the lens (d image is distance of image to lens and d object is distance of object to lens)

Then there appears to be something wrong with the numbers you specified.

I get for a) M obj = -I/f obj = -16.1 / 1.45 = -11.1
This differs from your outcome -11.5.
How did you arrive at this value?

Using your value of -11.5 for a), I'd get for b) M = 5.25 x 11.5 = 60.4.
The difference is presumably a rounding error.

Can it be that your tube length is actually 16.7 cm instead of 16.1 cm?

 

[I like Serena]

Note that with the numbers you provided and the answer of 1.58 cm for the objective distance, the objective would make an image at 17.6 cm, which is longer than the tube length.
That can't be right...

 

[dvolpe]

The tube length is 16.7 and the f of objective lens 1.45 cm

Per the solutions the distance for the objective distance is listed as 1.58. Help.

 

[I like Serena]

The tube length is 16.7 and the f of objective lens 1.45 cm

Per the solutions the distance for the objective distance is listed as 1.58. Help.

Well, according to my calculations the objective distance should be 1.637 cm.
Btw, I calculated this differently from you, but obviously this is not your given answer either.
Can it be that your textbook answer is wrong?

 

[dvolpe]

I do not know if the answer is wrong but this is for web assign and I am trying to understand why they might be right. Can you tell me how you solved the problem?

 

[I like Serena]

I do not know if the answer is wrong but this is for web assign and I am trying to understand why they might be right. Can you tell me how you solved the problem?

The objective lens projects the object (at distance d_obj) to an image within the tube.
Let's say at a distance obj_im from the focal plane of the objective lens.

Formula is: $$\frac 1 {d_{obj}} + \frac 1 {obj_{im}} = \frac 1 {f_{obj}}$$

This image has a distance of I - obj_im from the focal plane of the eye lens.
The eye lens projects it to a distance of -N (virtual).

Formula is: $$\frac 1 {I - obj_{im}} + \frac 1 {-N} = \frac 1 {f_{eye}}$$

So: $$obj_{im}=I - \frac 1 {\frac 1 {f_{eye}} - \frac 1 {-N}}$$

And: $$d_{obj} = \frac 1 {\frac 1 {f_{obj}} - \frac 1 {obj_{im}}}$$

Filling in the numbers gives d_obj = 1.637 cm.

 

[ehild]

The tube length is defined as the distance from the back focal plane of the objective to the primary image plane.
So the image distance from the objective is

di=I+f(objective)=16.1+1.45 (cm)

Substitute for di in the lens formula

1/do+1/di=1/f(objective)

ehild