Distance of the final image in a compound microscope

[moenste]

Homework Statement

A compound microscope has an objective of focal length 12 mm and an eyepiece of focal length 50 mm. The lenses have a separation of 90 mm and an object of height 0.3 mm is placed 15 mm from the objective. Calculate: (a) the distance of the intermediate image from the objective; (b) the distance of the final image from (i) the eyepiece, (ii) the objective; (c) the size of the intermediate image; (d) the angular magnification.

Answers: (a) 60 mm; (b) (i) 75 mm, (ii) 15 mm; (c) 1.2 mm; (d) 33.3

2. The attempt at a solution
(a) (1 / u) + (1 / v) = (1 / f)
u = 15 mm, f = 12 mm. Plug in and v = 60 mm

(c) (h / h_o) = (v / u)
h_o = 0.3 mm, v = 60 mm, u = 15 mm. Plug in and h = 1.2 mm

(d) M = β / α
α = h_o / D = 0.3 / 250 = 1.2 * 10^-3
β = h / (90 mm - 30 mm) = 0.04
M = 33.3

But (b)? I used this graph from the book for my question.

And according to this graph the distance of the final image (final virtual image) from the eyepiece is D and is 250 mm. And the distance from FVI to the objective should be more than 15 mm. Maybe I don't quite understand what shall I find. Really stuck on this one. Any help please?

 

[ehild]

You need to find the position of the final image, made by the eyepiece lens. The object is the first image. What is the object distance?

 

[moenste]

You need to find the position of the final image, made by the eyepiece lens. The object is the first image. What is the object distance?

By object you mean "intermediate real image"?

Is it something like this:
From (a) we have v = 60 mm for the obj. lens. So u = 30 for the eyepiece lens.

1 / 30 + 1 / v = 1 / 50
v = - 75 mm. So the distance from the eyepiece is 75 mm and the image is virtual.

Because S = 90 mm, 90 - 75 = 15 mm, distance from the objective.

Is this right? And why is it -75 but in the answer it's just 75? And is the 90-75 way a correct way to find 15 mm? And how come it is not negative?

Sorry for so many questions, a bit confused on this part.

 

[BvU]

You're doing fine. The -75 is correct, and it's a position on a coordinate axis with the objective at the origin.
The book answer is |-75| mm = 75 mm because they ask for a distance.
And yes, the virtual image is at position (+90) + (-75) wrt the objective (*), so at | +15| = 15 mm distance from the objective.

(*) relative position wrt objective = relative position of ocular wrt objective + relative position of virtual image wrt ocular

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