Focal length of the eyepiece lens in a compound microscope

[tellmesomething]

Homework Statement

In a compound microscope, the magnified virtual image is formed at a distance of 25 cm from the eye-piece. The focal length of its objective lens is 1 cm. If the magnification is 100 and the tube length of the microscope is 20 cm, then the focal length of the eye-piece lens (in cm) is __________.

Relevant Equations

None

$\frac{1}{v}−\frac{1}{u}=\frac{1}{f}$
For objective lens of focal length 1 cm
$\frac{1}{v}+\frac{1}{u}=1$
$\frac{1}{v}=\frac{u−1}{u}$
$v=\frac{u}{u−1}$

Distance from eyepiece = 20-v=u'

u′=$\frac{19u−20}{u-1}$

Magnification=(Angle subtended at eye by final image)/ (Angle subtended at eye without compound microscope when object is kept at 25 cm from eye)

Angle subtended at eye by final image= (size of final image)/25cm

=Angle subtended by the image of the objective lens=(size of image by objective lens)/(20−v)

Angle subtended at eye without compound microscope when object is kept at 25cm from eye= (size of object)/(25)

We know that
=(size of image by objective lens)/(Size of object)=$\frac{v}{u}$

=$\frac{1}{u-1}$
So accordingly magnification
=((size of image by objective lens) (25))/((20-v)(size of object))

We know that 20-v=$\frac{19u-20}{u-1}$
[LaTeX typo fixed by a Mentor at OP request]

So solving this we get u as 81/76

20-v=u'=19/5

So substituting this to lens formula for the eyepiece we get

$\frac{-1}{25}+\frac{5}{19}=\frac{1}{f}$

So f I get as 4.48 cm....

However the answer is 6.25, I don't know where I'm going wrong please help. Please refer to my solution and point out where I went wrong if possible

 

[TSny]

Your work looks correct to me. I did notice a typographical error where you wrote

We know that 20-v=$\frac{19-u}{u-1}$

So solving this we get u as 81/76

In the first line you should have 20 - v = $\frac{19u-20}{u-1}$, as you had obtained near the beginning of your calculations.

Nevertheless, I agree with u = 81/76. I get the same final answer of 4.48 cm for the eyepiece focal length.

 

[tellmesomething]

Your work looks correct to me. I did notice a typographical error where you wrote



In the first line you should have 20 - v = $\frac{19u-20}{u-1}$, as you had obtained near the beginning of your calculations.

Nevertheless, I agree with u = 81/76. I get the same final answer of 4.48 cm for the eyepiece focal length.

Thankyou, I will fix the error actually I had to write it twice since i lost the block of text the first time, so I got impatient.

This is a question from a competitive exam and the answer they released was 6.25, the supporting solution was this

So im unsure what different thing they did, till the first formula it makes sense, but then they equated vo/uo to L/fo

(O is for objective lens, e is for eyepiece)

 

[TSny]

We interpreted the "tube length" as the distance between the two lenses. But I did some searching. I found that there are different definitions of tube length. The tube length is sometimes defined as follows:

F₁ and F₂ are the focal points of the objective lens. The tube length is the distance between $F_2$ and the image of the objective lens. This length is sometimes called the "optical tube length". For example, see figure 3 here. Also, I found this definition in my old copy of Optics by Hecht and Zajac (1st edition).

I guess this definition is for convenience. You can show that with this definition the overall magnification of the microscope takes the simple form used in their solution: $M = -\dfrac {L}{f_o} \cdot (1+\dfrac D {f_e})$.

Anyway, your analysis is correct for your interpretation of tube length.

 

[Steve4Physics]

I believe the formula $M = -\dfrac {L}{f_o} \cdot (1+\dfrac D {f_e})$, with $L$ the lens-to-lens distance, can be derived using two (not that great IMO) approximations.

The approximations are explained in this (somewhat overlong) video derivation. In particular, watch from 20:50. The approximations (made at a late stage in the derivation) are:

a) $u_e \ll L$ so that $u_e$ can be neglected, i.e. the image formed by the objective is very close to the eye-piece;
b) $u_0 \approx f_o$, i.e. the object is positioned very close to (just outside) the objective's focal point.

If the formula comes from some official textbook related to the course, it might be worth checking the textbook. If the formula is not from an official textbook, it's hard to know how students are supposed to know it.

 

[TSny]

If the formula comes from some official textbook related to the course, it might be worth checking the textbook. If the formula is not from an official textbook, it's hard to know how students are supposed to know it.

Yes, I agree.

The formula $M = -\dfrac {L}{f_o} \cdot (1+\dfrac D {f_e})$ is "exact" if $L$ is defined as in post #4. (Here's another example using that definition). In this case, $f_e = 6.25$ cm is the accurate answer for the problem stated in the OP.

If $L$ is defined as the lens-to-lens distance, the accurate answer for $f_e$ is 4.48 cm. @tellmesomething got this result by working things out from scratch!

 

[tellmesomething]

I believe the formula $M = -\dfrac {L}{f_o} \cdot (1+\dfrac D {f_e})$, with $L$ the lens-to-lens distance, can be derived using two (not that great IMO) approximations.

The approximations are explained in this (somewhat overlong) video derivation. In particular, watch from 20:50. The approximations (made at a late stage in the derivation) are:

a) $u_e \ll L$ so that $u_e$ can be neglected, i.e. the image formed by the objective is very close to the eye-piece;
b) $u_0 \approx f_o$, i.e. the object is positioned very close to (just outside) the objective's focal point.

If the formula comes from some official textbook related to the course, it might be worth checking the textbook. If the formula is not from an official textbook, it's hard to know how students are supposed to know it.

Yes, I agree.

The formula $M = -\dfrac {L}{f_o} \cdot (1+\dfrac D {f_e})$ is "exact" if $L$ is defined as in post #4. (Here's another example using that definition). In this case, $f_e = 6.25$ cm is the accurate answer for the problem stated in the OP.

If $L$ is defined as the lens-to-lens distance, the accurate answer for $f_e$ is 4.48 cm. @tellmesomething got this result by working things out from scratch!

Thanks a lot for the help. Is there a way to guess when to use this approximation? In this question i didn't know if the image formed by the objective was close to the eyepiece or not, or if the object was kept at the focus of the objective, In the official prescribed textbook tube length is described as the distance between the eyepiece and objective as well,so considering that this approximation would have an error of 1.77cm, yet it was used :/

 

[Steve4Physics]

... Is there a way to guess when to use this approximation? In this question i didn't know if the image formed by the objective was close to the eyepiece or not, or if the object was kept at the focus of the objective, In the official prescribed textbook tube length is described as the distance between the eyepiece and objective as well,so considering that this approximation would have an error of 1.77cm, yet it was used :/

Does the textbook give the derivation of the formula?
- if so, the textbook should explain the approximations;
- if not, the approximations aren't, IMO, ones you could guess for yourself - in fact the approximations are not applicable in this problem and lead to the wrong answer.

Check-out what @TSny says in Posts #4 and #6. It sounds like two different definitions of 'tube length' are getting muddled.

Your official textbook definition of tube length is not consistent with the official answer. The official text book should be changed - or the official answer should be changed!