A compound microscope consists of an objective lens of focal length $2.0\ \text{cm}$ and an eyepiece of focal length $6.25\ \text{cm}$ separated by a distance of $15\ \text{cm}$. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision ($25\ \text{cm}$), and (b) at infinity? What is the magnifying power of the microscope in each case?

A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25

Exercises · 9.11

2.0\ \text{cm}

(a) the least distance of distinct vision (

25\ \text{cm}

), and

(b) at infinity? What is the magnifying power of the microscope in each case?

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Concept understanding — Microscope Magnification

Microscope Magnification

A compound microscope views very small, nearby objects using two lenses in sequence: the objective (near the object) and the eyepiece (near the eye). Its total magnifying power is the product of what each lens contributes.

How the two lenses work together

The object sits just beyond the focus of the short-focal-length objective ( $f_o$ ), which forms a real, inverted, magnified image inside the tube.
That real image falls just inside the focus of the eyepiece ( $f_e$ ), which acts as a simple magnifier, producing a large virtual, magnified final image for the eye.

Because each stage magnifies, the effects multiply:

M = m_o \times m_e

The objective's magnification

m_o = \frac{v_o}{u_o} \approx \frac{L}{f_o}

where $L$ is the tube length (roughly the distance between the objective's second focal point and the eyepiece's first focal point). The object sits close to $f_o$ , so this approximation holds for a well-designed microscope.

The eyepiece's magnification

The eyepiece behaves as a simple magnifier:

Final image at the near point ( $D = 25\ \text{cm}$ , largest magnification):

m_e = 1 + \frac{D}{f_e}

Final image at infinity (relaxed eye, "normal adjustment"):

m_e = \frac{D}{f_e}

Total magnifying power

M = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right)

Important

High magnification needs short $f_o$ and $f_e$ (both sit in denominators) and a large tube length $L$ — this is why a microscope objective is always a very short-focus lens.

Worked example

Objective $f_o = 1.0\ \text{cm}$ , eyepiece $f_e = 2.5\ \text{cm}$ , tube length $L = 20\ \text{cm}$ , near point $D = 25\ \text{cm}$ . Find $M$ with the final image at the near point.

m_o = \frac{20}{1.0} = 20, \qquad m_e = 1 + \frac{25}{2.5} = 11

$M = 20 \times 11 = 220$ …

Why this formula?

Microscope Magnification: Why the Formula Holds

Let's build the understanding from first principles — not just memorizing formulas, but seeing why they work.

1. What Does "Magnification" Mean in a Microscope?

A microscope creates a larger apparent image of a tiny object. The total magnification is the product of two stages:

Objective lens — creates a real, enlarged, inverted image of the specimen.
Eyepiece (ocular) — acts as a simple magnifier to view that real image.

So:

Total magnification = (magnification by objective) × (magnification by eyepiece)

2. The Key Formula

For a compound microscope in normal adjustment (final image at infinity, relaxed eye):

M = M_o \times M_e = \left( \frac{L}{f_o} \right) \times \left( \frac{D}{f_e} \right)

Where:

$f_o$ = focal length of objective
$f_e$ = focal length of eyepiece
$L$ = tube length (distance between second focal point of objective and first focal point of eyepiece)
$D$ = near point distance of the eye (usually 25 cm)

3. Derivation of Objective Magnification $M_o$

Step 1: How the objective works

The objective lens forms a real, inverted, enlarged image of the specimen. The specimen is placed just outside its focal point ( $f_o$ ).

Step 2: Using the lens formula

For a thin lens:

\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}

(Using sign convention: $u_o$ is negative, $v_o$ is positive)

Step 3: The tube length approximation

In a standard microscope, the specimen is placed very close to $f_o$ , so:

$u_o \approx -f_o$ (object just beyond focal point)
The image is formed at the first focal point of the eyepiece, which is at a distance $L$ from the second focal point of the objective.

Thus:

v_o \approx f_o + L

Step 4: Magnification formula

Lateral magnification by objective:

M_o = \frac{v_o}{|u_o|} \approx \frac{f_o + L}{f_o} = 1 + \frac{L}{f_o}

Since $L \gg f_o$ in practice, $1$ is negligible:

\boxed{M_o \approx \frac{L}{f_o}}

Why this makes sense: A shorter $f_o$ means the objective is more "powerful" — it bends light more sharply, creating a larger image at the fixed tube length.

4. Derivation of Eyepiece Magnification $M_e$

Step 1: The eyepiece as a simple magnifier

The eyepiece takes the real image from the objective and acts like a magnifying glass. For relaxed eye (final image at infinity), the real image must be placed at the focal point of the eyepiece.

Step 2: Angular magnification

Angular magnification is defined as:

M_e = \frac{\text{angle subtended by image at eye}}{\text{angle subtended by object at near point}}

For a simple magnifier with image at infinity:

M_e = \frac{D}{f_e}

Where $D = 25$ cm (standard near point). …

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Treating the objective and eyepiece as two lenses in sequence and working back from the required final-image position: (a) for the final image at the near point ( $25\ \text{cm}$ ) the object must be $2.5\ \text{cm}$ from the objective, giving magnifying power $20$ ;

(b) for the final image at infinity the object must be $\tfrac{70}{27} \approx 2.59\ \text{cm}$ from the objective, giving magnifying power $13.5$ .

How a compound microscope works

The objective ( $f_o = 2.0\ \text{cm}$ ) forms a real, enlarged, inverted intermediate image; the eyepiece ( $f_e = 6.25\ \text{cm}$ ) then acts as a simple magnifier on that image. The lenses are fixed $L = 15\ \text{cm}$ apart. We work backward from the eyepiece, since the required position of the final image fixes where the intermediate image must sit.

Case (a): final image at the least distance of distinct vision, $v_e = -25\ \text{cm}$

Eyepiece. Using $\dfrac{1}{v_e} - \dfrac{1}{u_e} = \dfrac{1}{f_e}$ with $v_e = -25\ \text{cm}$ , $f_e = 6.25\ \text{cm}$ :

\frac{1}{u_e} = \frac{1}{v_e} - \frac{1}{f_e} = \frac{1}{-25} - \frac{1}{6.25} = -0.04 - 0.16 = -0.20 \;\Rightarrow\; u_e = -5.0\ \text{cm}.

So the intermediate image is $5.0\ \text{cm}$ in front of the eyepiece, i.e. $15 - 5.0 = 10.0\ \text{cm}$ from the objective, giving $v_o = +10.0\ \text{cm}$ .

Objective. Using $\dfrac{1}{v_o} - \dfrac{1}{u_o} = \dfrac{1}{f_o}$ with $v_o = 10.0\ \text{cm}$ , $f_o = 2.0\ \text{cm}$ :

\frac{1}{u_o} = \frac{1}{v_o} - \frac{1}{f_o} = \frac{1}{10} - \frac{1}{2} = -0.4 \;\Rightarrow\; u_o = -2.5\ \text{cm}.

The object is placed $2.5\ \text{cm}$ from the objective (just beyond its focus $f_o = 2.0\ \text{cm}$ , as expected).

Magnifying power. …

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Method: Two-Step Image Formation (Ray Tracing by Lens Equations)

This problem treats the compound microscope as two lenses in series — the objective forms a real, inverted, enlarged image, and the eyepiece then magnifies that image further. We apply the thin lens formula to each lens in turn, using the fixed separation between the lenses.

Step 1 – Understand the layout

Objective: $f_o = 2.0\ \text{cm}$
Eyepiece: $f_e = 6.25\ \text{cm}$
Separation between lenses: $L = 15\ \text{cm}$
Final image distance from eyepiece:
- Case (a): $v_e = -25\ \text{cm}$ (least distance of distinct vision, virtual image)
- Case (b): $v_e = \infty$ (image at infinity)

We use the Cartesian sign convention with the lens formula $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ . The image formed by the objective acts as the object for the eyepiece; the two are linked by

|v_o| + |u_e| = L

Step 2 – Eyepiece, Case (a): final image at 25 cm

For the eyepiece, $v_e = -25\ \text{cm}$ and $f_e = +6.25\ \text{cm}$ :

\frac{1}{u_e} = \frac{1}{v_e} - \frac{1}{f_e} = \frac{1}{-25} - \frac{1}{6.25} = -0.04 - 0.16 = -0.20

u_e = -5\ \text{cm}

So the objective’s image lies $5\ \text{cm}$ in front of the eyepiece — a real object placed just inside the eyepiece’s focal length, which is exactly what produces a magnified virtual image at $25\ \text{cm}$ .

The image distance for the objective is therefore

v_o = L - |u_e| = 15 - 5 = 10\ \text{cm}

Step 3 – Objective, Case (a)

Lens formula for the objective ( $f_o = 2.0\ \text{cm}$ , $v_o = +10\ \text{cm}$ ):

\frac{1}{u_o} = \frac{1}{v_o} - \frac{1}{f_o} = \frac{1}{10} - \frac{1}{2} = 0.1 - 0.5 = -0.4

u_o = -2.5\ \text{cm}

The object must be placed 2.5 cm in front of the objective.

Step 4 – Magnifying power, Case (a)

With the final image at the least distance of distinct vision $D = 25\ \text{cm}$ :

M = \frac{v_o}{|u_o|}\left(1 + \frac{D}{f_e}\right)

\frac{v_o}{|u_o|} = \frac{10}{2.5} = 4, \qquad 1 + \frac{D}{f_e} = 1 + \frac{25}{6.25} = 1 + 4 = 5

M = 4 \times 5 = 20

Step 5 – Case (b): final image at infinity …

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Common Mistakes & How to Avoid Them

Mistake 1: Confusing the separation distance with image distances

The error: Students often take the given $15\ \text{cm}$ separation as $v_o$ or $u_e$ directly, without realising it is the distance between the two lenses ( $L = v_o + u_e$ ).

How to avoid:

Draw a clear ray diagram.
Label the objective-to-eyepiece distance as $L = v_o + u_e$ .
Never substitute $15\ \text{cm}$ for $v_o$ alone — it is the sum of two distances.

Mistake 2: Forgetting sign conventions for the eyepiece

The error: Treating the eyepiece as a converging lens but forgetting that the final image is virtual (on the same side as the object for the eyepiece). This leads to wrong signs in the lens formula.

How to avoid:

For the eyepiece, the final image distance $v_e$ $v_{e}$ is negative (virtual image).
- Case (a): $v_e = -D = -25\ \text{cm}$
- Case (b): $v_e = -\infty$
Use the lens formula with correct signs:

\frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e}

(where $u_e$ is negative because the object for eyepiece is real and on the opposite side).

Mistake 3: Using the wrong formula for magnifying power

The error: Mixing up the two cases — using the near-point formula when the final image is at infinity, or vice versa.

How to avoid:

Final image at near point ( $D = 25\ \text{cm}$ ):

M = M_o \times M_e = \left(-\frac{v_o}{u_o}\right) \times \left(1 + \frac{D}{f_e}\right)

Final image at infinity:

M = M_o \times M_e = \left(-\frac{v_o}{u_o}\right) \times \left(\frac{D}{f_e}\right)

Memorise: the $+1$ appears only when the eye is accommodating (near point).

Mistake 4: Forgetting the negative sign in magnification

The error: Reporting magnifying power as a positive number without indicating that the image is inverted relative to the object.

How to avoid:

The objective magnification $M_o = -v_o/u_o$ is negative (real, inverted image).
The eyepiece magnification $M_e$ is positive (virtual, erect relative to the intermediate image).
The total magnification $M = M_o \times M_e$ is negative, meaning the final image is inverted.
In exam answers, state: “Magnifying power = $|M|$ (magnitude)” or explicitly say “image is inverted.”

Mistake 5: Not checking if the intermediate image lies within the eyepiece focal length …

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How a compound microscope works

Case (a): final image at the least distance of distinct vision, ve=−25 cmv_e = -25\ \text{cm}ve​=−25 cm

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Case (a): final image at the least distance of distinct vision, $v_e = -25\ \text{cm}$

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