Q.A compound microscope has a magnification of 30. The focal length of the eyepiece is 5 cm. Assuming the final image to be at the least distance of distinct vision, find the magnification produced by the objective.

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):

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:

(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:

Step 4: Magnification formula

Lateral magnification by objective:

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

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:

For a simple magnifier with image at infinity:

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

Why this holds: The object (real image from objective) is at $f_e$, so it subtends an angle $\theta \approx h/f_e$ (where $h$ is object height). Without the eyepiece, the same object at the near point $D$ would subtend $\theta_0 \approx h/D$. The ratio gives $M_e = D/f_e$.

5. Putting It All Together

Key insights:

• Short $f_o$ → high objective magnification (but limited by lens aberrations)
• Short $f_e$ → high eyepiece magnification (but limited by eye relief)
• Longer tube length $L$ → higher magnification (but limited by mechanical constraints)

6. Exam-Relevant Notes

Final takeaway: The formula isn't arbitrary — it's a direct consequence of how two lenses work together, with the tube length acting as a "lever arm" for the objective and the near point as a reference for the eyepiece. Understanding this lets you derive it even if you forget the exact expression.