A compound microscope has a magnification of 30. The focal length of the eyepiece is 5 cm. Assuming the final

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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.

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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

The microscope magnifies about 220 times, and because the objective inverts the image, the final image is inverted relative to the object.

Key points

Total magnification is the product, not the sum, of the two lenses' contributions.
Short focal lengths for both lenses give higher magnifying power.
The final image is virtual and inverted.
Two working modes: image at the near point (maximum $M$ ) or at infinity (relaxed eye, slightly lower $M$ ).

Compound microscope magnifying power, M = (L/f₀)(1 + D/f_e), is a key numerical topic in the NCERT Class 12 Physics chapter on ray optics and optical instruments, tested in CBSE boards, JEE Main and NEET. Students searching "compound microscope magnification formula derivation class 12 physics" will find this objective-times-eyepiece derivation matches the NCERT-prescribed method.

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).

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

\boxed{M = M_o \times M_e = \frac{L}{f_o} \times \frac{D}{f_e}}

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

Condition	Formula	Why
Final image at infinity (relaxed eye)	$M = \frac{L}{f_o} \cdot \frac{D}{f_e}$	Most common in exams
Final image at near point (maximum strain)	$M = \frac{L}{f_o} \left(1 + \frac{D}{f_e}\right)$	Eyepiece acts as magnifier with image at $D$
Simple microscope (single lens)	$M = 1 + \frac{D}{f}$	Just the eyepiece alone

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.

Step 1. The eyepiece's own magnification, with the final image at the near point $D=25$ cm and $f_e=5$ cm, is $m_e=1+D/f_e=1+25/5=1+5=6$ .

Step 2. The compound microscope's total magnification is the product of the objective's and eyepiece's own magnifications: $m=m_o\times m_e$ .

Step 3. Substituting the given total magnification $m=30$ : $30=m_o\times6$ , so $m_o=30/6=5$ .

Step 4. So the objective lens alone contributes a magnification of 5, and the eyepiece (in near-point focusing) contributes the remaining factor of 6, multiplying together to give the stated overall power of 30.

✓Final answer

m_o=5

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

CBSE 2026Set A1 markMCQ
$Q. When the tube length of microscope is increased, its magnifying power (A) increases (B) decreases (C) becomes zero (D) remains unchanged$
›Reveal solutionSolution

The magnifying power of a compound microscope is proportional to the tube length, so increasing L increases the magnification.

For a compound microscope the magnifying power is approximately

$M = \frac{L}{f_o}\cdot\frac{D}{f_e}$

where $L$ is the tube length (separation between objective and eyepiece foci), $f_o$ and $f_e$ are the focal lengths of the objective and eyepiece, and $D$ is the least distance of distinct vision. Since $M \propto L$ , increasing the tube length increases the magnifying power.

✓Final answer
$(A) increases.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The image formed by a simple microscope is$
(a) imaginary and erect
(b) imaginary and inverted
(c) real and erect
(d) real and inverted
›Reveal solutionSolution

A simple microscope is just a convex lens used with the object inside its focal length, which always produces a virtual, erect, magnified image.

A simple microscope is a single convex lens. When the object is placed BETWEEN the lens and its focal point (object distance less than f), the convex lens produces an image that is on the SAME side as the object, virtual (cannot be caught on a screen - here called 'imaginary'), erect (same orientation as the object), and magnified. This is exactly how a magnifying glass works, letting the eye view a larger, upright image of a small object.

✓Final answer
$(a) imaginary (virtual) and erect.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion (A): The magnifying power of a simple microscope is inversely proportional to the focal length of the lens. Reason (R): Power of a lens is inversely proportional to the focal length.$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) and Reason (R) both are false.
›Reveal solutionSolution

Both statements are true, and since $M = D/f = D\times P$ , the fact that power $P \propto 1/f$ directly explains why magnifying power is inversely proportional to $f$ .

For a simple microscope, magnifying power (image at infinity) is $M = \dfrac{D}{f}$ , where $D$ is the least distance of distinct vision. This can be rewritten as $M = D \times \dfrac{1}{f} = D \times P$ , where $P = \dfrac{1}{f}$ is the power of the lens. So $M$ is directly proportional to the lens's power, and since power is inversely proportional to focal length, $M$ is also inversely proportional to focal length — Reason (R) correctly explains Assertion (A).

✓Final answer
$Both A and R are true and R is the correct explanation of A (option a).$
CBSE 2025Set X11 markMCQ
$Q. Final image of a real object formed by a compound microscope is __________ with respect to the object.$
(a) real, inverted and magnified
(b) virtual, erect and magnified
(c) virtual, erect and diminished
(d) virtual, inverted and magnified
›Reveal solutionSolution

(d) virtual, inverted and magnified. In a compound microscope the objective forms a real, inverted, magnified image, which acts as the object for the eyepiece. The eyepiece then forms a final imag

✓Final answer
$(d) virtual, inverted and magnified. In a compound microscope the objective forms a real, inverted, magnified image, which acts as the object for the eyepiece. The eyepiece then forms a final image that is virtual and magnified, and it remains inverted relative to the original object. Hence the final image is virtual, inverted and magnified.$
CBSE 2025Set IMPROVEMENT1 mark
$Q. Write the formula for the total magnification of a compound microscope when the final image is formed at infinity.$
›Reveal solutionSolution

For a compound microscope with the final image at infinity, the total magnifying power is the product of the objective's linear magnification and the eyepiece's angular magnification.

In a compound microscope, the objective lens forms a real, magnified image of the object; this image acts as the object for the eyepiece, which forms the final image. When the eyepiece is adjusted so that the final image is formed at infinity (normal/relaxed-eye viewing), the total magnifying power is:

$M = m_o \times m_e = \dfrac{L}{f_o}\times\dfrac{D}{f_e}$

where $L$ is the distance between the second focal point of the objective and the first focal point of the eyepiece (approximately the length of the microscope tube), $f_o$ is the focal length of the objective, $f_e$ is the focal length of the eyepiece, and $D = 25\,\text{cm}$ is the least distance of distinct vision.

✓Final answer
$M = \dfrac{L}{f_o}\times\dfrac{D}{f_e}$
CBSE 2024Set A1 markMCQ
$Q. Image formed in compound microscope is (A) real and erect (B) real and inverted (C) virtual and inverted (D) virtual and erect$
›Reveal solutionSolution

A compound microscope forms a final image that is virtual, magnified and inverted → option (C).

In a compound microscope the objective lens first forms a real, inverted, magnified image of the object. This intermediate image acts as the object for the eyepiece, which is used as a simple magnifier and forms the final image that is virtual, further magnified, and inverted relative to the original object. The observer sees this virtual image at (or beyond) the near point.

✓Final answer
$(C) virtual and inverted.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. If the magnification of objective and eyepiece in a compound microscope is 'm_o' and 'm_e' respectively, then the total magnifying power (m) of the microscope will be -$
(a) m_o + m_e
(b) m_o - m_e
(c) m_o . m_e
(d) m_o / m_e
›Reveal solutionSolution

In a compound microscope, the objective forms a magnified real image which the eyepiece further magnifies, so the two magnifications multiply.

In a compound microscope, the objective lens forms a real, magnified, inverted image of the object with magnification $m_o$ . This image acts as the object for the eyepiece, which further magnifies it (acting like a simple magnifier) with magnification $m_e$ .

Since the eyepiece magnifies the image already magnified by the objective, the overall (total) magnifying power is the product of the two:

$m = m_o \times m_e$

✓Final answer
$m_o \cdot m_e$
CBSE 2023Set ANNUAL1 markMCQ
$Q. The image formed by a simple microscope is$
(1) imaginary and erect
(2) imaginary and inverted
(3) real and erect
(4) real and inverted
›Reveal solutionSolution

A simple microscope is a single converging lens used with the object inside its focal length, producing a virtual, erect, magnified image.

When the object is placed within the focal length of a convex lens, the emergent rays diverge, and appear to come from a point behind the lens on the same side as the object — this is a virtual (imaginary), erect, and magnified image, as seen in a simple magnifying glass.

✓Final answer
$(1) imaginary and erect.$
CBSE 2022Set I1 markMCQ
$Q. Magnifying power of simple microscope is (A) M = 1 - D/f (B) M = 1 + D/f (C) M = 1 - f/D (D) M = 1 + f/D$
›Reveal solutionSolution

Simple microscope (image at near point): $M = 1 + \dfrac{D}{f}$ .

A simple microscope is a single convex lens of focal length $f$ used as a magnifier. When the final image is formed at the least distance of distinct vision $D$ ( $\approx 25$ cm), the angular magnification is

$M = 1 + \frac{D}{f}.$

(If the image is at infinity for relaxed viewing, $M = D/f$ .) The ' $+1$ ' term arises because the object is placed just inside the focus so the image is at $D$ . Options with a minus sign or $f/D$ are incorrect.

✓Final answer
$(B) M = 1 + D/f.$
CBSE 2022Set HE2171 markMCQ
$Q. The focal length of eye in compound microscope ________ the focal length of the objective.$
(i) is less than
(ii) is more than
(iii) is equal
(iv) None of these
›Reveal solutionSolution

The eyepiece focal length of a compound microscope is greater than the objective's focal length.

In a compound microscope, the objective lens is placed very close to the small object and has a very small focal length $f_o$ , so that it forms a highly magnified real, inverted, enlarged image close to the eyepiece. The eyepiece then acts like a simple magnifier and further magnifies this image; it is given a comparatively larger focal length $f_e$ so it can view the intermediate image comfortably while still giving good magnification. Because the overall magnifying power is roughly $M \approx \dfrac{L}{f_o}\left(1+\dfrac{D}{f_e}\right)$ (L = tube length), keeping $f_o$ small and $f_e$ larger than $f_o$ gives high total magnification.

✓Final answer
$(ii) is more than.$
CBSE 2022Set HE2171 mark
$Q. Match the following. Column A item: 'Magnifying power of Compound Microscope'. Choose its matching entry from Column B:$
(a) (1 - D/f)
(b) Hertz
(c) Electromagnetic induction
(d) (-v0/u0)(1 + D/fe)
(e) V.I.
(f) Hershel
(g) Instrument of measuring current
(h) Instrument of measuring potential
(i) Polarisation of light
(j) Unit of energy
›Reveal solutionSolution

The magnifying power of a compound microscope is $M = \left(-\dfrac{v_0}{u_0}\right)\left(1+\dfrac{D}{f_e}\right)$ , matching option (d).

The overall magnification of a compound microscope is the product of the linear magnification produced by the objective, $m_o = v_0/u_0$ , and the angular magnification produced by the eyepiece used as a simple magnifier with the final image at the near point, $m_e = 1 + D/f_e$ (D = least distance of distinct vision). This gives the combined magnifying power $M = \left(-\dfrac{v_0}{u_0}\right)\left(1+\dfrac{D}{f_e}\right)$ , which is exactly option (d).

✓Final answer
$\left(-\dfrac{v_0}{u_0}\right)\left(1+\dfrac{D}{f_e}\right)$
CBSE 2021Set A1 markMCQ
$Q. The image formed by simple microscope is (A) Virtual and erect (B) Virtual and inverted (C) Real and erect (D) Real and inverted$
›Reveal solutionSolution

A simple microscope produces a virtual, erect, magnified image.

A simple microscope is a single convex lens used as a magnifying glass. The object is placed within the focal length (between the lens and its focus), so the lens forms an image that is virtual, erect, and enlarged, on the same side as the object (typically at the near point, 25 cm, for maximum comfortable magnification). A real image is not formed for this object position.

✓Final answer
$(A) Virtual and erect.$

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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.

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