The near vision of an average person is 25cm. To view an object with an angular magnification of 10, what shou

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Q. The near vision of an average person is 25cm. To view an object with an angular magnification of 10, what should be the power of the microscope?

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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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Using the near-point magnification of a simple microscope, $M = 1 + D/f = 10$ gives $f = 25/9\ \text{cm}$ , so the power is $P = 36\ \text{D}$ .

Step 1 — Formula for a simple microscope

A single convex lens used as a magnifier, with the final image at the least distance of distinct vision $D = 25\ \text{cm}$ , has angular magnification

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

where $f$ is the focal length in the same units as $D$ .

Step 2 — Solve for the focal length

With $M = 10$ and $D = 25\ \text{cm}$ :

10 = 1 + \frac{25}{f} \;\Rightarrow\; \frac{25}{f} = 9 \;\Rightarrow\; f = \frac{25}{9} \approx 2.78\ \text{cm}

Step 3 — Convert focal length to power

The power in dioptres is $P = 1/f$ with $f$ in metres, i.e. $P = 100/f(\text{cm})$ : …

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

AHSEC Higher Secondary (HS) Final Examination 2020Set ANNUAL3 marks
$Q. Draw a ray diagram of a compound microscope forming an inverted and magnified image of an object. Which lens in the compound microscope acts as a simple microscope? If f0=1cm, fe=2cm and L=20cm respectively, calculate the total magnification of the microscope. OR You know the phenomenon of scattering of light by the atmospheric particles. Write a few lines about the blue colour of sky and reddish colour of sky in the morning as well as in the evening.$
›Reveal solutionSolution
Figure — Hard draw-gate on the first alternative: 'Draw a ray diagram of a compound microscope forming an inverted and

In a compound microscope the eyepiece works as a simple magnifier on the objective's real image; with the given data the overall magnification comes out to about 250.

Main question: A compound microscope has two converging lenses: the objective (near the object, short focal length) and the eyepiece/ocular (near the eye). The objective forms a real, inverted, magnified image of the tiny object just inside the focal length of the eyepiece; the eyepiece then acts exactly like a simple microscope (magnifying glass), further magnifying this intermediate image to give the final image, which is virtual, inverted (relative to the object), and highly magnified.

Using the standard approximate formula for overall magnification (normal adjustment, taking near point $D = 25$ cm, and tube length $L$ as the separation between the lenses):

$m \approx \frac{L}{f_o}\times\frac{D}{f_e}$

With $f_o = 1$ cm, $f_e = 2$ cm, $L = 20$ cm, $D = 25$ cm:

$m \approx \frac{20}{1}\times\frac{25}{2} = 20\times 12.5 = 250$

…

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Q.The near vision of an average person is 25cm. To view an object with an angular magnification of 10, what should be the power of the microscope?

Concept understanding — Microscope Magnification

Microscope Magnification

How the two lenses work together

The objective's magnification

The eyepiece's magnification

Total magnifying power

Worked example

Why this formula?

Microscope Magnification: Why the Formula Holds

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

2. The Key Formula

3. Derivation of Objective Magnification $M_o$

Step 1: How the objective works

Step 2: Using the lens formula

Step 3: The tube length approximation

Step 4: Magnification formula

4. Derivation of Eyepiece Magnification $M_e$

Step 1: The eyepiece as a simple magnifier

Step 2: Angular magnification

Step 1 — Formula for a simple microscope

Step 2 — Solve for the focal length

Step 3 — Convert focal length to power

Method: Solving for Power from a Given Angular Magnification (Simple Microscope)

Steps

Q.The near vision of an average person is 25cm. To view an object with an angular magnification of 10, what should be the power of the microscope?

Step 1 — Formula for a simple microscope

Step 2 — Solve for the focal length

Step 3 — Convert focal length to power

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