A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near p

NCERT Exemplar · Q15

Q. A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in

(a) a larger angle to be subtended by the object at the eye and hence viewed in greater detail.

(b) the formation of a virtual erect image.

(d) infinite magnification at the near point.

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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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A simple magnifier works by letting the object be placed closer than the unaided near point, so it subtends a larger angle at the eye, and it forms a magnified virtual, erect image. Correct options: (a) and (b).

Concept understanding. The unaided eye cannot focus on anything nearer than the near point $D \approx 25\ \text{cm}$ . The apparent size of an object is set by the angle it subtends at the eye, $\theta_0 \approx h/D$ for object height $h$ . A convex lens of short focal length placed near the eye lets the object be brought to a distance $u < f$ , forming a virtual, upright, enlarged image on which the eye can comfortably focus. The angular magnification is

M = \frac{\theta}{\theta_0} = 1 + \frac{D}{f}\quad(\text{image at near point}),\qquad M=\frac{D}{f}\quad(\text{image at infinity}),

both finite.

Testing each option.

(a) Because the object can be held nearer than $D$ , it subtends a larger angle at the eye, so finer detail becomes resolvable. This is precisely why the magnifier helps. Correct.
(b) With the object inside the focal length, the lens produces a virtual, erect, magnified image. Correct. …

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Q.A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in

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

Method: Evaluating Multi-Statement Conceptual MCQs About a Simple Magnifier

Steps

Q.A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in

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