(a) At what distance should the lens be held from the card sheet in Exercise 9.22 in order to view the squares

Exercises · 9.23

Q. (a) At what distance should the lens be held from the card sheet in Exercise 9.22 in order to view the squares distinctly with the maximum possible magnifying power?

(b) What is the magnification in this 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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With the $f = 9\ \text{cm}$ magnifier of Exercise 9.22, maximum magnifying power occurs with the image at the near point ( $v=-25\ \text{cm}$ ): the lens is held $\tfrac{225}{34}\approx6.6\ \text{cm}$ from the card, the linear magnification is $\tfrac{34}{9}\approx3.8$ , and it equals the magnifying power in this case.

(a) Distance for maximum magnifying power

A simple magnifier gives its largest angular magnification when the virtual image forms at the near point, $D=25\ \text{cm}$ , because the object can then sit closest to the lens while staying in focus. Setting $v=-25\ \text{cm}$ (virtual, same side as object) with $f=+9\ \text{cm}$ :

\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{-25} - \frac{1}{9} = -\frac{9 + 25}{225} = -\frac{34}{225} \Rightarrow u = -\frac{225}{34} \approx -6.6\ \text{cm}.

The lens should be held about $6.6\ \text{cm}$ from the card sheet.

(b) Magnification

m = \frac{v}{u} = \frac{-25}{-225/34} = \frac{25 \times 34}{225} = \frac{34}{9} \approx 3.8.

(c) Magnification versus magnifying power

The magnifying power with the image at the near point is

$M = 1 + \frac{D}{f} = 1 + \frac{25}{9} = \frac{34}{9} \approx 3.8.$ …

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Method: Angular Magnification (Magnifying Power) for a Simple Microscope (Single Convex Lens)

We use the lens formula and the definition of magnifying power for a simple microscope when the image is formed at the near point (D = 25 cm) — this gives the maximum possible magnifying power.

Step-by-step solution

Step 1: Identify given data (from Exercise 9.22 context)

Focal length of lens, $f = 5 \text{ cm}$ (assumed standard for such problems)
Least distance of distinct vision, $D = 25 \text{ cm}$
Object is a card sheet with squares of size $0.5 \text{ mm}$

Step 2: Condition for maximum magnifying power

The maximum magnifying power occurs when the final image is formed at the near point (D).

For a convex lens, the image is virtual and on the same side as the object.

Step 3: Apply lens formula

Lens formula:

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

Here:

$v = -D = -25 \text{ cm}$ (negative because image is virtual)
$f = +5 \text{ cm}$ (convex lens)

Substitute:

\frac{1}{-25} - \frac{1}{u} = \frac{1}{5}

\Rightarrow -\frac{1}{25} - \frac{1}{u} = \frac{1}{5}

\Rightarrow -\frac{1}{u} = \frac{1}{5} + \frac{1}{25} = \frac{5 + 1}{25} = \frac{6}{25}

\Rightarrow \frac{1}{u} = -\frac{6}{25}

\Rightarrow u = -\frac{25}{6} \approx -4.17 \text{ cm}

Step 4: Answer (a)

The lens should be held at a distance of $4.17$ cm from the card sheet.

(The negative sign indicates the object is on the same side as the incoming light — standard sign convention.)

Step 5: Calculate magnification (b)

Linear magnification for a simple microscope:

m = \frac{v}{u} = \frac{-25}{-25/6} = 6

Magnification = 6 (the image is 6 times larger linearly). …

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Common Mistakes: Microscope Magnification (Exercise 9.22)

Mistake 1: Confusing Magnification with Magnifying Power

The error: Students treat magnification ( $m$ ) and magnifying power ( $M$ ) as the same thing in all contexts.

Why it happens: Both involve "how big" the image appears, but they measure different things:

Magnification ( $m$ ) = $\frac{\text{image height}}{\text{object height}} = \frac{v}{u}$ (purely geometric)
Magnifying power ( $M$ ) = $\frac{\text{angle subtended by image at eye}}{\text{angle subtended by object at near point}}$ (visual perception)

How to avoid: Always check what the question asks. In part (c), the answer is No — they are equal only when the image is formed at the near point (25 cm). For relaxed eye (image at infinity), $M \neq m$ .

Mistake 2: Using the Wrong Formula for Maximum Magnifying Power

The error: Students use $M = 1 + \frac{D}{f}$ when the question asks for "maximum possible magnifying power."

Why it happens: There are two standard formulas:

Image at near point (maximum $M$ ): $M_{\text{max}} = 1 + \frac{D}{f}$
Image at infinity (relaxed eye): $M_{\infty} = \frac{D}{f}$

How to avoid: "Maximum possible magnifying power" always means the image is at the near point ( $v = -25$ cm). Use $M = 1 + \frac{D}{f}$ .

Mistake 3: Forgetting Sign Conventions in Lens Formula

The error: Plugging $u$ and $v$ without signs into $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ .

Why it happens: The Cartesian sign convention is often ignored.

How to avoid: For a convex lens used as a simple microscope:

$u$ is negative (object on left)
$v$ is negative (virtual image on same side as object)
$f$ is positive (convex lens)

So for image at near point: $v = -25$ cm, solve $\frac{1}{f} = \frac{1}{-25} - \frac{1}{u}$ .

Mistake 4: Using $D = 25$ cm Without Checking Units

The error: Mixing cm and m in calculations.

Why it happens: $D$ (least distance of distinct vision) is usually 25 cm, but $f$ might be given in cm or m.

How to avoid: Convert everything to the same unit before calculating. If $f = 5$ cm, keep $D = 25$ cm. If $f = 0.05$ m, use $D = 0.25$ m.

--- …

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

COMEDK 2026Set 2026-A1 markMCQ
$Q. The distance between the objective and eye piece of astronomical telescope in normal adjustment is 27 cm and its magnifying power is 8 . What is the focal length of the eye piece? (A) 12 cm (B) 3 cm (C) 6 cm (D) 24 cm$
›Reveal solutionSolution
[!TLDR]

From $f_o+f_e=27$ and $f_o/f_e=8$ , the eyepiece focal length is 3 cm — option (B).

Concept

For an astronomical telescope in normal adjustment, the objective and eyepiece are separated by $L=f_o+f_e$ , and the magnifying power is $m=f_o/f_e$ (CBSE/NCERT Class 12 Ray Optics).

Solution

Given $L=f_o+f_e=27$ cm and $m=\dfrac{f_o}{f_e}=8$ , so $f_o=8f_e$ . Substituting:

$8f_e+f_e=27\ \Rightarrow\ 9f_e=27\ \Rightarrow\ f_e=3\ \text{cm}.$ …
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COMEDK 2026Set 2026-M1 markMCQ
$(25 \mathrm{~cm})$
›Reveal solutionSolution

Working back from the eyepiece: object distance for the objective is $7.5\text{ cm}$ and the total magnification is $10$ — matching the choice "Object distance $=7.5$ cm; Total magnification $=10$ ", listed as option (D).

Step 1 — Eyepiece (final image at $D=25$ cm).

The final virtual image is on the same side as the object, so $v_e=-25\text{ cm}$ , $f_e=6.25\text{ cm}$ .
$\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}\ \Rightarrow\ \frac{1}{-25}-\frac{1}{u_e}=\frac{1}{6.25}$ $\frac{1}{u_e}=-0.04-0.16=-0.20\ \Rightarrow\ u_e=-5\text{ cm}$
So the intermediate image sits $5\text{ cm}$ in front of the eyepiece.

Step 2 — Locate the intermediate image relative to the objective.

The lenses are $20\text{ cm}$ apart, so the objective's real image is at
$v_o=20-5=15\text{ cm}.$
Step 3 — Objective (find the object distance). …
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COMEDK 2025Set 2025-E1 markMCQ
$\mathrm{F}_{\mathrm{O}}=28 \mathrm{~cm}, \mathrm{Fe}^{\mathrm{e}}=7 \mathrm{~cm}$
›Reveal solutionSolution
For an astronomical telescope in normal adjustment, the tube length equals the sum of the focal lengths, and the magnifying power equals the ratio of the objective focal length to the eyepiece focal length. Solving these two equations gives $f_o = 32\ \text{cm}$ and $f_e = 4\ \text{cm}$ , which corresponds to option (C).

The key idea is that in normal adjustment, the telescope is set so that the final image is at infinity. This means the eyepiece is positioned so that the intermediate image formed by the objective lies exactly at the focal point of the eyepiece. Consequently, the distance between the objective and the eyepiece (the tube length) is simply the sum of their focal lengths:
$L = f_o + f_e.$
The magnifying power $M$ of an astronomical telescope in normal adjustment is defined as the ratio of the angle subtended by the image to the angle subtended by the object, and it simplifies to
$M = \frac{f_o}{f_e}.$
We are given $L = 36\ \text{cm}$ and $M = 8$ . So we have two equations in two unknowns — a straightforward system.

Set up the equations From the magnifying power:

$\frac{f_o}{f_e} = 8 \quad \Rightarrow \quad f_o = 8 f_e.$
From the tube length:
$f_o + f_e = 36.$

Substitute and solve Replace $f_o$ in the second equation:

$8 f_e + f_e = 36 \quad \Rightarrow \quad 9 f_e = 36 \quad \Rightarrow \quad f_e = 4\ \text{cm}.$
Then
$f_o = 8 \times 4 = 32\ \text{cm}.$

Check against the options The pair $(f_o = 32\ \text{cm}, f_e = 4\ \text{cm})$ matches option (C). …
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COMEDK 2024Set 2024-E1 markMCQ
$60 \mathrm{~cm}$
›Reveal solutionSolution

The objective forms a real image ( $m_1=-\tfrac14$ ); the eyepiece throws the final image to the near point ( $m_2=+6$ ), giving a net magnification $M=-1.5$ .

The telescope is used on a near object: the object sits $300\,\text{cm}$ from the objective, and the final image is formed at the least distance of distinct vision, $D=25\,\text{cm}$ .

Objective ( $f_o=60\,\text{cm}$ , $u_1=-300\,\text{cm}$ ):
$\frac{1}{v_1}=\frac{1}{f_o}+\frac{1}{u_1}=\frac{1}{60}-\frac{1}{300}=\frac{4}{300}\;\Rightarrow\; v_1=+75\,\text{cm}$ $m_1=\frac{v_1}{u_1}=\frac{75}{-300}=-\frac14$
Eyepiece ( $f_e=5\,\text{cm}$ , final image $v_2=-25\,\text{cm}$ at the near point): …
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COMEDK 2024Set 2024-M1 markMCQ
$32 \mathrm{~cm}$
›Reveal solutionSolution
In normal adjustment, the telescope length equals the sum of the focal lengths, and the magnifying power equals the ratio of the objective focal length to the eyepiece focal length. Solving these two equations gives $f_o = 28\ \text{cm}$ and $f_e = 4\ \text{cm}$ , so the correct option is (D).

Concept & Intuition

An astronomical telescope in normal adjustment means the final image is formed at infinity — the eyepiece is set so that the light rays emerging from it are parallel. This happens when the image formed by the objective lies exactly at the first focal point of the eyepiece. Consequently, the distance between the two lenses (the tube length) is simply the sum of their focal lengths:
$L = f_o + f_e$
The magnifying power (angular magnification) in this setting is given by the ratio of the objective’s focal length to the eyepiece’s focal length:
$M = \frac{f_o}{f_e}$
We are given $L = 32\ \text{cm}$ and $M = 7$ . That gives us two equations in two unknowns — straightforward algebra.

Step-by-step solution

Write the two conditions From normal adjustment:

$f_o + f_e = 32 \quad \text{(1)}$
From magnifying power:
$\frac{f_o}{f_e} = 7 \quad \text{(2)}$

Express one variable in terms of the other From (2):

$f_o = 7 f_e$

Substitute into the length equation
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COMEDK 2023Set 2023-E1 markMCQ
$25 \mathrm{~cm}$
›Reveal solutionSolution

(The value D/f = 2.5 is the magnifying power for the image at INFINITY (relaxed eye), which is option (D) - not what is asked here.)

Concept: simple microscope (magnifying glass). When the final image is formed at the least distance of distinct vision D (image at the near point), the magnifying power is

M = 1 + D/f.

Given D = 25 cm, f = 10 cm:

M = 1 + 25/10 = 1 + 2.5 = 3.5. …
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KCET 2022Set B-31 markMCQ
$\dfrac{(m+1)^2 x}{m}$
›Reveal solutionSolution
Split the fixed object–screen distance $x$ into $u$ and $v$ using the magnification ratio, then feed both into $f = \dfrac{uv}{u+v}$ .

1. Set up the geometry

The lens sits between the object and the screen, and it forms a real image on the screen. Working with numerical (unsigned) distances:

object distance $= u$

image distance $= v$

they must add up to the object–screen separation:

$u + v = x \qquad (1)$
2. Use the magnification

For a thin lens the numerical magnification of a real image is
$m = \frac{v}{u} \quad\Longrightarrow\quad v = m\,u \qquad (2)$
3. Solve for $u$ and $v$

Substituting (2) into (1):
$u + mu = x \;\Longrightarrow\; u(1+m) = x \;\Longrightarrow\; \boxed{u = \frac{x}{m+1}}$ $v = m u = \boxed{\frac{mx}{m+1}}$
4. Apply the lens formula

In magnitudes, the thin-lens relation for a real object and real image is
$\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{u+v}{uv} \quad\Longrightarrow\quad f = \frac{uv}{u+v}$
Now substitute, noting the denominator $u+v$ is just $x$ :
$f = \frac{\left(\dfrac{x}{m+1}\right)\left(\dfrac{mx}{m+1}\right)}{x} = \frac{\dfrac{m x^{2}}{(m+1)^{2}}}{x}$ $\therefore\quad f = \frac{m\,x}{(m+1)^{2}}$
5. Sanity checks …
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COMEDK 2021Set 20211 markMCQ
$Q. The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm. The focal length of lenses are (A) 10 cm, 10 cm (B) 15 cm, 5 cm (C) 18 cm, 2 cm (D) 11 cm, 9 cm$
›Reveal solutionSolution

Focal lengths: 18 cm (objective) and 2 cm (eyepiece).

Concept: astronomical telescope in normal adjustment (parallel rays / relaxed eye).

Magnifying power m = f_o/f_e = 9

Tube length L = f_o + f_e = 20 cm

Substituting f_o = 9 f_e: …
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KCET 2020Set A-11 markMCQ
$x$
›Reveal solutionSolution

A beam already converging towards a point beyond the lens means that point is a virtual object ( $u$ positive); apply $\frac1v - \frac1u = \frac1f$ with $u = +12\ \text{cm}$ , $f = -16\ \text{cm}$ .

Step 1 — Read the geometry

From the figure: light converges towards $P$ , and the concave lens is inserted at the dotted line through $O$ , with $OP = 12\ \text{cm}$ measured along the axis, on the far (outgoing) side of the lens.

Step 2 — Identify the virtual object

If the lens were absent, the rays would meet at $P$ . Because the rays are already converging when they strike the lens, they never actually diverge from a real object point. The point $P$ — where they would have met — acts as a virtual object.

In the Cartesian sign convention (light travelling left → right, distances measured from the optical centre, rightward positive):
$u = +12\ \text{cm} \quad (\text{virtual object, on the outgoing side})$ $f = -16\ \text{cm} \quad (\text{concave / diverging lens})$
The positive $u$ is the whole trick of the question — a real object would give a negative $u$ .

Step 3 — Apply the lens formula

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ …
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Q.(a) At what distance should the lens be held from the card sheet in Exercise 9.22 in order to view the squares distinctly with the maximum possible magnifying power?

(a) Distance for maximum magnifying power

(b) Magnification

(c) Magnification versus magnifying power

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