In many experimental set-ups the source and screen are fixed at a distance say $D$ and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.

> [!TLDR] > For a fixed object-screen separation $D > 4f$, a thin lens has two distinct positions that form a sharp image on the screen — these positions are symmetric about the midpoint, separated by $\sqrt{D(D-4f)}$, and the image sizes are in the ratio $1:1$ (they are reciprocals of each other). This is the classic **displacement method** for finding the focal length of a lens. The key insight is that the lens formula is symmetric: if $u$ and $v$ are the object and image distances, swapping them gives the same $f$. When the total distance $D = u + v$ is fixed and larger than $4f$, the quadratic in $u$ has two real roots — these correspond to the two lens positions. Let’s work through it step by step. --- 1. **Set up the geometry.** The object and screen are fixed, separated by distance $D$. The lens is placed between them. Let $u$ be the distance from the object to the lens, and $v$ the distance from the lens to the screen. Then $$u + v = D.$$ 2. **Apply the thin lens formula.** For a real image on the screen, both $u$ and $v$ are positive. The lens formula is $$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}.$$ Substitute $v = D - u$: $$\frac{1}{u} + \frac{1}{D-u} = \frac{1}{f}.$$ 3. **Obtain the quadratic in $u$.** Combine the fractions: $$\frac{D-u + u}{u(D-u)} = \frac{D}{u(D-u)} = \frac{1}{f}.$$ Cross-multiply: $$Df = u(D-u) \quad \Rightarrow \quad u^2 - Du + Df = 0.$$ 4. **Two real positions exist when $D > 4f$.** The discriminant is $$\Delta = D^2 - 4Df = D(D-4f).$$ For a real image to form, we need $\Delta > 0$, i.e. $D > 4f$. Then the two roots are $$u_1 = \frac{D - \sqrt{D(D-4f)}}{2}, \quad u_2 = \frac{D + \sqrt{D(D-4f)}}{2}.$$ Notice that $u_1$ and $u_2$ are symmetric about $D/2$, and $u_1 + u_2 = D$, $u_1 u_2 = Df$. > [!WARNING] > If $D = 4f$, the two positions coincide at $u = D/2$ — only one position works. If $D < 4f$, no real image forms on the screen at all. 5. **Fin

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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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For a fixed object-screen separation $D > 4f$ , a thin lens has two distinct positions that form a sharp image on the screen — these positions are symmetric about the midpoint, separated by $\sqrt{D(D-4f)}$ , and the image sizes are in the ratio $1:1$ (they are reciprocals of each other).

This is the classic displacement method for finding the focal length of a lens. The key insight is that the lens formula is symmetric: if $u$ and $v$ are the object and image distances, swapping them gives the same $f$ . When the total distance $D = u + v$ is fixed and larger than $4f$ , the quadratic in $u$ has two real roots — these correspond to the two lens positions.

Let’s work through it step by step.

Set up the geometry. The object and screen are fixed, separated by distance $D$ . The lens is placed between them. Let $u$ be the distance from the object to the lens, and $v$ the distance from the lens to the screen. Then

u + v = D.

Apply the thin lens formula. For a real image on the screen, both $u$ and $v$ are positive. The lens formula is

\frac{1}{u} + \frac{1}{v} = \frac{1}{f}.

Substitute $v = D - u$ :

\frac{1}{u} + \frac{1}{D-u} = \frac{1}{f}.

Obtain the quadratic in $u$ . Combine the fractions:

\frac{D-u + u}{u(D-u)} = \frac{D}{u(D-u)} = \frac{1}{f}.

Cross-multiply:

Df = u(D-u) \quad \Rightarrow \quad u^2 - Du + Df = 0.

Two real positions exist when $D > 4f$ . The discriminant is

\Delta = D^2 - 4Df = D(D-4f).

For a real image to form, we need $\Delta > 0$ , i.e. $D > 4f$ . Then the two roots are

u_1 = \frac{D - \sqrt{D(D-4f)}}{2}, \quad u_2 = \frac{D + \sqrt{D(D-4f)}}{2}.

Notice that $u_1$ and $u_2$ are symmetric about $D/2$ , and $u_1 + u_2 = D$ , $u_1 u_2 = Df$ .

Watch out

If $D = 4f$ , the two positions coincide at $u = D/2$ — only one position works. If $D < 4f$ , no real image forms on the screen at all.

Find the distance between the two lens positions. The lens positions differ by

|u_2 - u_1| = \sqrt{D(D-4f)}.

This is the separation between the two points where the lens can be placed.

Find the ratio of image sizes. Magnification for a thin lens is $m = v/u$ . For the first position: $m_1 = \frac{v_1}{u_1} = \frac{D - u_1}{u_1}.$ …

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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.In many experimental set-ups the source and screen are fixed at a distance say $D$ and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.

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: The Displacement Method for Two Conjugate Lens Positions

Steps

Concept

Solution

1. Set up the geometry

2. Use the magnification

3. Solve for $u$ and $v$

4. Apply the lens formula

5. Sanity checks …

Step 1 — Read the geometry

Step 2 — Identify the virtual object

Step 3 — Apply the lens formula

Q.In many experimental set-ups the source and screen are fixed at a distance say DDD and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.

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Q.In many experimental set-ups the source and screen are fixed at a distance say $D$ and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.

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