(a)(i) Draw a ray diagram to show the image formation by a compound microscope. Obtain the expression for the

Question

Q. (a)(i) Draw a ray diagram to show the image formation by a compound microscope. Obtain the expression for the total magnification of the microscope when the final image is formed at infinity.

(ii) In a compound microscope, an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. The eyepiece has a focal length of 5 cm. The final image is formed at infinity. Calculate the distance between the objective and the eyepiece.

(OR)

(b)(i) Using Huygens' principle, show the refraction of a plane wavefront propagating in air at a plane interface between two media and hence verify Snell's law.

(ii) Use mirror formula to deduce that a convex mirror always produces a virtual image of an object kept in front of it.

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Part (a)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). …

Part (b)Concept understanding — Spherical Mirror Equation

The Spherical Mirror Equation: From Intuition to Formula

Imagine you're standing in front of a concave mirror — the kind that makes your face look bigger when you're close, but flips everything upside down when you step far back. That change isn't magic; it's geometry. The spherical mirror equation is the single relationship that predicts exactly where an image will form, and whether it's real or virtual, for any spherical mirror.

The Core Idea

Every point on an object sends out light rays in all directions. A mirror redirects those rays. The mirror equation tells you: given the mirror's curvature and the object's distance, where will those rays meet again (or appear to meet)?

There are only three quantities you need:

$u$ — object distance (from the mirror's pole)
$v$ — image distance (from the mirror's pole)
$f$ — focal length (a property of the mirror's curvature)

The equation is:

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

The power is in the sign convention, because every distance can point in one of two directions.

The Sign Convention (New Cartesian Sign Convention)

This is where most students slip. The equation works for all spherical mirrors — concave and convex — only if you follow the convention used throughout NCERT and CBSE:

All distances are measured from the mirror's pole.
The incident light is taken to travel left to right, so distances measured in that same direction (to the right) are positive, and distances measured against it (to the left) are negative.
Heights above the principal axis are positive; heights below are negative.

Because a real object is always placed in front of the mirror (to the left, where the incident light originates), its distance $u$ is always negative.

Important

Under this convention, the focal length of a concave mirror is negative (its focus $F$ sits in front of the mirror, on the same side as the object), and the focal length of a convex mirror is positive (its focus lies behind the mirror). This is one of the most frequently tested facts in CBSE board exams.

Watch out

A very common mistake is writing $f$ as positive for a concave mirror because "it converges light." Convergence tells you the type of mirror, not the sign — the sign comes purely from where the focus physically sits relative to the pole, under the convention above.

Where Does the Formula Come From?

For a concave mirror, parallel rays from a distant object converge at the focus, a point at (signed) distance $f$ from the mirror. The derivation uses similar triangles from a ray diagram.

›Proof

Consider an object of height $h_o$ in front of a concave mirror. Draw the ray parallel to the axis: it reflects through the focus $F$ . Draw the ray through the centre of curvature $C$ : it strikes the mirror normally and reflects straight back on itself. These two reflected rays cross to form the image, of height $h_i$ .

From similar triangles formed by the ray through $C$ :

$\frac{h_o}{h_i} = \frac{u - R}{R - v}$

where $R = 2f$ is the radius of curvature (with the same sign convention as $f$ ).

From similar triangles formed by the ray through $F$ :

$\frac{h_o}{h_i} = \frac{u - f}{f}$

Equating the two ratios and simplifying (using $R = 2f$ ) gives:

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

What the Equation Tells You

Rearranging for $v$ :

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

Because $u$ is negative for a real object, and $v$ takes the sign the geometry dictates:

$v$ negative → the image forms in front of the mirror → real image (can be projected on a screen).
$v$ positive → the image forms behind the mirror → virtual image.

For a concave mirror ( $f$ negative), using the magnitude of the object distance $|u|$ measured from the pole:

$|u| > 2|f|$ → real, inverted, diminished image between $f$ and $2f$
$|u| = 2|f|$ → real, inverted, same-size image at $2f$
$|f| < |u| < 2|f|$ → real, inverted, magnified image beyond $2f$
$|u| = |f|$ → image at infinity …

Why this formula?

Spherical Mirror Equation: Why the Formula Holds

The spherical mirror equation — also called the mirror formula — relates the object distance ( $u$ ), image distance ( $v$ ), and focal length ( $f$ ) of a spherical mirror. Let's build the reasoning step by step.

1. The Key Formula

For a spherical mirror (concave or convex):

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

Where:

$f$ = focal length (positive for concave, negative for convex)
$u$ = object distance from pole (always negative by sign convention)
$v$ = image distance from pole (sign depends on image location)

2. Why This Formula Holds — The Derivation

Step 1: Start with a ray diagram

Consider a concave mirror with:

Pole $P$
Centre of curvature $C$ (radius $R$ )
Focus $F$ (midpoint of $PC$ , so $f = R/2$ )

Take an object placed beyond $C$ . Draw two rays from the object's tip:

A ray parallel to the principal axis → reflects through $F$
A ray through $C$ → reflects back along itself

These rays meet at the image point.

Step 2: Use similar triangles

Let the object height be $h_o$ and image height be $h_i$ .

From the geometry of the ray through $C$ :

Triangle formed by object, $C$ , and axis is similar to triangle formed by image, $C$ , and axis.

This gives:

\frac{h_o}{h_i} = \frac{u - R}{R - v}

(Here $u$ and $v$ are distances from $P$ , with sign conventions applied later.)

Step 3: Use the parallel ray

From the ray parallel to the axis:

Triangle formed by object, $F$ , and axis is similar to triangle formed by image, $F$ , and axis.

This gives:

\frac{h_o}{h_i} = \frac{u - f}{f}

Step 4: Equate the two ratios

Since both ratios equal $h_o/h_i$ :

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

Step 5: Substitute $R = 2f$

For a spherical mirror, the focal length is half the radius of curvature:

R = 2f

Substitute:

\frac{u - 2f}{2f - v} = \frac{u - f}{f}

Step 6: Cross-multiply and simplify

Cross-multiply:

f(u - 2f) = (u - f)(2f - v)

Expand:

fu - 2f^2 = 2fu - uv - 2f^2 + fv

Cancel $-2f^2$ on both sides:

fu = 2fu - uv + fv

Bring all terms to one side:

0 = fu - uv + fv

Rearrange:

uv = fu + fv

Step 7: Divide by $uvf$

Divide both sides by $uvf$ :

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

This is the mirror formula.

3. Why the Sign Convention Matters

The derivation above used distances as positive magnitudes. In actual problem-solving, we use the Cartesian sign convention:

Distances measured against incident light are negative
Distances measured along incident light are positive

For a concave mirror:

$u$ is negative (object in front)
$f$ is negative (focus in front)
$v$ is negative for real images (in front) …

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A compound microscope at infinity has $M=\dfrac{L}{f_o}\cdot\dfrac{D}{f_e}$ ; for the given lenses the objective–eyepiece separation is $12.5$ cm.
Huygens' construction gives Snell's law $n_1\sin i=n_2\sin r$ , and the mirror formula shows a convex mirror always gives a virtual ( $v>0$ ) image.

Ray diagram of a compound microscope with the final image at infinity: the objective forms a real, inverted, magnified intermediate image A'B' exactly at the eyepiece's own focal point, so the two construction rays leaving the eyepiece emerge genuinely parallel to each other, sending the final image to infinity for a relaxed eye.

Part (a)

Ray diagram and magnification. Place the object just beyond the objective focus $F_o$ . The objective forms a real, inverted, magnified intermediate image $A'B'$ . For the final image at infinity, $A'B'$ is placed exactly at the eyepiece's first focus, so parallel rays emerge and the eye views the image at infinity (relaxed). The objective's linear magnification is $m_o=\dfrac{v_o}{|u_o|}\approx\dfrac{L}{f_o}$ , where $L$ is the tube length; the eyepiece's angular magnification (image at infinity) is $m_e=\dfrac{D}{f_e}$ with $D=25$ cm. Hence $M=m_o\,m_e=\frac{L}{f_o}\cdot\frac{D}{f_e}.$
Numerical. Objective: $f_o=1.25$ cm, object distance $u_o=-1.5$ cm. Lens formula: $\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}\ \Rightarrow\ \frac{1}{v_o}=\frac{1}{1.25}-\frac{1}{1.5}=\frac{6-5}{7.5}=\frac{1}{7.5},$ so $v_o=7.5$ cm. For the final image at infinity the intermediate image lies at the eyepiece's focus, so the eyepiece is $f_e=5$ cm beyond it. The objective–eyepiece distance is …

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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 26 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

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

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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 orienta …

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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 inver …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. The location of the image is$
(a) formed at 6.67 cm behind the mirror.
(b) formed at 67 cm behind the mirror.
(c) formed at 70 cm same side of the mirror.
(d) formed at 5.57 cm same side of the mirror.
›Reveal solutionSolution

Using the mirror formula with the correct sign convention, the convex mirror forms a virtual image 6.67 cm behind the mirror.

Given: Needle height $h=4.5$ cm, object distance $u = -12$ cm (object in front, so negative by convention), convex mirror so focal length $f = +15$ cm (behind the mirror, positive).

Mirror formula:

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

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - \frac{1}{-12} = \frac{1}{15} + \frac{1}{12}$

Taking LCM (60): $\dfrac{1}{v} = \dfrac{4}{60} + \dfrac{5}{60} = \dfrac{9}{60} = \dfrac{3}{20}$

$v = \frac{20}{3} = 6.67\ \text{cm}$ …

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CBSE 2025Set 55/4/11 markMCQ
$-2.0$
›Reveal solutionSolution
A magnification of $-2.0$ means the image is inverted (negative sign) and magnified (magnitude > 1). Only a concave mirror can produce an inverted, magnified image, and such an image is always real. So the mirror is concave and the image is real — option (B).

Concept and Intuition

The magnification $m$ of a spherical mirror tells you two things at once: the sign tells you orientation, and the magnitude tells you size.

If $m$ is positive, the image is virtual and erect (upright).

If $m$ is negative, the image is real and inverted (upside down).

The magnitude $|m|$ tells you relative size:

$|m| > 1$ → image is magnified (larger than object)

$|m| < 1$ → image is diminished (smaller)

$|m| = 1$ → same size

Here $m = -2.0$ means the image is inverted (negative) and twice as large as the object ( $|m| = 2$ ).

Now, which mirror can produce an inverted, magnified image? A convex mirror always gives a virtual, erect, and diminished image — so it can never produce a negative magnification. A concave mirror, however, can produce both real (inverted) and virtual (erect) images depending on where the object is placed. The real image from a concave mirror is always inverted, and when the object is between the centre of curvature and the focus, that real image is also magnified.

So the only mirror that fits $m = -2.0$ is a concave mirror, and the image must be real.

Step-by-step reasoning

Interpret the sign of $m$

$m = -2.0$ is negative. For spherical mirrors, a negative magnification always means the image is inverted relative to the object. An inverted image formed by a single mirror is always real (it can be projected on a screen). So the image is real.

Interpret the magnitude of $m$

$|m| = 2.0 > 1$ , so the image is magnified — larger than the object.

Eliminate convex mirror

A convex mirror always produces a virtual, erect, and diminished image for any real object. That means $m$ is always positive and $|m| < 1$ . Since our $m$ is negative and $|m| > 1$ , a convex mirror is impossible. This eliminates options (A) and (D).

Check concave mirror possibilities

A concave mirror can produce:

A real, inverted, magnified image when the object is placed between $F$ and $C$ (focus and centre of curvature).

A virtual, erect, magnified image when the object is placed between $P$ and $F$ (pole and focus). In that case $m$ is positive.

Since our $m$ is negative, the image cannot be virtual. So the only possibility is the real, inverted, magnified case — which is exactly what a concave mirror gives for an object between $F$ and $C$ . …
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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 …

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CBSE 2025Set IMPROVEMENT1 markMCQ
$Q. Assertion (A): The radius of curvature of a concave mirror is 20 cm. If an object is placed in front of the mirror at a distance of 10 cm from its pole, its image is formed at infinity. Reason (R): The image of an object placed at the focus of a spherical mirror is formed at infinity. Select the correct option.$
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is incorrect.
(d) Both A and R are incorrect.
›Reveal solutionSolution

Both statements are true, and the reason correctly explains why the assertion is true.

For a concave mirror of radius of curvature $R = 20\,\text{cm}$ , the focal length is $f = R/2 = 10\,\text{cm}$ . In Assertion (A), the object is placed at a distance of 10 cm from the pole — exactly at the focus. Using the mirror formula $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ , when $u = f$ , we get $\dfrac{1}{v} = \dfrac{1}{f}-\dfrac{1}{f} = 0$ , so $v \to \infty$ — the image is indeed formed at infinity. This is exactly the general principle stated in Reason (R): rays from an object placed at the f …

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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:

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

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Focal length of a concave mirror in air is 25 cm. Its focal length in water will be -$
(a) 50 cm
(b) 12.5 cm
(c) $\infty$
(d) 25 cm
›Reveal solutionSolution

A mirror's focal length depends only on its radius of curvature, not on the surrounding medium.

For a spherical mirror, $f = R/2$ , where $R$ is the radius of curvature -- a purely geometrical quantity. Since reflection (unlike refraction) does not depend on the refractive index of the surrounding medium, the focal length …

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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 magnifie …

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CBSE 2024Set A1 markMCQ
$Q. The correct relationship between the radius of curvature (R) and focal length$
(f) of a spherical mirror is ______.
(a) R = 2f
(b) f = 2R
(c) R = f/2
(d) R = 1/f
›Reveal solutionSolution

For a spherical mirror, R = 2f because the focal point lies midway between the pole and the centre of curvature.

For a spherical mirror (concave or convex), a ray parallel to the principal axis, after reflection, passes through (or appears to diverge from) the focus F. Using the mirror geometry, for paraxial rays, the focal length f is related to the radius of curvature R by: …

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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 $m_e$ .

…

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Q.(a)(i) Draw a ray diagram to show the image formation by a compound microscope. Obtain the expression for the total magnification of the microscope when the final image is formed at infinity.

Part (a)

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