A $4.5\ \text{cm}$ needle is placed $12\ \text{cm}$ away from a convex mirror of focal length $15\ \text{cm}$. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

A 4.5cm needle is placed 12cm away from a convex mirror of focal length 15cm. Give the location of the image a

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4.5\ \text{cm}

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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
$|u| < |f|$ → virtual, erect, magnified image behind the mirror (the "shaving mirror" case)

For a convex mirror ( $f$ positive), any real object gives a virtual, erect, diminished image behind the mirror — the familiar "rear-view mirror" case.

The Magnification Link

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

A negative $m$ means the image is inverted relative to the object; a positive $m$ means it is erect.

A Worked Example

A concave mirror has focal length of magnitude $20\ \text{cm}$ , so $f = -20\ \text{cm}$ . An object is placed $30\ \text{cm}$ in front, so $u = -30\ \text{cm}$ .

\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-20} - \frac{1}{-30} = -\frac{1}{20} + \frac{1}{30} = \frac{-3+2}{60} = -\frac{1}{60}

v = -60\ \text{cm}

$v$ is negative, so the image is real, $60\ \text{cm}$ in front of the mirror. Magnification: $m = -v/u = -(-60)/(-30) = -2$ — the image is twice the object's size and inverted, matching the $|f| < |u| < 2|f|$ case above.

The Big Picture

The spherical mirror equation is one instance of a pattern that recurs across optics: the lens formula, the refraction-at-a-spherical-surface formula, and even more advanced optical-system equations share the same reciprocal-distance structure. Master the mirror equation together with its sign convention, and the rest of ray optics — telescopes, microscopes, your own eye — follows the same logic.

The spherical mirror equation, 1/v + 1/u = 1/f, together with the Cartesian sign convention, is one of the most heavily tested formulas in the NCERT Class 12 Physics chapter on ray optics, appearing in nearly every CBSE board paper and in JEE Main/NEET. Searches for "mirror formula sign convention numericals class 12 physics" will find this concave-versus-convex-mirror derivation matches the NCERT textbook precisely.

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)

The formula $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ remains valid with these signed values.

4. Key Insight — Why It's Not Just a Formula

The mirror equation is not an arbitrary rule. It emerges from:

Geometry (similar triangles from ray paths)
Physics (law of reflection: angle of incidence = angle of reflection)
Approximation (paraxial rays — rays close to the axis, so $\sin\theta \approx \theta$ )

For rays far from the axis (marginal rays), spherical mirrors show spherical aberration — the formula breaks down.

5. Quick Summary

Step	What we did
Drew two special rays	Parallel ray → through F; Ray through C → reflects back
Used similar triangles	Two pairs of similar triangles from geometry
Equated height ratios	$h_o/h_i$ from both pairs
Substituted $R = 2f$	Key relation for spherical mirrors
Simplified algebra	Cross-multiplied, cancelled, rearranged
Divided by $uvf$	Got $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

Bottom line: The mirror formula is a direct consequence of the law of reflection applied to a spherical surface, under the paraxial approximation. It's geometry + physics, not magic.

For this convex mirror ( $f=+15\ \text{cm}$ ), a needle at $u=-12\ \text{cm}$ forms a virtual, erect, diminished image at $v\approx+6.67\ \text{cm}$ behind the mirror, with magnification $m\approx+0.56$ . As the needle moves farther away, the image shrinks further and creeps toward the focal point, always staying virtual and erect.

Setting up the sign convention

For a convex mirror, the Cartesian sign convention gives: focal length $f=+15\ \text{cm}$ (focus is behind the mirror), object distance $u=-12\ \text{cm}$ (real object in front), object height $h_o=4.5\ \text{cm}$ .

Applying the mirror formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \quad\Longrightarrow\quad \frac{1}{v} = \frac{1}{f}-\frac{1}{u} = \frac{1}{15}-\frac{1}{-12} = \frac{1}{15}+\frac{1}{12}.

Using LCM $60$ : $\dfrac{1}{15}=\dfrac{4}{60}$ , $\dfrac{1}{12}=\dfrac{5}{60}$ , so

\frac{1}{v} = \frac{4+5}{60} = \frac{9}{60} \quad\Longrightarrow\quad v = \frac{60}{9} \approx 6.67\ \text{cm}.

Since $v$ is positive, the image forms behind the mirror - it is virtual.

Watch out

Plugging in $u=+12$ (forgetting the sign) would give $v=-60\ \text{cm}$ , incorrectly suggesting a real image in front of a convex mirror - something a convex mirror can never do for a real object.

Magnification

m = -\frac{v}{u} = -\frac{60/9}{-12} = \frac{60}{108} = \frac{5}{9} \approx +0.56.

Positive $m$ means the image is erect; $|m|<1$ means it is diminished. Image height: $h_i = m\, h_o = \dfrac{5}{9}\times4.5 \approx 2.5\ \text{cm}$ .

Tip

A quick check using $m=\dfrac{f}{f-u}=\dfrac{15}{15-(-12)}=\dfrac{15}{27}=\dfrac{5}{9}$ confirms the same value with less arithmetic.

As the needle moves farther away

As $|u|\to\infty$ , $\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\to\dfrac{1}{f}$ , so $v\to f=15\ \text{cm}$ - the image creeps toward the focal point from below, but for a convex mirror it never quite reaches or passes it. Correspondingly $m=\dfrac{f}{f-u}\to0$ as $|u|\to\infty$ , so the image keeps shrinking, while remaining virtual and erect at every step - this is the defining, unique behaviour of a convex mirror: for any real object, the image is always virtual, erect, diminished, and confined between the pole and the focus.

✓Final answer

v\approx6.67\ \text{cm}

Method: Mirror Formula & Magnification (Cartesian Sign Convention)

We use the mirror formula and magnification formula with the Cartesian sign convention — the standard for board exams.

Step 1: Assign signs using Cartesian convention

Convex mirror → focal length $f$ is positive.
Object distance $u$ is always negative (object in front of mirror).

Given:

$f = +15\ \text{cm}$
$u = -12\ \text{cm}$
Object height $h_o = +4.5\ \text{cm}$ (positive, upright)

Step 2: Apply mirror formula

Mirror formula:

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

Substitute:

\frac{1}{+15} = \frac{1}{v} + \frac{1}{-12}

\frac{1}{15} = \frac{1}{v} - \frac{1}{12}

\frac{1}{v} = \frac{1}{15} + \frac{1}{12}

Take LCM (60):

\frac{1}{v} = \frac{4 + 5}{60} = \frac{9}{60}

v = \frac{60}{9} = \frac{20}{3} \approx +6.67\ \text{cm}

Image location: $v = +6.67\ \text{cm}$ → behind the mirror (virtual image).

Step 3: Find magnification

Magnification formula:

m = -\frac{v}{u}

Substitute:

m = -\frac{+20/3}{-12} = +\frac{20}{3 \times 12} = +\frac{20}{36} = +\frac{5}{9} \approx +0.556

Magnification: $m = +0.556$ → image is erect and diminished.

Image height:

h_i = m \times h_o = \frac{5}{9} \times 4.5 = 2.5\ \text{cm}

Step 4: What happens as needle moves farther?

As $|u|$ increases (needle moved away):

$v$ increases but remains positive and less than $f$ .
Image stays virtual, erect, and behind the mirror.
Magnification $m = -\frac{v}{u}$ decreases (image becomes smaller).
As $u \to -\infty$ , $v \to f$ (image approaches focus), and $m \to 0$ .

Final Answer

Quantity	Value
Image location	$6.67\ \text{cm}$ behind the mirror
Magnification	$+\frac{5}{9}$ (erect, diminished)
Image height	$2.5\ \text{cm}$

As the needle moves farther, the virtual image moves toward the focus and becomes smaller.

Common Mistakes & How to Avoid Them

Mistake 1: Using the wrong sign for $u$ in the mirror formula

The error: Students often forget that for a convex mirror, the object distance $u$ is negative according to the Cartesian sign convention (object is placed in front of the mirror).

How to avoid:

Always draw a quick ray diagram first — object is on the left of the mirror, so $u$ is negative.
Memorise: $u$ is always negative for real objects (placed in front of the mirror).
For this problem: $u = -12\ \text{cm}$ .

Mistake 2: Forgetting that $f$ is positive for a convex mirror

The error: Some students treat $f$ as negative because they associate "convex" with diverging — but the sign convention says:

Convex mirror → $f$ is positive (focus is behind the mirror).

How to avoid:

Remember: $f > 0$ for convex mirrors, $f < 0$ for concave mirrors.
Quick check: For a convex mirror, the image is always virtual and behind the mirror — so $v$ comes out positive.

Mistake 3: Incorrectly applying the mirror formula

The error: Plugging in $u$ and $f$ without checking signs leads to wrong $v$ .

Correct approach:

Mirror formula:

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

Substitute:

\frac{1}{15} = \frac{1}{v} + \frac{1}{(-12)}

\frac{1}{v} = \frac{1}{15} + \frac{1}{12} = \frac{4 + 5}{60} = \frac{9}{60}

v = \frac{60}{9} = 6.67\ \text{cm}

Key: $v$ is positive → image is behind the mirror (virtual).

Mistake 4: Confusing magnification formula sign

The error: Using $m = -\frac{v}{u}$ without checking sign convention.

Correct:

m = -\frac{v}{u} = -\frac{6.67}{(-12)} = +0.556

Positive $m$ → image is erect (virtual image).
$|m| < 1$ → image is diminished.

How to avoid:

For convex mirrors, $m$ is always positive and less than 1.
If you get a negative $m$ , recheck your signs.

Mistake 5: Misinterpreting "as needle moves farther"

The error: Students think the image size keeps decreasing indefinitely or the image disappears.

Correct description:

As the needle moves farther from the mirror:

$|u|$ increases (more negative)
$v$ approaches $f$ from behind (i.e., $v \to 15\ \text{cm}$ )
$|m|$ decreases and approaches zero
The image remains virtual, erect, and diminished, moving closer to the focus behind the mirror.

How to avoid:

Think of the limiting case: $u \to -\infty$ → $v \to f$ and $m \to 0$ .
The image never disappears — it just becomes a tiny point at the focus.

Mistake 6: Not stating the nature of the image

The error: Giving only numerical values without describing the image.

Always mention:

Position: $6.67\ \text{cm}$ behind the mirror
Nature: Virtual, erect, diminished
Magnification: $0.556$ (size = $4.5 \times 0.556 = 2.5\ \text{cm}$ )

Quick Checklist to Avoid All Mistakes

Step	What to do	Common Pitfall
1	Set $u = -12\ \text{cm}$	Using $+12$
2	Set $f = +15\ \text{cm}$	Using $-15$
3	Use $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$	Wrong sign in formula
4	Solve for $v$ → positive	Forgetting to invert
5	$m = -\frac{v}{u}$ → positive	Using $m = \frac{v}{u}$
6	Describe image: virtual, erect, diminished	Only giving numbers

Final answer for the problem:

Image location: $6.67\ \text{cm}$ behind the mirror
Magnification: $+0.556$
As needle moves farther: Image moves toward the focus ( $15\ \text{cm}$ behind mirror) and becomes smaller, always remaining virtual and erect.

←Previous question9.1 — A small candle, 2.5cm in size is placed at 27cm in front of a concave …→Next question9.3 — A tank is filled with water to a height of 12.5cm. The apparent depth …

Q.A 4.5 cm4.5\ \text{cm}4.5 cm needle is placed 12 cm12\ \text{cm}12 cm away from a convex mirror of focal length 15 cm15\ \text{cm}15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Setting up the sign convention

Applying the mirror formula

Magnification

As the needle moves farther away

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Q.A $4.5\ \text{cm}$ needle is placed $12\ \text{cm}$ away from a convex mirror of focal length $15\ \text{cm}$ . Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

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