A small candle, $2.5\ \text{cm}$ in size is placed at $27\ \text{cm}$ in front of a concave mirror of radius of curvature $36\ \text{cm}$. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

A small candle, 2.5cm in size is placed at 27cm in front of a concave mirror of radius of curvature 36cm. At w

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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.

Using the mirror formula $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ with $f = -18\ \text{cm}$ (concave mirror) and $u = -27\ \text{cm}$ , we find $v = -54\ \text{cm}$ . The screen must be placed 54 cm in front of the mirror. The image is real, inverted, and $5.0\ \text{cm}$ tall. Moving the candle closer requires moving the screen away from the mirror until the object reaches the focal point.

1. Understanding the physics: Why the mirror formula works

A concave mirror converges light. When an object is placed beyond the centre of curvature ( $C$ ), the image forms between $C$ and $F$ — real and inverted. When the object is between $F$ and the pole, the image is virtual and erect. The mirror formula ties object distance $u$ , image distance $v$ , and focal length $f$ :

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

Sign convention (Cartesian): distances measured from the pole. For a concave mirror, $f$ is negative, $u$ is negative (object in front), and $v$ is negative for a real image (in front of the mirror).

2. Step-by-step solution

Step 1: Find the focal length.

Radius of curvature $R = 36\ \text{cm}$ . For any spherical mirror, $f = R/2$ .

f = \frac{36}{2} = 18\ \text{cm}

Since it's concave, $f = -18\ \text{cm}$ .

Step 2: Write the object distance.

Object is placed $27\ \text{cm}$ in front of the mirror.

u = -27\ \text{cm}

Step 3: Apply the mirror formula.

\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-18} - \frac{1}{-27} = -\frac{1}{18} + \frac{1}{27}

Find a common denominator (LCM = 54):

-\frac{3}{54} + \frac{2}{54} = -\frac{1}{54}

Thus:

v = -54\ \text{cm}

Watch out

A common mistake is forgetting the negative signs. If you plug $u = 27$ and $f = 18$ without signs, you get $v = 54\ \text{cm}$ — but that would be for a convex mirror. Always apply the sign convention.

Step 4: Interpret $v$ .

The negative sign means the image is formed in front of the mirror — real and inverted. The screen must be placed $54\ \text{cm}$ from the mirror on the same side as the object.

Step 5: Find the magnification and image size.

Magnification $m = -\frac{v}{u}$ :

m = -\frac{(-54)}{(-27)} = -\frac{54}{27} = -2

The negative sign indicates inversion. Image height $h_i = m \times h_o$ :

h_i = (-2) \times 2.5\ \text{cm} = -5.0\ \text{cm}

The magnitude $5.0\ \text{cm}$ tells the size; the negative sign confirms inversion.

Tip

Magnification $|m| > 1$ means the image is enlarged. Here $|m| = 2$ , so the image is twice the object size.

Step 6: Describe the image.

Real (can be projected on a screen)
Inverted (upside down)
Magnified ( $5.0\ \text{cm}$ tall)
Formed 54 cm in front of the mirror

Step 7: What happens when the candle is moved closer?

If the object moves toward the mirror (i.e., $|u|$ decreases), the image distance $v$ changes. Let's examine two cases:

Object beyond $C$ ( $|u| > 36\ \text{cm}$ ): Image between $C$ and $F$ , real and smaller.
Object between $C$ and $F$ ( $18 < |u| < 36$ ): Image beyond $C$ , real and enlarged.
Object at $F$ ( $|u| = 18\ \text{cm}$ ): Image at infinity — no sharp image on any screen.
Object between $F$ and pole ( $|u| < 18\ \text{cm}$ ): Image virtual, behind the mirror — cannot be caught on a screen.

In our problem, the candle is at $27\ \text{cm}$ (between $C$ and $F$ ). Moving it closer to the mirror means $|u|$ decreases from $27$ toward $18\ \text{cm}$ . From the mirror formula:

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

As $|u|$ decreases, $\frac{1}{|u|}$ increases, so $\frac{1}{v}$ becomes more negative — meaning $|v|$ increases. The screen must be moved farther away from the mirror.

When the candle reaches $18\ \text{cm}$ (the focal point), $v \to \infty$ — no image on any screen. Beyond that, the image becomes virtual and the screen is useless.

Important

For a concave mirror, as the object moves from infinity toward the focus, the real image moves from the focus toward infinity. The screen must be moved away from the mirror to keep the image sharp — until the object reaches the focus, after which no real image forms.

✓Final answer

The screen must be placed 54 cm in front of the concave mirror to obtain a sharp, real, inverted image of size 5.0 cm . If the candle is moved closer to the mirror, the screen must be moved farther away until the candle reaches the focal point, beyond which no real image is formed.

Method: Mirror Formula and Magnification Approach

This problem uses the Mirror Formula combined with Magnification relations — the standard method for all spherical mirror image-location problems in ray optics.

Step 1: Identify given data and sign convention

Sign convention: Cartesian sign convention (all distances measured from pole, direction of incident light is positive).

Object size, $h_o = +2.5\ \text{cm}$
Object distance, $u = -27\ \text{cm}$ (negative: object is in front of mirror)
Radius of curvature, $R = -36\ \text{cm}$ (negative for concave mirror)

Step 2: Find focal length

For a spherical mirror:

f = \frac{R}{2}

f = \frac{-36}{2} = -18\ \text{cm}

Step 3: Apply mirror formula to find image distance

Mirror formula:

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

Substitute $u = -27\ \text{cm}$ , $f = -18\ \text{cm}$ :

\frac{1}{v} + \frac{1}{-27} = \frac{1}{-18}

\frac{1}{v} - \frac{1}{27} = -\frac{1}{18}

\frac{1}{v} = -\frac{1}{18} + \frac{1}{27}

Take LCM (54):

\frac{1}{v} = \frac{-3 + 2}{54} = -\frac{1}{54}

\boxed{v = -54\ \text{cm}}

Screen should be placed 54 cm in front of the mirror.

Step 4: Find magnification and image size

Magnification:

m = -\frac{v}{u} = -\frac{(-54)}{(-27)} = -\frac{54}{27} = -2

Image size:

h_i = m \times h_o = (-2) \times (2.5) = \boxed{-5\ \text{cm}}

Step 5: Describe the nature of the image

Real (since $v$ is negative, image is in front of mirror)
Inverted (since $m$ is negative)
Magnified (since $|m| > 1$ )
Size: 5 cm (inverted)

Step 6: Effect of moving candle closer

If the candle is moved closer to the mirror (i.e., $|u|$ decreases):

For a concave mirror, as object moves from beyond $C$ toward $F$ , the image moves farther away from the mirror (beyond $C$ ).
The screen must be moved away from the mirror to capture the sharp image.

Rule: For a concave mirror, object moving toward mirror → image moves away from mirror (until object reaches focus, after which image becomes virtual).

Final Answer Summary

Quantity	Value
Screen distance	54 cm in front of mirror
Image nature	Real, inverted, magnified
Image size	5 cm
If candle moves closer	Move screen away from mirror

Here are the common mistakes students make with this exact problem, and how to avoid each one.

1. Confusing Radius of Curvature ( $R$ ) with Focal Length ( $f$ )

The Mistake:

Students plug $R = 36\ \text{cm}$ directly into the mirror formula as $f$ .

Why it’s wrong:

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

f = \frac{R}{2}

How to avoid:

Always write the relation first:

f = \frac{R}{2} = \frac{36}{2} = 18\ \text{cm}

Then use $f = -18\ \text{cm}$ (concave mirror → negative focal length as per Cartesian sign convention).

2. Forgetting the Sign Convention (Cartesian)

The Mistake:

Using $u = 27\ \text{cm}$ as positive, or treating $f$ as positive.

Why it’s wrong:

In the Cartesian sign convention:

Distances measured against the incident light direction are negative.
For a concave mirror, both $u$ and $f$ are negative.

How to avoid:

Always draw a quick ray diagram and label directions. Then:

u = -27\ \text{cm}, \quad f = -18\ \text{cm}

3. Using the Lens Formula Instead of the Mirror Formula

The Mistake:

Writing $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ (lens formula) instead of the mirror formula.

Why it’s wrong:

The mirror formula is:

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

How to avoid:

Memorise separately:

Mirror: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
Lens: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

Write the correct one before substituting.

4. Sign Error When Solving for $v$

The Mistake:

After substituting $u = -27$ and $f = -18$ , students get the algebra wrong and end up with $v$ positive when it should be negative (or vice versa).

How to avoid:

Do the algebra step-by-step:

\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-18} - \frac{1}{-27} = -\frac{1}{18} + \frac{1}{27}

Find LCM = 54:

\frac{1}{v} = \frac{-3 + 2}{54} = -\frac{1}{54}

Thus:

v = -54\ \text{cm}

Check: Negative $v$ means the image is in front of the mirror (real image), which matches the problem’s requirement for a screen.

5. Misinterpreting Magnification Sign

The Mistake:

Using $m = \frac{v}{u}$ without signs, or forgetting that $m$ negative means inverted image.

How to avoid:

Use the signed formula:

m = -\frac{v}{u} = -\frac{(-54)}{(-27)} = -\frac{54}{27} = -2

Negative $m$ → image is inverted.
$|m| = 2$ → image is magnified (twice the object size).

6. Calculating Image Size Incorrectly

The Mistake:

Using $h_i = m \times h_o$ but forgetting the sign of $m$ or $h_o$ .

How to avoid:

Object size $h_o = +2.5\ \text{cm}$ (always positive). Then:

h_i = m \times h_o = (-2) \times (2.5) = -5.0\ \text{cm}

$|h_i| = 5.0\ \text{cm}$ → size.
Negative sign → image is inverted relative to object.

7. Describing the Image Incorrectly

The Mistake:

Saying “real, inverted, diminished” when it’s actually magnified.

How to avoid:

Check $|m|$ :

$|m| > 1$ → magnified
$|m| < 1$ → diminished
$|m| = 1$ → same size

Here $|m| = 2$ , so the image is real, inverted, and magnified.

8. Answering the “Move the Screen” Part Wrong

The Mistake:

Saying “move the screen away” when the candle is moved closer, without checking the new $u$ .

How to avoid:

If the candle is moved closer (say $u$ becomes less negative, e.g., $-20\ \text{cm}$ ), recalculate $v$ :

\frac{1}{v} = \frac{1}{-18} - \frac{1}{-20} = -\frac{1}{18} + \frac{1}{20} = \frac{-10 + 9}{180} = -\frac{1}{180}

So $v = -180\ \text{cm}$ — farther from the mirror.

Key rule: For a concave mirror, as the object moves from infinity toward the focus, the image moves away from the mirror (from focus to infinity). So the screen must be moved farther from the mirror.

Quick Summary Checklist

Step	Common Mistake	Correct Approach
$f$	Use $R$ directly	$f = R/2 = -18\ \text{cm}$
Signs	$u$ positive	$u = -27\ \text{cm}$
Formula	Lens formula	$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$v$	Algebra slip	Solve stepwise → $v = -54\ \text{cm}$
$m$	Forget sign	$m = -v/u = -2$
$h_i$	Wrong sign	$h_i = m \times h_o = -5.0\ \text{cm}$
Nature	Wrong description	Real, inverted, magnified
Screen move	Wrong direction	Move away from mirror

Final Answer for the problem:

Screen distance: $54\ \text{cm}$ in front of the mirror
Image: real, inverted, $5.0\ \text{cm}$ tall
If candle is moved closer: move screen farther from the mirror

←Previous section · pointsPoints to Ponder →Next question9.2 — A 4.5cm needle is placed 12cm away from a convex mirror of focal lengt…

1. Understanding the physics: Why the mirror formula works

2. Step-by-step solution

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