Derive the mirror equation and the equation for lateral magnification.

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Q. Derive the mirror equation and the equation for lateral magnification.

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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.

Step 1. Consider an object $AB$ beyond the centre of curvature of a concave mirror. Trace three paraxial rays from $B$ : one parallel to the axis, striking the mirror near the pole at $D$ and reflecting through the focus $F$ ; one through the pole $P$ , reflecting symmetrically; and one through the centre of curvature $C$ , striking at $E$ and retracing its own path. These three reflected rays meet at $B'$ , locating the real, inverted image $A'B'$ .

Step 2. By the law of reflection, $\angle BPA=\angle B'PA'$ , so $\triangle BPA\sim\triangle B'PA'$ , giving $\dfrac{A'B'}{AB}=\dfrac{PA'}{PA}$ .

Step 3. Also $\triangle DPF\sim\triangle B'A'F$ (since $PD\approx AB$ for a paraxial ray near the pole), giving $\dfrac{A'B'}{AB}=\dfrac{A'F}{PF}$ . Combining with Step 2, $\dfrac{PA'}{PA}=\dfrac{A'F}{PF}$ , and writing $A'F=PA'-PF$ gives $\dfrac{PA'}{PA}=\dfrac{PA'-PF}{PF}$ .

Step 4. Applying the sign convention $PA=-u$ , $PA'=-v$ , $PF=-f$ and simplifying the algebra (cross-multiplying and dividing through by $uvf$ ) leads step by step to $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ -- the mirror equation.

Step 5. For magnification: from $\triangle BPA\sim\triangle B'PA'$ , $m=\dfrac{A'B'}{AB}=\dfrac{PA'}{PA}$ ; applying the sign convention ( $A'B'=-h'$ , $AB=h$ , $PA'=-v$ , $PA=-u$ ) gives $m=\dfrac{h'}{h}=-\dfrac{v}{u}$ . Using the mirror equation to eliminate $v$ or $u$ gives the equivalent forms $m=\dfrac{f}{f-u}=\dfrac{f-v}{f}$ .

✓Final answer

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

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

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

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

Since $v$ is positive, the image is formed BEHIND the mirror (virtual), as expected for a convex mirror with a real object — convex mirrors always give a virtual, erect, diminished image for real objects.

✓Final answer
$(a) The image is formed at 6.67 cm behind the mirror.$
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 $C$ .

Confirm the mirror type

Only a concave mirror can produce a real, inverted, magnified image. Therefore the mirror is concave and the image is real.

Watch out

A common mistake is to think that a negative magnification always means a concave mirror — but a convex mirror can never give a negative $m$ at all. Also, don't confuse "magnified" with "virtual": a concave mirror gives a magnified virtual image only when the object is very close (between pole and focus), and that image is erect (positive $m$ ). So $m = -2.0$ cannot be virtual.

Tip

You can remember the four cases for concave mirrors with a simple table:

| Object position | Image type | $m$ sign | $|m|$ |

|----------------|------------|----------|-------|

| Beyond $C$ | Real, inverted, diminished | negative | $< 1$ |

| At $C$ | Real, inverted, same size | negative | $= 1$ |

| Between $C$ and $F$ | Real, inverted, magnified | negative | $> 1$ |

| Between $F$ and $P$ | Virtual, erect, magnified | positive | $> 1$ |

Our $m = -2.0$ fits the third row perfectly.

✓Final answer
$The mirror is concave and the image is real, so the correct option is (B) .$
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 focus of a spherical mirror emerge parallel after reflection and meet only at infinity. So R is a true, general statement, and it directly and correctly explains why A is true.

✓Final answer
$(a) Both A and R are correct and R is the correct explanation of A.$
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 of a mirror does not change when it is placed in a different medium such as water. So the focal length remains 25 cm.

✓Final answer
$(d) 25 cm$
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:

$f = \frac{R}{2} \quad \Rightarrow \quad R = 2f$

This is because the focus F lies exactly midway between the pole (P) of the mirror and the centre of curvature (C).

✓Final answer
$(a) R = 2f.$
CBSE 2024Set ANNUAL1 mark
$Q. The radius of curvature of a concave mirror is 24 cm. The value of its focal length will be __________ cm.$
›Reveal solutionSolution

For a spherical mirror, f = R/2, a direct geometric consequence of paraxial ray reflection.

For any spherical mirror (concave or convex), the focal length is related to the radius of curvature by:

$f = \dfrac{R}{2}$

Given $R = 24\ \text{cm}$ :

$f = \dfrac{24}{2} = 12\ \text{cm}$

✓Final answer
$12 cm.$
CBSE 2023Set 55/1/11 markMCQ
$f$
(a) zero
(b) $f$
(c) $2f$
(d) $4f$
›Reveal solutionSolution
For a concave mirror, the object and its real image can be brought arbitrarily close together, but the minimum possible distance between them is zero — achieved when the object is at the centre of curvature and the image coincides with it.

The question asks for the minimum distance between an object and its real image formed by a concave mirror. This is a classic problem that tests your understanding of the mirror formula and the concept of real images.

The core idea

A real image is formed when rays actually converge after reflection. For a concave mirror, a real image is formed only when the object is placed beyond the focus (i.e., $u > f$ ). The image distance $v$ is then positive (real) and given by the mirror formula:
$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
The distance between the object and its image is $|u - v|$ . We want to find the smallest possible value of this distance for real images.

Step-by-step reasoning

Set up the mirror formula For a concave mirror, $f$ is positive. Let $u$ be the object distance (positive, measured from the mirror). For a real image, $v$ is also positive. The mirror formula gives:

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{u - f}{uf}$
So:
$v = \frac{uf}{u - f}$

Write the distance between object and image Let $D = |u - v|$ . Since both $u$ and $v$ are positive and measured from the mirror, the object and image lie on the same side of the mirror. The distance between them is:

$D = |u - v| = \left| u - \frac{uf}{u - f} \right|$

Simplify the expression Factor $u$ :

$D = u \left| 1 - \frac{f}{u - f} \right| = u \left| \frac{u - f - f}{u - f} \right| = u \left| \frac{u - 2f}{u - f} \right|$
Since for real images $u > f$ , the denominator $u - f > 0$ . The sign of $u - 2f$ depends on $u$ . So:
$D = u \cdot \frac{|u - 2f|}{u - f}$

Analyse the behaviour

If $u > 2f$ , then $u - 2f > 0$ , so $D = u \cdot \frac{u - 2f}{u - f}$ .

If $f < u < 2f$ , then $u - 2f < 0$ , so $D = u \cdot \frac{2f - u}{u - f}$ .

In both cases, $D$ is positive. The question is: can $D$ become zero?

When does $D = 0$ ?

$D = 0$ when $|u - 2f| = 0$ , i.e., when $u = 2f$ .

At $u = 2f$ , the mirror formula gives:

$v = \frac{2f \cdot f}{2f - f} = \frac{2f^2}{f} = 2f$
So the image is also at $2f$ — the object and image coincide at the centre of curvature. The distance between them is $0$ .

Is this a real image? Yes. At $u = 2f$ , the image is real, inverted, and of the same size as the object. It is formed at the same point as the object. So the distance is indeed zero.

Watch out

A common mistake is to think the minimum distance is $4f$ (which is the minimum distance between an object and its real image when the image is not at the same location — that's a different problem). Here, the object and image can actually coincide, so the distance can be zero.

Tip

The condition $u = 2f$ gives $v = 2f$ , so $D = 0$ . This is the only case where the object and its real image are at the same point. For any other $u$ , $D > 0$ .

✓Final answer
$The minimum distance is zero, so the correct option is (a) .$
CBSE 2023Set F1 markMCQ
$Q. A spherical mirror is immersed in water. Its focal length will (A) decrease (B) increase (C) remain same (D) none of these$
›Reveal solutionSolution

A mirror's focal length depends only on its radius, so immersing it in water does not change $f$ .

Unlike a lens (whose focal length depends on the refractive index of the medium through the lens-maker's formula), a spherical mirror forms images purely by reflection. Its focal length is

$f = \frac{R}{2},$

where $R$ is the radius of curvature. This is a geometric property of the mirror and is independent of the medium in which the mirror is placed. Hence in water the focal length remains the same.

✓Final answer
$(C) remain same.$
CBSE 2023Set ANNUAL1 mark
$Q. The radius of curvature of a concave mirror is 28 cm, its focal length will be?$
›Reveal solutionSolution

For a spherical mirror, the focal length is half the radius of curvature: f = R/2.

For a concave mirror, the focus lies midway between the pole and the centre of curvature, so $f = \dfrac{R}{2}$ .

Given $R = 28\ \text{cm}$ : $f = \dfrac{28}{2} = 14\ \text{cm}$ .

✓Final answer
$f = 14 cm.$
CBSE 2023Set TERM21 markMCQ
$Q. Focal length of Plane mirror is :$
(a) Infinite
(b) Zero
(c) 10 cm
(d) 20 cm
›Reveal solutionSolution

A plane mirror can be thought of as a spherical mirror whose radius of curvature is infinite, so its focal length is also infinite.

For a spherical mirror, $f = R/2$ , where $R$ is the radius of curvature. A plane mirror is the limiting case of a spherical mirror with $R \to \infty$ (an infinitely large sphere looks flat locally). Hence its focal length is also infinite — parallel rays incident on a plane mirror remain parallel after reflection and never converge to a real focal point (nor a virtual one at a finite distance).

✓Final answer
$(a) Infinite.$
CBSE 2019Set ANNUAL1 mark
$Q. A convex mirror is placed inside water. Will its focal length change? (Write 'Yes' or 'No')$
›Reveal solutionSolution

A mirror's focal length depends only on its radius of curvature (f = R/2), not on the medium, so it does not change in water.

Reflection at a mirror obeys the law of reflection regardless of the medium in which it is placed. The focal length of a spherical mirror is fixed purely by its geometry:

f = R/2,

where R is the radius of curvature. There is no refractive index in this relation. This is unlike a lens, whose focal length depends on the refractive index of the surrounding medium. Hence a convex mirror's focal length is unchanged when placed in water.

✓Final answer
$No.$

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Q.Derive the mirror equation and the equation for lateral magnification.

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