Q.Draw a ray diagram showing the image formation when a concave mirror produces a real, inverted and magnified image of an object and hence obtain the mirror formula.

Concept: Mirror formula derivation using ray geometry for a concave mirror when the object is placed between $F$ and $C$ (real, inverted, magnified image).

Reasoning:

1. Draw a concave mirror with principal axis, focus $F$ (at distance $f$), and centre of curvature $C$ (at distance $R = 2f$). Place object $AB$ between $F$ and $C$. A ray from $B$ parallel to the axis reflects through $F$; another ray from $B$ through $C$ reflects back along itself. These rays meet in front of the mirror, forming a real, inverted, magnified image $A'B'$.

2. From the ray diagram, two similar triangles give the relation. $\triangle ABC \sim \triangle A'B'C$ (both right-angled, share angle at $C$), so: …

The Concept: Why Geometry Gives the Formula

A concave mirror's behaviour is governed by the law of reflection, but the relationship between object distance $u$, image distance $v$, and focal length $f$ comes from pure geometry. When we draw two characteristic rays from the object, their intersection after reflection locates the image. The triangles formed by these rays and the principal axis are similar, and that similarity yields the mirror formula.

For a real, inverted, and magnified image, the object must be placed between $C$ and $F$. The image then forms beyond $C$, is real (can be projected on a screen), inverted, and larger than the object. This is the classic case used in shaving mirrors or for projection.

Step-by-Step Derivation

1. Draw the ray diagram

Place a concave mirror with its pole $P$, principal axis, focus $F$, and centre of curvature $C$. Let $PF = f$ and $PC = 2f$.

Position the object $AB$ (an upright arrow) between $F$ and $C$, perpendicular to the principal axis. The tip $A$ is on the axis; the tip $B$ is above it.

Now draw two rays from point $B$ (the top of the object):

• Ray 1: A ray parallel to the principal axis. After reflection, it passes through $F$.
• Ray 2: A ray passing through $C$. Since it strikes the mirror normally (along the radius), it reflects back along the same path through $C$.

These two reflected rays intersect at $B'$ beyond $C$, forming the real, inverted image $A'B'$. The image is larger than the object.

2. Label distances on the diagram

Mark the following on the axis:

• $P$: pole of the mirror
• $F$: focus, at distance $f$ from $P$
• $C$: centre of curvature, at distance $2f$ from $P$
• $A$: foot of the object, at distance $u$ from $P$ (so $PA = u$)
• $A'$: foot of the image, at distance $v$ from $P$ (so $PA' = v$)

Here $u$, $v$ and $f$ are treated simply as the lengths $PA$, $PA'$ and $PF$ measured along the axis — this keeps the triangle geometry below clean. A real object and the real image it forms both actually lie in front of the mirror, so in the New Cartesian sign convention their signed values are both negative ($u \to -u$, $v \to -v$), and the focal length of a concave mirror is negative too ($f \to -f$). This does not change the derivation below: negating $u$, $v$ and $f$ together leaves the relation $\frac1u+\frac1v=\frac1f$ exactly as it is (both sides simply pick up an overall minus sign, which cancels) — so the identical equation is obtained whether you work with these magnitudes first, or with their signed Cartesian values from the start.

3. Identify similar triangles

Look at two pairs of right-angled triangles in the diagram:

Pair 1: $\triangle ABP$ and $\triangle A'B'P$

• $\angle APB = \angle A'PB'$ (vertically opposite angles)
• $\angle BAP = \angle B'A'P = 90^\circ$
• Therefore $\triangle ABP \sim \triangle A'B'P$ (AA similarity)

From this similarity:

Pair 2: $\triangle B' A' F$ and $\triangle P F D$ (where $D$ is the point where the parallel ray hits the mirror)

• The ray parallel to the axis meets the mirror at $D$, then passes through $F$ to $B'$.
• $\angle B'A'F = \angle PFD = 90^\circ$
• $\angle B'FA' = \angle PFD$ (common angle)
• So $\triangle B'A'F \sim \triangle PFD$

From this similarity:

But $PF = f$, and $PD = AB$ (since the incident ray from $B$ is parallel to the axis, the segment $PD$ on the mirror equals the object height $AB$). Also $A'F = PA' - PF = v - f$.

Therefore:

4. Combine the two similarity relations

From Pair 1: $\frac{AB}{A'B'} = \frac{u}{v}$, so $\frac{A'B'}{AB} = \frac{v}{u}$.

From Pair 2: $\frac{A'B'}{AB} = \frac{v - f}{f}$.

Equating the two expressions for $\frac{A'B'}{AB}$:

5. Rearrange to get the mirror formula

Cross-multiply:

Bring terms involving $uv$ to one side:

Factor $f$:

$f(u + v) = uv$ …