Reflection of Light by Spherical Mirrors

What this subtopic is examined on

• Definitions: focal length, principal focus, focal plane, paraxial rays, real and virtual image.
• The derivation $f = R/2$ with a ray diagram (a full 5-mark question in 2020).
• The derivation of the mirror equation from similar triangles, for a real image and for a virtual image, plus the expression for linear magnification.
• Numericals on the mirror formula and magnification, including "how far must the object be moved".
• Reasoning questions: why a convex mirror always gives a virtual image; why the magnification of an object lying along the axis is not uniform; what happens if half the mirror is covered.

Laws of reflection and the basic terms

• The angle of reflection equals the angle of incidence, and the incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane. These hold at every point of any surface, plane or curved.
• For a spherical mirror the normal at a point is the normal to the tangent plane there, which is along the radius — the line joining the centre of curvature to the point of incidence.
• Pole $P$: the geometric centre of the mirror. Centre of curvature $C$ and radius of curvature $R$: the centre and radius of the sphere of which the mirror is a part. Principal axis: the line $PC$.
• Principal focus $F$: the point on the principal axis at which a parallel paraxial beam converges after reflection (concave), or from which it appears to diverge (convex).
• Focal length: the distance between the pole and the principal focus of the mirror. That one sentence is the definition CBSE wants.
• Focal plane: the plane through $F$ perpendicular to the principal axis. A parallel beam incident at a small angle to the axis converges to a point in this plane.
• Paraxial rays: rays incident close to the pole and making small angles with the axis. Every formula below assumes them, i.e. a mirror of small aperture.
• Real image: the rays actually meet there. Virtual image: the rays only appear to diverge from there when produced backwards.

Sign convention

• Distances are measured from the pole; along the incident light positive, against it negative; heights above the axis positive, below negative.
• Hence for a concave mirror $f$ and $R$ are negative, for a convex mirror they are positive, and a real object always has $u$ negative.

Derivation: $f = R/2$

Diagram in words

• Draw a concave mirror of small aperture with pole $P$, centre of curvature $C$ and focus $F$ on the principal axis. Draw a ray parallel to the principal axis striking the mirror at a point $M$ close to $P$. Join $CM$: this is the normal at $M$. The reflected ray from $M$ crosses the axis at $F$. Drop the perpendicular $MD$ from $M$ onto the principal axis.

Steps

• Let $\theta$ be the angle of incidence at $M$. Since $CM$ is the normal, the angle of reflection is also $\theta$.
• Because $MC$ is a transversal of the parallel incident ray and the axis, $\angle MCP = \theta$; and from the exterior-angle result at $M$, $\angle MFP = 2\theta$.
• In the right triangles $MCD$ and $MFD$:

$$\tan\theta = \dfrac{MD}{CD}, \qquad \tan 2\theta = \dfrac{MD}{FD}$$

• For paraxial rays $\theta$ is small, so $\tan\theta \approx \theta$ and $\tan 2\theta \approx 2\theta$. Dividing,

$$\dfrac{MD}{FD} = 2\,\dfrac{MD}{CD} \qquad\Rightarrow\qquad FD = \dfrac{CD}{2}$$

• Since $M$ is close to $P$, the foot $D$ is close to $P$, so $FD \approx FP = f$ and $CD \approx CP = R$. Hence

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

• The same construction with a convex mirror gives the same result, with $f$ and $R$ both positive.

Derivation: the mirror equation (real image)

Diagram in words

• An object $AB$ stands on the principal axis beyond the centre of curvature $C$ of a concave mirror, with $B$ on the axis and $A$ at the top.
• Ray 1: from $A$ parallel to the principal axis, striking the mirror at $M$, reflected through the focus $F$.
• Ray 2: from $A$ through the pole $P$, reflected making an equal angle with the axis on the other side.
• The two reflected rays meet at $A'$ below the axis; the perpendicular $A'B'$ dropped onto the axis is the real, inverted image of $AB$.

Steps

• For paraxial rays, $MP$ may be taken as a straight line perpendicular to the axis with $MP = AB$.
• Triangles $A'B'F$ and $MPF$ are similar, so

$$\dfrac{A'B'}{MP} = \dfrac{B'F}{FP} \qquad\Rightarrow\qquad \dfrac{A'B'}{AB} = \dfrac{B'F}{FP}$$

• Since $\angle APB = \angle A'PB'$ (law of reflection at $P$), triangles $A'B'P$ and $ABP$ are similar, so

$$\dfrac{A'B'}{AB} = \dfrac{B'P}{BP}$$

• Comparing the two,

$$\dfrac{B'P - FP}{FP} = \dfrac{B'F}{FP} = \dfrac{B'P}{BP}$$

• Now apply the sign convention. Light travels from the object to the mirror, so that direction is positive; the object, the image and the focus all lie on the other side of $P$, so $B'P = -v$, $FP = -f$, $BP = -u$. Substituting,

$$\dfrac{-v+f}{-f} = \dfrac{-v}{-u} \qquad\Rightarrow\qquad \dfrac{v-f}{f} = \dfrac{v}{u} \qquad\Rightarrow\qquad uv = fv + fu$$

• Dividing throughout by $uvf$:

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

• An equivalent route, which CBSE also accepts, uses the pair of similar triangles $A'B'C \sim ABC$ and $A'B'P \sim ABP$ with $CP = -R = -2f$:

$$\dfrac{A'B'}{AB} = \dfrac{-2f+v}{-u+2f} = \dfrac{v}{u}$$

• Cross-multiplying, $-2fu + uv = -uv + 2fv$, so $uv = fv + fu$ and again $\dfrac1f = \dfrac1u + \dfrac1v$.

Derivation: the mirror equation for a virtual image

• CBSE has set this as a separate 5-mark question. Take the object $AB$ between the pole $P$ and the focus $F$ of a concave mirror.
• Ray 1 from $A$ parallel to the axis reflects so that the reflected ray, produced backwards, passes through $F$. Ray 2 from $A$ strikes the pole and reflects at an equal angle. The two reflected rays diverge; produced backwards they meet at $A'$ behind the mirror, giving the virtual, erect, magnified image $A'B'$.
• The same two pairs of similar triangles ($A'B'C \sim ABC$ and $MPF \sim A'B'F$) give, after substituting $BP = -u$, $PB' = v$, $FP = -f$, $CP = -2f$:

$$\dfrac{-2f+u}{-2f+v} = \dfrac{-f}{-f+v}$$

• Cross-multiplying and cancelling the $2f^2$ terms leaves $uv = fv + fu$, hence again

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

• The point to make in the answer: the same formula, with the same sign convention, covers concave and convex mirrors and real and virtual images alike.

Magnification

• Linear (transverse) magnification is the ratio of the height of the image to the height of the object:

$$m = \dfrac{h'}{h}$$

• From the similar triangles $A'B'P$ and $ABP$, $\dfrac{A'B'}{AB} = \dfrac{B'P}{BP}$, and with the sign convention $\dfrac{-h'}{h} = \dfrac{-v}{-u}$, so

$$m = \dfrac{h'}{h} = -\dfrac{v}{u}$$

• $m$ negative means a real, inverted image; $m$ positive means a virtual, erect image. $\lvert m\rvert>1$ magnified, $\lvert m\rvert<1$ diminished.
• Two forms worth remembering for numericals: $m = \dfrac{f}{f-u}$ and $m = \dfrac{f-v}{f}$.

Ray-diagram rules

• A ray parallel to the principal axis is reflected through $F$ (concave) or appears to come from $F$ (convex).
• A ray through $C$, or directed towards $C$, retraces its path.
• A ray through $F$, or directed towards $F$, is reflected parallel to the principal axis.
• A ray striking the pole is reflected making an equal angle with the principal axis on the other side.
• Any two of these suffice; drawing three is safest in an exam.

Image formation table (concave mirror)

• A convex mirror always forms a virtual, erect, diminished image between $P$ and $F$, whatever the object distance.

Reasoning answers CBSE has asked for

• Why does a convex mirror always give a virtual image? For a convex mirror $f>0$, and a real object gives $u<0$. From the mirror formula, $\dfrac1v = \dfrac1f - \dfrac1u = \dfrac1f + \dfrac{1}{-u}$. Both terms are positive, so $\dfrac1v>0$, i.e. $v>0$ always — the image is always behind the mirror, hence always virtual.
• Object between $f$ and $2f$ of a concave mirror gives a real image beyond $2f$. With $f<0$ and $2f<u<f$, the mirror formula $\dfrac1v = \dfrac1f - \dfrac1u$ gives $\dfrac{1}{2f} < \dfrac1v < 0$, so $v$ is negative (real image, in front of the mirror) and $\lvert v\rvert > \lvert 2f\rvert$ — the image lies beyond $2f$, and is magnified and inverted.
• A mobile phone lying along the principal axis. Different parts of the phone are at different object distances, and $m = -v/u$ depends on $u$, so each part is magnified differently. The image is therefore distorted along its length, even though each individual point is imaged correctly. The part of the phone perpendicular to the axis is imaged in the same plane and at the same size.
• Lower half of a concave mirror covered with opaque material. The laws of reflection still hold at every point of the uncovered half, so the image of the whole object is still formed, of the same size and at the same place. Only the reflecting area has halved, so the intensity (brightness) of the image is reduced, roughly to half.
• Rear-view mirror numericals. Use $v = \dfrac{fu}{u-f}$ and find $v$ at two object positions one second apart; the difference gives the average speed of the image. The image speed grows sharply as the object approaches the mirror, though the object moves at constant speed.

Standard results and worked patterns

• $R = 60$ cm concave mirror, object at 20 cm: $f = -30$ cm, $u = -20$ cm, $\dfrac1v = \dfrac{1}{-30} + \dfrac{1}{20} = \dfrac{1}{60}$, so $v = +60$ cm — a virtual, erect, magnified image behind the mirror.
• Concave mirror, $f = -15$ cm, real image three times the object: $m = -3 = -v/u$ so $v = 3u$; the mirror formula gives $u = -10$ cm.
• Concave mirror giving a two-times enlarged virtual image at $u = -10$ cm: $m = +2$ gives $v = +20$ cm and $f = -20$ cm. For a two-times enlarged real image, $m = -2$ gives $v = 2u$, and the mirror formula gives $u = -30$ cm — the object must be moved 20 cm further away from the mirror.
• Concave mirror of $R = 20$ cm ($f = -10$ cm) with a real image of magnification 2: $v = 2u$ leads to $u = -15$ cm and $v = -30$ cm.
• Object at (i) 10 cm and (ii) 5 cm from a concave mirror of $R = 15$ cm ($f = -7.5$ cm): (i) $v = -30$ cm, $m = -3$, real, inverted, magnified; (ii) $v = +15$ cm, $m = +3$, virtual, erect, magnified.