Reflection at Plane and Spherical Surfaces

PhysicsRay Optics And Optical InstrumentsFor JEE aspirants

Reflection at plane and spherical surfaces follows two simple laws: the incident ray, the normal and the reflected ray lie in one plane, and the angle of incidence equals the angle of reflection. A plane mirror forms a virtual, erect, same-size image as far behind the mirror as the object is in front; a spherical mirror obeys $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ with $f = \frac{R}{2}$ . This page covers reflection at plane and spherical surfaces for JEE Main, JEE Advanced and NEET: rotation of mirrors, multiple images, sign convention and image formation.

On this page1Laws of reflection2Plane mirror3Two mirrors4Spherical mirrors5Image cases6Combinations7Solved examples

Key Formulas - Quick Reference

★ Must learnLaws of reflection: $∠ i = ∠ r$ ; incident ray, normal and reflected ray are coplanar
★ Must learnDeviation by one plane mirror: $δ = 18 0^{\circ} - 2 i$
Two mirrors inclined at $θ$ : net deviation $= 36 0^{\circ} - 2 θ$ (independent of $i$ )
★ Must learnMirror rotated by $θ$ (incident ray fixed): reflected ray turns by $2 θ$
Plane mirror, velocity normal to mirror: $v_{I} = 2 v_{M} - v_{O}$ ; components along the mirror are equal
★ Must learnNumber of images, $n = \frac{36 0 ^{\circ}}{θ}$ : $n$ even gives $n - 1$ ; $n$ odd gives $n - 1$ (object on bisector) or $n$ (off bisector)
Minimum mirror height to see full image $= \frac{h}{2}$ (any distance)
Vector law: $\overset{r}{^} = \overset{e}{^} - 2 (\overset{e}{^} \cdot \overset{n}{^}) \overset{n}{^}$
★ Must learnSpherical mirror: $f = \frac{R}{2}$ and $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ (New Cartesian signs)
Concave: $f < 0$ , converging. Convex: $f > 0$ , diverging

1. Light and Ray Optics: The Basics

Light is an electromagnetic wave: oscillating electric and magnetic fields travelling at $c \approx 3 \times 1 0^{8} m s^{- 1}$ in vacuum. It carries energy and momentum, and the wave relation $c = f λ$ holds. Some experiments (photoelectric effect) need a particle picture and others (interference) need a wave picture, so optics is split into ray (geometrical) optics and wave optics.

Ray optics is valid when the obstacles and openings are much larger than the wavelength of light. Light is then treated as travelling in straight lines (rectilinear propagation), and images are found with geometry alone.

Region of the spectrum	Approximate wavelength
Radio waves	$> 0.1 m$
Microwaves	$0.1 m$ to $1 mm$
Infrared	$1 mm$ to $700 nm$
Visible light	$700 nm$ (red) to $400 nm$ (violet)
Ultraviolet	$400 nm$ to $1 nm$
X-rays	$1 nm$ to $1 0^{- 3} nm$
Gamma rays	$< 1 0^{- 3} nm$

1.1 Ray and beam

A ray is an imaginary line along the direction in which light energy travels. A beam is a bundle of rays:

Parallel beam: from a very distant source such as the Sun, a searchlight or a headlight.
Divergent beam: rays spreading out from a point source.
Convergent beam: rays meeting at a point, for example a parallel beam after a convex lens.

2. Reflection and Its Laws

When light strikes a surface separating two media, part (or all) of it is sent back into the first medium. This is reflection. Surfaces made to reflect well are mirrors: plane or curved (spherical).

Angle of incidence $i$ : angle between the incident ray and the normal at the point of incidence.
Angle of reflection $r$ : angle between the reflected ray and the normal.
Glancing angle $g$ : angle between the incident ray and the mirror surface, $g = 9 0^{\circ} - i$ .
Deviation $δ$ : angle between the original direction of the incident ray and the reflected ray.

Laws of reflection (valid for plane and curved surfaces):
1. The incident ray, the reflected ray and the normal at the point of incidence lie in the same plane.
2. The angle of incidence equals the angle of reflection:

∠ i = ∠ r

Figure 1: Laws of reflection. The deviation of the ray is

δ = 18 0^{\circ} - 2 i

; normal incidence (

i = 0

) gives

δ = 18 0^{\circ}

, so the ray retraces its path.

Normal incidence: if $i = 0$ then $r = 0$ and the ray retraces its path. For a spherical mirror this happens for any ray through the centre of curvature $C$ , because the radius is the normal.
What does not change: frequency, wavelength and speed stay the same after reflection. Intensity (and hence amplitude, since $I \propto A^{2}$ ) usually decreases.
Diffuse reflection: on a rough surface each tiny patch obeys the laws, but the normals point in random directions, so a parallel beam scatters in all directions. This is why we see non-shiny objects from every angle.

Key idea

Reflection keeps

i = r

with the incident ray, normal and reflected ray in one plane; every mirror result here follows from that.

3. Plane Mirror

3.1 Image of a point and of an extended object

Drop a perpendicular from the object to the mirror (extended if needed). The image lies on this line, as far behind the mirror as the object is in front. Similar triangles formed by any two reflected rays prove this.

Plane mirror image: virtual, erect, same size, same distance behind the mirror, and laterally inverted (left and right swapped). Only the coordinate normal to the mirror changes sign:

x_{I} = - x_{O}

y_{I} = y_{O}

z_{I} = z_{O}

(mirror in the

y

z

plane).

For an extended object, image its end points and join them:

Size of image = size of object, always.
Object parallel to the mirror: image erect.
Object perpendicular to the mirror: image points the opposite way along the normal (the end nearer the mirror stays nearer).
Horizontal object in front of a mirror inclined at $4 5^{\circ}$ to the horizontal: image is vertical, because horizontal rays reflect vertically. This is the principle of the periscope.

Exam Trick

Mirror clock = 11:60 minus actual time. For a clock with no numbers, the time shown by the image is $11 : 60$ minus the real time (for times from 1:00 to 11:59). Example: $11 : 60 - 8 : 12 = 3 : 48$ . If the result exceeds 12, subtract 12. Or simply trace the dial on paper and look from the back.

3.2 Field of view

The field of view is the region in which reflected rays exist. An observer inside it sees the image; outside it the image still exists but no reflected ray reaches the eye. It is bounded by the rays through the two edges of the mirror, which look as if they come from the image.

Figure 2: Field of view. Only rays reflected between the mirror's edges reach the observer; on a line

2 d

from the mirror the image is seen over a length

3 ℓ

(similar triangles).

For an extended object, the field of view is the common region of the fields of view of its end points.

3.3 Velocity of the image in a plane mirror

Differentiating $x_{I} = - x_{O}$ (both measured from the mirror) gives the rule for the component perpendicular to the mirror, measured from the ground:

v_{I} - v_{M} = - (v_{O} - v_{M}) \Rightarrow v_{I} = 2 v_{M} - v_{O}

Components parallel to the mirror are simply equal: $v_{I, ∥} = v_{O, ∥}$ (the mirror's motion along its own plane changes nothing).

Exam Trick

Mirror speed $v$ gives image speed $2 v$ . For a fixed object and a mirror moving along its normal with speed $v$ , $v_{I} = 2 v_{M} - 0 = 2 v$ relative to the ground (and $v$ relative to the mirror). Moving the mirror along its own plane does nothing to the image.

3.4 Deviation by one and two plane mirrors

One reflection deviates a ray by $δ = 18 0^{\circ} - 2 i$ (Figure 1). For two mirrors inclined at $θ$ , with the ray reflecting once from each:

Deviations at the two mirrors: $δ_{1} = 18 0^{\circ} - 2 i_{1}$ and $δ_{2} = 18 0^{\circ} - 2 i_{2}$ , both in the same sense.
Angle sum in the triangle formed by the two mirrors and the ray between them: $θ + (9 0^{\circ} - i_{1}) + (9 0^{\circ} - i_{2}) = 18 0^{\circ}$ , so $i_{1} + i_{2} = θ$ .
Net deviation: $δ = 36 0^{\circ} - 2 (i_{1} + i_{2}) = 36 0^{\circ} - 2 θ$ , the same as $2 θ$ in the opposite sense.

Net deviation after two reflections

= 36 0^{\circ} - 2 θ

, independent of the angle of incidence. For

θ = 9 0^{\circ}

the ray is sent straight back (

δ = 18 0^{\circ}

): the corner (retro) reflector used on bicycles and road signs.

3.5 Real and virtual objects and images

Term	Meaning
Real object	Incident rays actually diverge from the point (on the incident side)
Virtual object	Incident rays converge towards a point behind the mirror; they are intercepted before meeting
Real image	Reflected rays actually meet; can be caught on a screen
Virtual image	Reflected rays only appear to diverge from the point (backward extensions meet)

A plane mirror turns a real object into a virtual image, and a virtual object into a real image.

3.6 Rotation of a plane mirror

Keep the incident ray fixed and rotate the mirror by $θ$ about an axis lying in the mirror and perpendicular to the plane of incidence. The normal also turns by $θ$ , so the angle of incidence changes by $θ$ and the reflected ray turns by $2 θ$ in the same sense.

Figure 3: Rotate the mirror by

θ

and the normal turns by

θ

, so the reflected ray turns by

2 θ

in the same sense (incident ray fixed).

Proof from the figure: the angle of incidence falls from $i$ to $i - θ$ (measured from $N_{2}$ ), so the reflected ray sits at $i - θ$ beyond $N_{2}$ . Measured from $N_{1}$ that is $θ - (i - θ) = 2 θ - i$ , compared with $- i$ before: a turn of $2 θ$ . Consequently, a mirror spinning with angular speed $ω$ makes the reflected ray spin at $2 ω$ . A light spot on a circular screen of radius $R$ centred on the mirror moves at $2 ω R$ ; on a flat wall it moves faster away from the foot of the perpendicular (Solved Example 5).

3.7 Images formed by two plane mirrors

Rays reflected by one mirror may strike the other. The image formed by the first mirror then acts as the object for the second, and so on. Name images by the order of reflection: $I_{12}$ means reflection at $M_{1}$ first, then $M_{2}$ .

Parallel mirrors: images form repeatedly on both sides; in principle infinitely many, each fainter than the last (see Solved Example 6).
Perpendicular mirrors: object at $(x, y)$ gives images at $(x, - y)$ , $(- x, y)$ and $(- x, - y)$ ; the last one is formed by two reflections in either order, so the two coincide. Total 3.
Circle concept: all images of two inclined mirrors lie on a circle centred at the line of intersection, with radius equal to the object's distance from it.

Figure 4: Perpendicular mirrors give

\frac{36 0 ^{\circ}}{9 0 ^{\circ}} - 1 = 3

images. The third image

I_{3}

comes from two reflections, and all images lie on a circle centred at the junction.

Quick Recall: tap to check

A plane mirror turns through

1 0^{\circ}

with the incident ray fixed. Through what angle does the reflected ray turn?

2 0^{\circ}

How many images do two plane mirrors at

6 0^{\circ}

form of an object between them?

\frac{36 0 ^{\circ}}{6 0 ^{\circ}} = 6

is even, so

6 - 1 = 5

An object approaches a fixed plane mirror at

2 m s^{- 1}

. How fast does it approach its image?

4 m s^{- 1}

Value of $n = \frac{36 0 ^{\circ}}{θ}$	Object position	Number of images
Even integer	Anywhere	$n - 1$
Odd integer	On the angle bisector	$n - 1$
Odd integer	Off the bisector	$n$
Not an integer	Anywhere	integer part of $n$ in most positions; count exactly with the circle method

Worth memorising: $θ = 9 0^{\circ} \to 3$ , $6 0^{\circ} \to 5$ , $4 5^{\circ} \to 7$ , $3 0^{\circ} \to 11$ , $7 2^{\circ} \to 4$ or $5$ , parallel ( $θ = 0$ ) $\to$ infinite.

JEE Advanced

Counting images without the formula. An image lies behind a mirror at the same angle as its object in front. An image lying at angle $x$ behind $M_{2}$ is at angle $x + θ$ in front of $M_{1}$ . So build two chains, adding $θ$ at every step: images formed by $M_{1}$ (angles measured from $M_{1}$ ) and images formed by $M_{2}$ (angles measured from $M_{2}$ ). Stop a chain when the next angle would exceed $18 0^{\circ}$ : that object is behind the mirror's extension and no image forms. If the last angles of the two chains plus $θ$ make exactly $36 0^{\circ}$ , the two final images coincide: subtract one. Solved Example 7 does this.

3.8 Minimum length of mirror to see the full image

The ray from the head reflects at a point midway (in height) between the head and the eye; the ray from the feet reflects midway between the eye and the feet. The part of the mirror between these points is all that is needed.

Figure 5: The useful part of the mirror is

M_{1} M_{2} = \frac{h}{2}

, whatever the distance from the mirror. Its lower edge sits at half the eye's height above the floor.

Minimum mirror length

= \frac{h}{2}

, independent of the distance from the mirror. Its lower edge must be at half the eye's height above the floor, and its top edge midway between eye and top of the head.

JEE Advanced

Vector form of the law of reflection. If $\overset{e}{^}$ is the unit vector along the incident ray and $\overset{n}{^}$ the unit normal to the mirror, the reflected ray is along

\overset{r}{^} = \overset{e}{^} - 2 (\overset{e}{^} \cdot \overset{n}{^}) \overset{n}{^}

The component of $\overset{e}{^}$ along the mirror is unchanged and the normal component flips. This form handles 3D problems and mirrors that are not along the axes.

Key idea

Plane mirror: virtual, erect, same-size image as far behind as the object is in front; turning the mirror by

θ

turns the reflected ray by

2 θ

4. Spherical Mirrors

A spherical mirror is a part of a hollow sphere. If the inner (hollow) surface reflects it is concave; if the outer (bulging) surface reflects it is convex.

Term	Meaning
Pole $P$	Centre of the mirror surface; origin for all distances
Centre of curvature $C$	Centre of the sphere of which the mirror is a part
Radius of curvature $R$	Radius of that sphere, $P C$
Principal axis	Line through $P$ and $C$
Principal focus $F$	Point where paraxial rays parallel to the axis meet (concave, real) or appear to diverge from (convex, virtual)
Focal length $f$	Distance $P F$
Aperture	Size (diameter) of the reflecting surface
Focal plane	Plane through $F$ perpendicular to the principal axis
Paraxial rays	Rays close to the axis making small angles with it

4.1 Why $f = R /2$ , and only for paraxial rays

A ray parallel to the axis strikes the mirror at angle of incidence $i$ (the normal is the radius). It reflects at $i$ and crosses the axis at a point $Q$ . The triangle formed with $C$ is isosceles, giving

C Q = \frac{R}{2 cos i}

For paraxial rays $i \to 0$ , $cos i \to 1$ and $C Q \to \frac{R}{2}$ : all such rays meet at the midpoint of $P C$ . This is the focus, so $f = \frac{R}{2}$ . Rays far from the axis (large $i$ ) cross closer to the mirror. This spreading of the focus is spherical aberration, and it is why the mirror formula needs small apertures.

Figure 6: Spherical aberration, drawn with exact reflection (upper half only; the lower half is the mirror image). Rays near the axis meet at

F

, midway between

C

and

P

; the marginal ray (

i \approx 4 4^{\circ}

) crosses at

C Q = \frac{R}{2 cos i} \approx 0.70 R

. A parabolic mirror (car headlight, telescope) brings all parallel rays to one point.

Parallel paraxial rays inclined at a small angle $θ$ to the axis meet on the focal plane at a height $h = f tan θ \approx f θ$ from the axis.

Concave mirror

Reflecting surface curves inward; converging; $f < 0$ . Gives real inverted images (object beyond $F$ ) or a virtual enlarged image (object inside $F$ ). Used in shaving mirrors, headlights, solar furnaces.

Convex mirror

Reflecting surface bulges outward; diverging; $f > 0$ . Always a virtual, erect, diminished image between $P$ and $F$ for a real object. Wide field of view: rear-view mirrors.

4.2 Rules for ray diagrams

A ray parallel to the principal axis passes (concave) or appears to pass (convex) through $F$ after reflection.
A ray through (or directed towards) $F$ becomes parallel to the axis (reversibility of light).
A ray through (or directed towards) $C$ retraces its path, since it strikes along the normal.
A ray striking the pole reflects symmetrically about the principal axis.

Any two of these rays from the top of an object locate the top of the image.

Figure 7: Concave mirror, object beyond

C

(

f = - 20 cm

u = - 50 cm

). The parallel, focal and pole rays meet at

v = - 33.3 cm

with

m = - 0.67

: real, inverted, diminished, between

F

and

C

. Distances are measured from the pole

P

along the incident light (New Cartesian convention).

4.3 New Cartesian sign convention

All distances are measured from the pole along the principal axis.
Distances in the direction of the incident light are positive; against it, negative. (Draw light travelling left to right, so "right of $P$ " is positive.)
Heights above the axis are positive; below, negative.

Mirror	$f$ and $R$	Real object $u$	Real image $v$	Virtual image $v$
Concave	negative	negative	negative	positive
Convex	positive	negative	not possible for a real object	positive

The sign of $f$ belongs to the mirror, not to the problem. Put

f = - 20 cm

for a concave mirror of focal length

20 cm

every time, then let the algebra give the sign of

v

4.4 Derivation of the mirror formula

Figure 8: Mirror formula geometry. With

α \approx \frac{A M}{P O}

β \approx \frac{A M}{P C}

γ \approx \frac{A M}{P I}

(paraxial rays),

α + γ = 2 β

becomes

\frac{1}{v} + \frac{1}{u} = \frac{2}{R}

Object $O$ on the axis; ray $O A$ reflects along $A I$ ; $C A$ is the normal, so $∠ O A C = ∠ C A I = θ$ .
Exterior angle of triangle $O A C$ : $β = α + θ$ . Exterior angle of triangle $O A I$ : $γ = α + 2 θ$ .
Eliminate $θ$ : $α + γ = 2 β$ .
Paraxial rays: $A$ is close to $P$ , so $α \approx \frac{A M}{P O}$ , $β \approx \frac{A M}{P C}$ , $γ \approx \frac{A M}{P I}$ . Hence $\frac{1}{P O} + \frac{1}{P I} = \frac{2}{P C}$ .
Signs: $P O = - u$ , $P I = - v$ , $P C = - R$ (all to the left of $P$ ). Substituting,
$\frac{1}{v} + \frac{1}{u} = \frac{2}{R} = \frac{1}{f}$

Mirror formula

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

holds for concave and convex mirrors, real and virtual objects, provided every quantity carries its sign. Lateral magnification

m = - \frac{v}{u}

(treated fully in the Magnification concept).

4.5 Image formation: all cases

Mirror	Object position	Image position	Nature and size
Concave	At infinity	At $F$	Real, inverted, highly diminished
Concave	Beyond $C$	Between $F$ and $C$	Real, inverted, diminished
Concave	At $C$	At $C$	Real, inverted, same size
Concave	Between $C$ and $F$	Beyond $C$	Real, inverted, enlarged
Concave	At $F$	At infinity	Real, inverted, highly enlarged
Concave	Between $F$ and $P$	Behind the mirror	Virtual, erect, enlarged
Convex	At infinity	At $F$ (behind)	Virtual, erect, highly diminished
Convex	Anywhere in front	Between $P$ and $F$ (behind)	Virtual, erect, diminished

Figure 9: The two virtual-image cases. (a) Concave mirror, object inside

F

v = + 30 cm

m = + 2.5

(shaving mirror). (b) Convex mirror:

v = + 12 cm

m = + 0.4

, image always between

P

and

F

behind the mirror (rear-view mirror). Dashed lines are the backward extensions that locate the image.

Figure 10:

v

against

u

for a concave mirror (exact plot of

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

f < 0

). The branches never cross

u = f

v = f

; object at

F

sends the image to infinity.

Exam Trick

Real goes with inverted, virtual goes with erect (for a real object and a single mirror). A convex mirror always gives a virtual, erect, diminished image of a real object, which is why it is the rear-view mirror: wide field of view. Only a concave mirror can magnify, and it gives an erect magnified image only when the object is inside $F$ (shaving and dentist's mirrors).

Virtual objects widen the picture: a concave mirror forms a real, erect image of a virtual object, and a convex mirror forms a real image of a virtual object placed between $P$ and $F$ . The general pairing (single mirror) is:

Object	Image	Orientation
Real	Real	Inverted
Real	Virtual	Erect
Virtual	Real	Erect
Virtual	Virtual	Inverted

4.6 Cutting a mirror

Cut a spherical mirror into pieces and keep them in place: every piece is part of the same sphere, with the same $C$ and $F$ , so there is still one image (only its brightness falls). If the pieces are displaced, each piece has its own shifted centre of curvature and forms its own image: two displaced halves give two images, each shifted in the direction the half was shifted.

Quick Recall: tap to check

Where must an object be for a concave mirror to form a real image of the same size?

At the centre of curvature

C

(

u = 2 f

What kind of image does a convex mirror give of a real object?

Virtual, erect, diminished, between

P

and

F

Key idea

Use

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

with signs: concave

f < 0

, convex

f > 0

;

m = - \frac{v}{u}

5. Combinations of Mirrors and Intensity

Find the image formed by the first mirror the light meets.
Treat that image as the object for the next mirror. Re-measure $u$ from the new pole, with signs set by the direction in which the light is now travelling.
If the rays hit the next mirror before meeting, the object is virtual ( $u > 0$ for that mirror).
Repeat for each reflection in the stated order.

Intensity with a mirror. The mirror collects the power falling on its aperture and redirects it. Intensity at a point = (direct light) + (reflected power divided by the area over which the reflected beam spreads there). A source at the focus of a concave mirror, for example, produces a parallel reflected beam whose intensity never falls with distance.

6. Flowchart and Mind Map

Decide first whether the mirror is plane or spherical; the flowchart picks the rule, the map lists everything on this page.

Figure 11: Problem-solving flowchart. Plane mirror: image, motion and rotation rules, or the two-mirror formulas. Spherical mirror: signs first, then

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

, image by image for several mirrors.

Figure 12: Reflection on one page. Revise from the map, then test yourself on the solved examples.

7. Solved Examples

Solved Example 1

A wall clock without numbers shows 8:12. What time does its image in a plane mirror show?

Solution:

Lateral inversion reflects the hands about the 12-6 line. Use the rule: image time $= 11 : 60 - 8 : 12 = 3 : 48$ . Check: the minute hand at 12 minutes appears at 48 minutes, and the hour hand just past 8 appears just before 4.

Answer: 3:48.

Solved Example 2

A small object

O

is at a distance

d

in front of a plane mirror of length

ℓ

, opposite its midpoint. A man walks along a straight road

A B

parallel to the mirror, at a further distance

d

beyond

O

. Over what length of the road can he see the image?
(A)

ℓ

(B)

3 ℓ

(C)

1.5 ℓ

(D)

2 ℓ

Solution:

The image $I$ is $d$ behind the mirror, so the road is $3 d$ from $I$ (Figure 2). Rays reaching the road seem to come from $I$ through the mirror's edges. By similar triangles the visible stretch is $ℓ \times \frac{3 d}{d} = 3 ℓ$ .

Answer: (B). Most geometric-optics problems are similar-triangle problems in disguise.

Solved Example 3

A plane mirror lies in the

y

z

plane (normal along

x

). It moves with

5 m s^{- 1}

in a direction making

3 0^{\circ}

with the

- x

axis. An object in front of it moves with

10 m s^{- 1}

6 0^{\circ}

to the

+ x

axis. Find the velocity of the image.

Solution:

Normal ( $x$ ) components: $v_{M} = - 5 cos 3 0^{\circ} = - \frac{5 3}{2}$ , $v_{O} = 10 cos 6 0^{\circ} = 5$ .

v_{I, x} = 2 v_{M} - v_{O} = - 53 - 5 = - 5 (1 + 3) m s^{- 1}

Parallel ( $y$ ) component: same as the object's, $v_{I, y} = 10 sin 6 0^{\circ} = 53 m s^{- 1}$ . The mirror's own $y$ -motion does not matter.

Answer: $v_{I} = - 5 (1 + 3) \hat{i} + 53 \hat{j} m s^{- 1}$ , about $16.2 m s^{- 1}$ .

Solved Example 4

A ray falls on a horizontal plane mirror at

3 0^{\circ}

to the normal, travelling down and to the right. Through what angle, and in which sense, must the mirror be turned so that the reflected ray becomes vertical?

Solution:

The reflected ray initially points up at $3 0^{\circ}$ to the vertical (direction $6 0^{\circ}$ above the $+ x$ axis). A mirror rotation $ϕ$ (anticlockwise positive) turns the reflected ray by $2 ϕ$ : new direction $6 0^{\circ} + 2 ϕ$ .

Vertically up ( $9 0^{\circ}$ ): $2 ϕ = 3 0^{\circ}$ , $ϕ = 1 5^{\circ}$ anticlockwise.
Vertically down ( $27 0^{\circ}$ ): $2 ϕ = 21 0^{\circ}$ , $ϕ = 10 5^{\circ}$ anticlockwise; or $2 ϕ = - 15 0^{\circ}$ , $ϕ = 7 5^{\circ}$ clockwise.

Check which the ray can still reach on the silvered face: the incident ray meets the front face only while $ϕ$ lies between $- 6 0^{\circ}$ and $12 0^{\circ}$ . The $7 5^{\circ}$ clockwise turn fails (the ray would hit the back).

Answer: $1 5^{\circ}$ anticlockwise (ray vertically up) or $10 5^{\circ}$ anticlockwise (ray vertically down).

Solved Example 5

A plane mirror at

O

rotates with angular speed

ω

about a vertical axis. A fixed ray falls on it and the reflected spot moves on a flat wall at perpendicular distance

d

from

O

. Find the spot's speed when the reflected ray makes angle

θ

with the perpendicular to the wall.

Solution:

The reflected ray rotates at $2 ω$ , so $\frac{d θ}{d t} = 2 ω$ . The spot is at $x = d tan θ$ from the foot of the perpendicular:

v = \frac{d x}{d t} = d sec^{2} θ \frac{d θ}{d t} = 2 ω d sec^{2} θ

Answer: $v = 2 ω d sec^{2} θ$ . Minimum $2 ω d$ at the foot; it grows without limit as $θ \to 9 0^{\circ}$ . On a circular screen of radius $R$ centred at $O$ the speed would be a constant $2 ω R$ .

Solved Example 6

A point object is between two parallel plane mirrors

M_{1}

and

M_{2}

that are

10 cm

apart,

2 cm

from

M_{1}

. Find the distances of the first four images formed when light reflects first at

M_{1}

Solution:

Reflection at	Object	Object distance	Image	Image distance
$M_{1}$	$O$	$2 cm$ from $M_{1}$	$I_{1}$	$2 cm$ behind $M_{1}$
$M_{2}$	$I_{1}$	$12 cm$ from $M_{2}$	$I_{12}$	$12 cm$ behind $M_{2}$
$M_{1}$	$I_{12}$	$22 cm$ from $M_{1}$	$I_{121}$	$22 cm$ behind $M_{1}$
$M_{2}$	$I_{121}$	$32 cm$ from $M_{2}$	$I_{1212}$	$32 cm$ behind $M_{2}$

Answer: $2, 12, 22, 32 cm$ , alternately behind $M_{1}$ and $M_{2}$ (each step adds twice the separation, $20 cm$ , every two reflections). Starting at $M_{2}$ gives a second chain: $8, 18, 28, \dots cm$ . Infinitely many images in all.

Solved Example 7

Two plane mirrors are inclined at

3 0^{\circ}

. An object makes

1 0^{\circ}

with

M_{1}

(so

2 0^{\circ}

with

M_{2}

). Find the number of images (i) by formula and (ii) by counting.

Solution:

(i) $n = \frac{36 0 ^{\circ}}{3 0 ^{\circ}} = 12$ , even, so images $= 12 - 1 = 11$ .

(ii) Each new image's angle = previous image's angle from the other mirror $+ 3 0^{\circ}$ ; stop before $18 0^{\circ}$ :

Chain	Image angles	Count
Formed by $M_{1}$ (from $M_{1}$ )	$1 0^{\circ}, 5 0^{\circ}, 7 0^{\circ}, 11 0^{\circ}, 13 0^{\circ}, 17 0^{\circ}$	6
Formed by $M_{2}$ (from $M_{2}$ )	$2 0^{\circ}, 4 0^{\circ}, 8 0^{\circ}, 10 0^{\circ}, 14 0^{\circ}, 16 0^{\circ}$	6

Last angles: $17 0^{\circ} + 16 0^{\circ} + 3 0^{\circ} = 36 0^{\circ}$ , so the two final images coincide. Total $= 6 + 6 - 1 = 11$ .

Answer: 11 images.

Solved Example 8

A ray parallel to the principal axis of a concave mirror reflects and passes through the pole

P

. Find its angle of incidence.

Solution:

Let the ray hit at $A$ with incidence angle $θ$ . Since the ray is parallel to $C P$ , alternate angles give $∠ A C P = θ$ . By the law of reflection $∠ C A P = θ$ . $C A = C P = R$ , so triangle $C A P$ is isosceles and $∠ C P A = ∠ C A P = θ$ . All three angles are equal: $3 θ = 18 0^{\circ}$ .

Answer: $6 0^{\circ}$ (a marginal ray, far from paraxial).

Solved Example 9

A concave mirror has

R = 20 cm

. Find the position and nature of the image of a point object on its axis at (a)

30 cm

and (b)

5 cm

from the pole.

Solution:

$f = - 10 cm$ .

(a) $u = - 30$ : $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = - \frac{1}{10} + \frac{1}{30} = - \frac{1}{15}$ , so $v = - 15 cm$ : real, $15 cm$ in front (between $F$ and $C$ ).

(b) $u = - 5$ : $\frac{1}{v} = - \frac{1}{10} + \frac{1}{5} = + \frac{1}{10}$ , so $v = + 10 cm$ : virtual, $10 cm$ behind the mirror.

Answer: (a) $15 cm$ in front, real; (b) $10 cm$ behind, virtual. Same mirror, opposite natures: the object crossed $F$ .

Solved Example 10

A concave mirror

m_{1}

(

R = 20 cm

) faces a plane mirror

40 cm

away. A point object on the common axis is

15 cm

from

m_{1}

. Find the final image after three reflections, the first at

m_{1}

Solution:

At $m_{1}$ : $u = - 15$ , $f = - 10$ : $\frac{1}{v} = - \frac{1}{10} + \frac{1}{15} = - \frac{1}{30}$ , $v = - 30 cm$ . Image $30 cm$ from $m_{1}$ , i.e. $10 cm$ in front of the plane mirror.
At the plane mirror: image $10 cm$ behind it, i.e. $50 cm$ from $m_{1}$ .
At $m_{1}$ again: $u = - 50$ : $\frac{1}{v} = - \frac{1}{10} + \frac{1}{50} = - \frac{4}{50}$ , $v = - 12.5 cm$ .

Answer: real image $12.5 cm$ in front of $m_{1}$ .

Solved Example 11

A point source is

30 cm

in front of a concave mirror of focal length

20 cm

. A screen perpendicular to the axis is

10 cm

beyond the source. Without the mirror the intensity at the centre of the screen is

I

. Find it with the mirror present (small aperture, perfect reflection).

Solution:

Image of the source: $u = - 30$ , $f = - 20$ gives $v = - 60 cm$ . The reflected beam converges towards a point $60 cm$ from the mirror, but the screen is at $40 cm$ .
Intensity reaching the mirror (distance $30 cm$ instead of $10 cm$ ): $I (\frac{10}{30})^{2} = \frac{I}{9}$ . Power collected by an aperture of radius $a$ : $\frac{I}{9} π a^{2}$ .
At the screen, $20 cm$ short of the convergence point, the beam radius is $a \times \frac{20}{60} = \frac{a}{3}$ . Reflected intensity $= \frac{( I /9 ) π a ^{2}}{π a ^{2} /9} = I$ .

Answer: $I + I = 2 I$ .

Solved Example 12

A person

1.80 m

tall has eyes

10 cm

below the top of the head. Find the minimum height of a vertical plane mirror in which the person sees the full image, and how high its lower edge must be.

Solution:

Minimum length $= \frac{h}{2} = 0.90 m$ . Eye height $= 1.70 m$ . Lower edge at half the eye height $= 0.85 m$ ; top edge midway between eye and head, $1.75 m$ (check: $1.75 - 0.85 = 0.90$ ).

Answer: $0.90 m$ mirror, lower edge $0.85 m$ above the floor, at any distance from the mirror.

Practice Questions

An object moves at $5 m s^{- 1}$ towards the right and a plane mirror in front of it moves at $1 m s^{- 1}$ towards the left. Find the velocity of the image.Answer: $7 m s^{- 1}$ towards the left ( $v_{I} = 2 v_{M} - v_{O} = - 2 - 5$ )
Two plane mirrors are inclined at $4 5^{\circ}$ . Find the deviation of a ray reflected once from each.Answer: $9 0^{\circ}$ (that is, $36 0^{\circ} - 9 0^{\circ} = 27 0^{\circ}$ in the other sense)
Two plane mirrors are perpendicular. A ray reflects once from each. Find the deviation.Answer: $18 0^{\circ}$ : the ray returns antiparallel (retro-reflector)
A mirror at the centre of a spherical screen of radius $R$ rotates at $ω$ . Find the speed of the reflected light spot on the screen.Answer: $2 ω R$
A point object is midway between two parallel mirrors $2 cm$ apart. Locate the images.Answer: At $1, 3, 5, 7, \dots cm$ behind each mirror (an arithmetic progression)
A convex mirror $M_{1}$ ( $f = 10 cm$ ) and a concave mirror $M_{2}$ ( $f = 20 cm$ ) face each other $55 cm$ apart. An object is $10 cm$ from $M_{1}$ . Find the final image after reflection at $M_{1}$ and then $M_{2}$ .Answer: $M_{1}$ : $v = + 5 cm$ ; $M_{2}$ : $u = - 60$ , $v = - 30 cm$ , real, $30 cm$ in front of $M_{2}$
A concave mirror $M_{1}$ ( $f = 20 cm$ ) and a convex mirror $M_{2}$ ( $f = 10 cm$ ) face each other $45 cm$ apart. An object is $30 cm$ from $M_{1}$ . Find the final image after reflection at $M_{1}$ and then $M_{2}$ .Answer: $M_{1}$ : $v = - 60 cm$ , so a virtual object $15 cm$ behind $M_{2}$ ; $M_{2}$ : $v = + 30 cm$ , virtual, $30 cm$ behind $M_{2}$
A point source is at the focus of a concave mirror ( $f = 10 cm$ ) and a screen is $10 cm$ beyond the source. The intensity at the screen without the mirror is $I$ . Find it with the mirror.Answer: $2 I$ (the reflected beam is parallel and carries the intensity $I$ found at the mirror)

Common Mistakes to Avoid

Watch out

Measuring the angle of incidence from the mirror surface instead of from the normal. The surface angle is the glancing angle $g = 9 0^{\circ} - i$ .
Writing image speed $=$ mirror speed. A moving mirror moves the image at twice its speed (fixed object), and only the normal component doubles.
Saying the reflected ray turns by $θ$ when the mirror turns by $θ$ . It turns by $2 θ$ .
Using $n - 1$ images for every angle. For odd $\frac{36 0 ^{\circ}}{θ}$ with the object off the bisector the answer is $n$ .
Putting $f = + 20 cm$ for a concave mirror. In the New Cartesian convention a concave mirror always has $f < 0$ , a convex mirror $f > 0$ .
Forgetting the sign of $u$ : a real object in front of any mirror has $u < 0$ .
In multi-mirror problems, measuring the new object distance from the old pole, or not noticing that rays hit the second mirror before meeting (virtual object, $u > 0$ ).
Believing the mirror must be as tall as you, or that standing farther away lets a smaller mirror work. It is $\frac{h}{2}$ at every distance.

Frequently Asked Questions

What are the laws of reflection?

First, the incident ray, the reflected ray and the normal at the point of incidence lie in one plane. Second, the angle of incidence equals the angle of reflection, both measured from the normal. The laws hold for plane and curved mirrors; for a curved mirror the normal is the radius at that point.

Why is a plane mirror image laterally inverted?

A plane mirror reverses only the coordinate perpendicular to it; points along the mirror keep their positions. Your right hand's image is directly opposite it, so when you face the image the sides appear swapped. This front-back reversal is what we call lateral inversion, and it is why AMBULANCE is written reversed.

How many images are formed by two inclined plane mirrors?

Find n equal to 360 degrees divided by the angle between the mirrors. If n is even, there are n minus 1 images. If n is odd, there are n minus 1 images when the object is on the angle bisector and n images otherwise. Parallel mirrors give infinitely many images.

Why does the reflected ray turn by twice the angle when a mirror rotates?

Rotating the mirror by an angle theta rotates its normal by theta. The angle of incidence changes by theta, and the angle of reflection changes by the same amount on the other side of the normal, so the reflected ray turns by two theta. Mirror galvanometers use this doubling to magnify small rotations.

What is the minimum length of mirror needed to see your full image?

Half your height. The ray from your head reflects at the level midway between head and eyes, and the ray from your feet reflects midway between eyes and feet. The mirror section between these two points is exactly half your height, and it does not depend on how far you stand.

Why is the focal length of a spherical mirror half the radius of curvature?

A ray parallel to the axis meets the axis at a distance R divided by twice the cosine of the angle of incidence from the centre of curvature. For paraxial rays that cosine is almost 1, so all such rays meet at R over 2 from the centre, the midpoint between centre and pole. Hence f equals R over 2.

Is reflection at plane and spherical surfaces important for JEE Main and JEE Advanced?

Yes. JEE Main regularly asks mirror formula numericals, number of images and rotation of mirrors. JEE Advanced adds image velocity, multiple reflections between mirrors, field of view and combinations with lenses. Master the New Cartesian sign convention first, because most errors in these questions are sign errors.

Which mirror questions are common in NEET?

NEET mostly asks direct mirror formula numericals, nature and position of images for concave and convex mirrors, uses of mirrors, number of images between inclined mirrors and the minimum mirror length. Memorise the image formation table and the rule that real images are inverted and virtual images erect for a real object.

Previous year questions on Reflection at Plane and Spherical Surfaces

4 questions from past papers, each with a step-by-step solution.

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1. Light and Ray Optics: The Basics

1.1 Ray and beam

2. Reflection and Its Laws

3. Plane Mirror

3.1 Image of a point and of an extended object

3.2 Field of view

3.3 Velocity of the image in a plane mirror

3.4 Deviation by one and two plane mirrors

3.5 Real and virtual objects and images

3.6 Rotation of a plane mirror

3.7 Images formed by two plane mirrors

3.8 Minimum length of mirror to see the full image

4. Spherical Mirrors

4.1 Why f=R/2, and only for paraxial rays

4.2 Rules for ray diagrams

4.3 New Cartesian sign convention

4.4 Derivation of the mirror formula

4.5 Image formation: all cases

4.6 Cutting a mirror

5. Combinations of Mirrors and Intensity

6. Flowchart and Mind Map

7. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What are the laws of reflection?

Why is a plane mirror image laterally inverted?

How many images are formed by two inclined plane mirrors?

Why does the reflected ray turn by twice the angle when a mirror rotates?

What is the minimum length of mirror needed to see your full image?

Why is the focal length of a spherical mirror half the radius of curvature?

Is reflection at plane and spherical surfaces important for JEE Main and JEE Advanced?

Which mirror questions are common in NEET?

Previous year questions on Reflection at Plane and Spherical Surfaces

Ready to master Ray Optics And Optical Instruments?

4.1 Why $f = R /2$ , and only for paraxial rays