Ray Optics · Study notes

PHYReflection by Spherical Mirrors

Reflection at Spherical Mirrors

Reflection at spherical mirrors deals with how concave and convex mirrors form images, located using the mirror formula $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ together with the magnification relation. The core idea is that a single sharp focus (the principal focus, at a focal length $f=\dfrac{R}{2}$ ) plus the Cartesian sign convention let you predict an image's position, size, and nature for any object distance.

This is examined heavily in ISC numericals because almost every question reduces to substituting signed values, solving for $v$ , and then interpreting $m$ — the sign and magnitude of the answers carry all the physics (real/virtual, erect/inverted, magnified/diminished).

Mirror formula

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

u

= object distance,

v

= image distance,

f

= focal length, all measured from the pole

P

with signs from the Cartesian convention. For a concave mirror

f<0

; for a convex mirror

f>0

Focal length and radius of curvature

f = \dfrac{R}{2}

R

= radius of curvature,

f

= focal length. Both are negative for a concave mirror and positive for a convex mirror under the sign convention; valid in the paraxial approximation.

Linear magnification

m = \dfrac{h_2}{h_1} = -\dfrac{v}{u}

m

= magnification,

h_1

= object height,

h_2

= image height,

v

u

as above.

m>0

means an erect (virtual) image,

m<0

means an inverted (real) image;

|m|>1

magnified,

|m|<1

diminished.

Magnification in terms of focal length

m = \dfrac{f}{f - u} = \dfrac{f - v}{f}

Same symbols as above. Useful when the magnification (or image size) is given and you must find

u

v

without first solving the full mirror equation.

Object position	Image position	Nature	Size
At infinity	At $F$	Real, inverted	Highly diminished
Beyond $C$	Between $F$ and $C$	Real, inverted	Diminished
At $C$	At $C$	Real, inverted	Same size
Between $C$ and $F$	Beyond $C$	Real, inverted	Magnified
At $F$	At infinity	Real, inverted	Highly magnified
Between $F$ and $P$	Behind the mirror	Virtual, erect	Magnified

A convex mirror needs no such table: for every real object the image is virtual, erect and diminished — which is exactly why it is the driver's mirror.

Sign convention (ISC/NCERT Cartesian): all distances are measured from the pole $P$ along the principal axis, with the incident light travelling in the $+x$ direction; distances measured against the incident light are negative, and heights above the axis are positive. A real object is always at $u<0$ .
For a concave mirror $f$ and $R$ are negative; for a convex mirror they are positive. Substitute these signs at the start — do not 'add the negative later.'
Interpret the answer purely from signs: $v<0$ means a real image (same side as the object, in front of the mirror); $v>0$ means a virtual image (behind the mirror). $m<0$ inverted, $m>0$ erect.
A convex mirror always gives a virtual, erect, diminished image ( $0<m<1$ ) for any real object, which is why it is used as a rear-view (driver's) mirror.
When magnification is specified only by size, e.g. 'three times the size,' the image may be real ( $m=-3$ ) or virtual ( $m=+3$ ), giving two possible object positions — consider both cases unless the question fixes the nature.
The relations $f=\dfrac{R}{2}$ and $m=-\dfrac{v}{u}$ hold only in the paraxial (small-angle) approximation, where rays stay close to the principal axis; otherwise spherical aberration spoils the single focus.
For 'displacement of image' problems, find $v$ at each object position separately, then take the difference $\Delta v = v_2 - v_1$ ; its magnitude is the displacement and its sign shows the direction of shift.

Worked example · 1 mark

An object is placed

10\,\text{cm}

in front of a concave mirror of focal length

20\,\text{cm}

. Where is the image formed, and what is its nature?

Apply the sign convention first: $u = -10\,\text{cm}$ , and for a concave mirror $f = -20\,\text{cm}$ .
Mirror formula: $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{-20} - \dfrac{1}{-10} = \dfrac{-1+2}{20} = \dfrac{1}{20}$ .
So $v = +20\,\text{cm}$ . A positive $v$ means the image is behind the mirror, so it is virtual.
Magnification $m = -\dfrac{v}{u} = -\dfrac{20}{-10} = +2$ — erect and twice the size.

AnswerVirtual and erect,

20\,\text{cm}

behind the mirror, magnified two times. The object lies inside the focal length, which is the only concave-mirror case that gives a virtual image.

Where the marks go

Forgetting to make $u$ , $f$ and $R$ negative for a concave mirror (or forgetting that $R=2f$ ), then getting a positive $v$ and wrongly calling a real image 'virtual.'
Using $R$ in place of $f$ in the mirror formula — you must convert with $f=\dfrac{R}{2}$ first; e.g. a radius of curvature of magnitude $24\,\text{cm}$ gives $f$ of magnitude $12\,\text{cm}$ .
When magnification is given without specifying nature, taking only one case. 'Image three times the object' allows both $m=+3$ (virtual) and $m=-3$ (real), so there are two valid object distances.
Reading the sign of $m$ as just the size and ignoring it for the nature: $m=-2$ is inverted and magnified, not 'minus two times big.' The sign gives the orientation, the magnitude gives the scaling.

How the board asks it

Numericalmirror formula and magnification relation1 mkAsked 2023
An object is placed $20\,\text{cm}$ in front of a concave mirror of focal length $15\,\text{cm}$ . Calculate the position and magnification of the image, and state its nature.
Numerical $f=\dfrac{R}{2}$ applied before the mirror formula
The radius of curvature of a concave mirror is $24\,\text{cm}$ . An object is placed $16\,\text{cm}$ from the mirror. Find the focal length and hence calculate the position and nature of the image.
Numericalmagnification specified by size only, giving two cases
A concave mirror of focal length $12\,\text{cm}$ forms an image three times the size of the object. Find the two possible positions of the object.
Give reasonsconvex mirror gives a virtual, erect, diminished image with wider field of view
Give reasons why a convex mirror is preferred over a plane or concave mirror as a rear-view (driver's) mirror in vehicles.
Diagram / graphimage character versus object position for a concave mirror
Draw a labelled ray diagram to show the formation of the image when an object is placed between the centre of curvature $C$ and the focus $F$ of a concave mirror, and state the nature of the image.
Assertion–Reasona mirror's focal length does not depend on the surrounding medium1 mkAsked 2026
Assertion: the focal length of a convex mirror increases when it is placed in water. Reason: for a convex mirror of radius $R$ , $f = R/2$ . Decide whether each statement is true and whether the reason explains the assertion.
Diagram / graphreading $f$ off a magnification-versus-image-distance graph2 mk
A graph of $|m|$ against $|v|$ is given for a mirror. Name the type of mirror and find its focal length from the point where $|m| = 1$ .

Practise this topic

NextRefraction at Plane Surfaces & Total Internal Reflection

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.