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Reflection of Light and Spherical Mirrors



Reflection Of Light

Reflection of light is the phenomenon where light bounces off a surface. When a ray of light strikes a boundary between two different media and returns into the same medium, it is said to be reflected.

Reflection is responsible for how we see objects that do not emit their own light. Light from a source (like the sun or a bulb) falls on an object, is reflected from its surface, and when this reflected light enters our eyes, we perceive the object.


Laws of Reflection

Reflection of light from a smooth, polished surface (like a mirror) obeys the following two laws, known as the laws of reflection:

  1. The angle of incidence is equal to the angle of reflection.

    $ \angle i = \angle r $

    where $\angle i$ is the angle between the incident ray and the normal to the reflecting surface at the point of incidence, and $\angle r$ is the angle between the reflected ray and the normal.
  2. The incident ray, the reflected ray, and the normal to the reflecting surface at the point of incidence all lie in the same plane.
Diagram illustrating the laws of reflection from a plane mirror.

(Image Placeholder: A diagram showing a flat line representing a plane mirror. An incident ray strikes the mirror. A perpendicular line to the mirror at the point of incidence is the normal. A reflected ray bounces off the mirror. Label the incident ray, reflected ray, normal, angle of incidence (i), and angle of reflection (r). Show that i = r, and all three lines are in the same plane, which is the plane of the diagram.)


Types of Reflection

The laws of reflection apply to both regular and diffused reflection at a microscopic level (at each point of incidence), but the macroscopic appearance is different due to the orientation of the surface normals.

We see images in mirrors due to regular reflection. Plane mirrors produce virtual, erect, and laterally inverted images of the same size as the object, located as far behind the mirror as the object is in front.



Spherical Mirrors

Spherical mirrors are mirrors whose reflecting surface is a part of a hollow sphere. They are classified into two types based on whether the inner or outer surface is polished:


Terms Related to Spherical Mirrors


Image Formation By Spherical Mirrors

The type and characteristics of the image formed by a spherical mirror depend on the position of the object relative to the mirror and the type of mirror (concave or convex).

Real images can be formed on a screen, while virtual images cannot. Real images formed by single mirrors are always inverted, and virtual images are always erect.


Representation Of Images Formed By Spherical Mirrors Using Ray Diagrams

Ray diagrams are graphical tools used to determine the position, size, and nature of the image formed by spherical mirrors. Drawing ray diagrams helps in understanding the process of image formation. We use a few standard rays whose paths after reflection are known:

  1. A ray parallel to the principal axis, after reflection, passes through the principal focus (F) in the case of a concave mirror, or appears to diverge from the principal focus (F) in the case of a convex mirror.
  2. A ray passing through the principal focus (F) of a concave mirror, or directed towards the principal focus (F) of a convex mirror, after reflection, becomes parallel to the principal axis.
  3. A ray passing through the centre of curvature (C) of a concave mirror, or directed towards the centre of curvature (C) of a convex mirror, after reflection, is reflected back along the same path. This is because the ray strikes the mirror along the normal at that point.
  4. A ray incident towards the pole (P) of the mirror is reflected according to the law of reflection ($\angle i = \angle r$), with the principal axis acting as the normal.

To find the position of the image, we draw at least two of these rays from a point on the object. The point where the reflected rays intersect (for a real image) or appear to intersect (for a virtual image) is the corresponding point on the image. The image of an extended object can be found by locating the images of a few key points (like the top and bottom) of the object.

Ray diagram for image formation by a concave mirror.

(Image Placeholder: Ray diagram for a concave mirror. Principal axis, Pole P, Focus F, Centre of Curvature C are marked. An object is placed beyond C. Ray 1 parallel to axis passes through F after reflection. Ray 2 passing through C reflects back along C. The intersection of reflected rays between F and C forms a real, inverted, diminished image.)

Ray diagram for image formation by a convex mirror.

(Image Placeholder: Ray diagram for a convex mirror. Principal axis, Pole P, Focus F (behind mirror), Centre of Curvature C (behind mirror) are marked. An object is placed in front of the mirror. Ray 1 parallel to axis appears to diverge from F after reflection. Ray 2 directed towards C reflects back along the same path. The intersection of the *extended* reflected rays behind the mirror, between P and F, forms a virtual, erect, diminished image.)


Sign Convention For Reflection By Spherical Mirrors

To use the mirror formula consistently, a sign convention is necessary to assign positive or negative values to distances and heights. The most widely used convention is the New Cartesian Sign Convention:

  1. Origin: The pole (P) of the mirror is taken as the origin (0,0).
  2. Principal Axis as X-axis: The principal axis of the mirror is taken as the X-axis.
  3. Distances Measured from Pole: All distances are measured from the pole of the mirror.
  4. Direction of Incident Light: Distances measured in the direction of the incident light are taken as positive.
  5. Opposite to Incident Light: Distances measured in the direction opposite to the direction of incident light are taken as negative.
  6. Heights Above Principal Axis: Heights measured upwards, perpendicular to the principal axis, are taken as positive.
  7. Heights Below Principal Axis: Heights measured downwards, perpendicular to the principal axis, are taken as negative.

Unless otherwise specified, incident light is usually assumed to travel from left to right. Following this convention:



Reflection Of Light By Spherical Mirrors

Sign Convention

As discussed in the previous section, consistent use of a sign convention is crucial when applying the mirror formula and magnification formula. We will use the convention where distances of real objects and real images are positive, virtual objects and virtual images are negative, and focal length is positive for concave and negative for convex mirrors. Heights above the axis are positive, below are negative.

Example: A real object is placed in front of a concave mirror. By this convention, $u$ is positive, $f$ is positive. The image position $v$ will depend on $u$ (can be positive or negative). If the image is real, $v$ will be positive, and the image will be inverted ($h_i$ negative, $m$ negative). If the image is virtual, $v$ will be negative, and the image will be erect ($h_i$ positive, $m$ positive).


Focal Length Of Spherical Mirrors

The principal focus (F) of a spherical mirror is a specific point on the principal axis related to the reflection of parallel rays. The focal length ($f$) is the distance between the pole (P) and the principal focus (PF).

For spherical mirrors of small aperture, the principal focus is located midway between the pole and the centre of curvature (C).

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

where $R$ is the radius of curvature.

Following our adopted sign convention:

This convention aligns with the converging nature of concave mirrors (real focus) and the diverging nature of convex mirrors (virtual focus).


The Mirror Equation ($ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $)

The mirror equation is a mathematical relationship that links the object distance ($u$), the image distance ($v$), and the focal length ($f$) of a spherical mirror. It can be derived using geometry and the laws of reflection for paraxial rays (rays close to the principal axis and making small angles with it).


Derivation Sketch (Concave Mirror, Real Image)

Consider a concave mirror, an object AB placed beyond the centre of curvature C. Let the pole be P, focus F, and centre of curvature C. Object height $h_o$, image height $h_i$. Object distance $u$ (positive), image distance $v$ (positive), focal length $f$ (positive).

Ray diagram for derivation of mirror formula for a concave mirror forming a real image.

(Image Placeholder: Detailed ray diagram for concave mirror with object AB beyond C, forming real image A'B' between F and C. Show rays from B: parallel to axis through F, through C back along C, to P reflecting at angle i=r. Identify similar triangles like A'B'P ~ ABP and A'B'F ~ MPF (where M is a point on the mirror near P, and MP is considered a straight line perpendicular to axis).)

Using similar triangles formed by rays from the object, reflecting from the mirror, and forming the image, and applying the sign convention, one can arrive at the mirror formula.

Consider $\triangle ABP$ and $\triangle A'B'P$. These are similar (angles are equal due to law of reflection at P, and right angles at A and A').

$ \frac{A'B'}{AB} = \frac{PA'}{PA} \implies \frac{-h_i}{h_o} = \frac{v}{u} \implies m = \frac{h_i}{h_o} = -\frac{v}{u} $ (This gives the magnification formula)

Consider a ray from B parallel to the principal axis, reflecting through F. Let this ray strike the mirror at M, close to P. Draw a perpendicular from M to the principal axis. For a small aperture, M is very close to P, and MP can be considered perpendicular to the axis, length approximately equal to the height of the ray's point of incidence from the axis. Let's use a simpler triangle involving F.

Consider $\triangle A'B'F$ and $\triangle MPF$. These are similar (if M is very close to P, MP is nearly perpendicular to the axis). $MP \approx AB = h_o$.

$ \frac{A'B'}{MP} = \frac{FA'}{FP} \implies \frac{-h_i}{h_o} = \frac{v - f}{f} $ (using distances from P and F)

Equating the two expressions for magnification:

$ -\frac{v}{u} = \frac{v - f}{f} $

$ -\frac{v}{u} = \frac{v}{f} - 1 $

Divide by $v$ (assuming $v \ne 0$):

$ -\frac{1}{u} = \frac{1}{f} - \frac{1}{v} $

$ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} $

$ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} $

This derivation uses $u, v, f$ as positive magnitudes. If we use the sign convention where $u$ is positive for real object, $v$ positive for real image, $f$ positive for concave, then the formula holds as stated in the subheading $ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $. For the derivation to yield the formula with plus signs, the distances must be defined relative to a fixed direction from the pole (e.g. distances measured against incident light are negative). Let's re-evaluate the similar triangles with proper signed distances from the pole P.

Let incident light travel from left to right. P is origin (0). Object is at $u$ (negative). Image is at $v$ (real is negative, virtual is positive). Concave $f$ is negative. Convex $f$ is positive.

$\triangle ABP$ and $\triangle A'B'P$: $\frac{A'B'}{AB} = \frac{PA'}{PA}$. Let P be the origin. A is at $u$, A' at $v$. B is at $(u, h_o)$, B' at $(v, h_i)$. $\frac{h_i}{h_o} = \frac{v}{u}$. This gives $m=v/u$. But magnification formula uses $m=-v/u$. This suggests the ratio of heights should be equal to the ratio of distances with a negative sign if distances are measured from the pole in a consistent direction. The standard $m=-v/u$ and $1/f = 1/v + 1/u$ using the sign convention as described (real distances +ve, virtual -ve for v, real object +ve for u, concave f +ve, convex f -ve) seems to have inconsistencies if $u$ for real object is +ve. Let's revert to the consistent New Cartesian convention where distances measured in the direction of incident light are positive.

New Cartesian Sign Convention (re-revisited): Incident light from Left to Right. Pole P at Origin. Distances to the right of P are positive, to the left are negative. Heights above axis are positive, below are negative.

Let's re-derive $m = -v/u$ and $1/f = 1/v + 1/u$ using this convention.

$\triangle ABP$ and $\triangle A'B'P$: $\frac{A'B'}{AB} = \frac{PA'}{PA}$. Coordinates: A is $(u, 0)$, B is $(u, h_o)$, A' is $(v, 0)$, B' is $(v, h_i)$. P is $(0,0)$. $PA = u$, $PA' = v$. $AB$ is height $h_o$, $A'B'$ is height $h_i$. $\frac{h_i}{h_o} = \frac{v-0}{u-0} = \frac{v}{u}$. But image might be inverted... Let's use vector heights. $\vec{AB}$ is $(0, h_o)$. $\vec{A'B'}$ is $(0, h_i)$. $\triangle ABP$ and $\triangle A'B'P$ are similar, using angles. Angle of incidence = angle of reflection at P. The ratio of corresponding sides $|A'B'|/|AB| = |PA'|/|PA|$. The sign comes from comparing directions. If A'B' is inverted relative to AB, $h_i/h_o$ should be negative. By geometry $\frac{h_i}{h_o} = -\frac{v}{u}$. This seems standard and consistent with magnification definition $m=h_i/h_o$. So $m = -v/u$. Using the New Cartesian convention, if $h_o$ is positive, $h_i$ is negative for inverted real image, $v$ is negative for real image, $u$ is negative for real object. $m = (-ve)/(+ve) = -ve$. $-v/u = -(-ve)/(-ve) = -ve$. Signs work.

For the mirror formula, consider the ray parallel to axis through M reflecting through F. Slope of incident ray is 0. Slope of reflected ray from M to F is related to coordinates of M and F. This derivation is more involved. A simpler approach relates similar triangles formed by rays from the object and reflecting through F or C.

Let's use triangles formed by the ray from B passing through C and the ray parallel to the axis from B reflecting through F. Object AB, Image A'B'. Concave mirror.

$\triangle ABC$ and $\triangle A'B'C$ are similar. $\frac{A'B'}{AB} = \frac{A'C}{AC}$. Let distances be from P. $AC = PA - PC = u - R$. $A'C = PC - PA' = R - v$. $\frac{|h_i|}{h_o} = \frac{|R-v|}{|u-R|}$. With sign convention, $h_i/h_o = -(v-R)/(u-R)$. Or $\frac{h_i}{h_o} = \frac{R-v}{u-R}$ magnitude wise. For concave mirror, $u, v, R$ are negative in New Cartesian. $\frac{h_i}{h_o} = \frac{-v - (-R)}{-u - (-R)} = \frac{R-v}{R-u}$. Also $m = h_i/h_o = -v/u$. So $\frac{R-v}{R-u} = -\frac{v}{u}$. $u(R-v) = -v(R-u)$. $uR - uv = -vR + uv$. $uR + vR = 2uv$. Divide by $uvR$: $\frac{1}{v} + \frac{1}{u} = \frac{2}{R}$. Since $f = R/2$, $2/R = 1/f$.

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

This formula is derived using the New Cartesian sign convention where distances to the left of the pole are negative and distances to the right are positive. And $f=R/2$ relation also holds with signs (concave $f, R$ negative; convex $f, R$ positive).

Summary of Mirror Formula and Magnification:

These formulas, combined with the New Cartesian sign convention (origin at P, left is negative, right is positive, up is positive, down is negative, incident light direction usually from left to right), are used to solve problems involving image formation by spherical mirrors.

Example 1. An object is placed at a distance of 20 cm in front of a concave mirror of focal length 15 cm. Find the position, nature, and size of the image. (Take object height $h_o = 2$ cm).

Answer:

Mirror type: Concave. Focal length $f = 15$ cm. By convention, concave mirror's focus is to the left of the pole, so $f = -15$ cm in New Cartesian convention.

Object distance $u = 20$ cm. Object is in front of the mirror, to the left of the pole, so $u = -20$ cm.

Using the Mirror Formula: $ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} $

$ \frac{1}{-15} = \frac{1}{-20} + \frac{1}{v} $

$ \frac{1}{v} = \frac{1}{-15} - \frac{1}{-20} = -\frac{1}{15} + \frac{1}{20} $

$ \frac{1}{v} = \frac{-4}{60} + \frac{3}{60} = \frac{-4+3}{60} = \frac{-1}{60} $

$ v = -60 $ cm.

Position: The image is formed at a distance of 60 cm from the pole. Since $v$ is negative, the image is formed on the same side as the object, i.e., 60 cm in front of the mirror.

Nature: Since the image is formed in front of the mirror ($v$ is negative in this convention, corresponding to a real image), it is a real image. Real images formed by mirrors are always inverted.

Size (Magnification): Calculate the magnification $m = -\frac{v}{u}$.

$ m = -\frac{(-60 \text{ cm})}{(-20 \text{ cm})} = -\frac{60}{20} = -3 $.

The magnification is -3. The negative sign confirms that the image is inverted. The magnitude $|m| = 3 > 1$ indicates that the image is magnified (3 times larger than the object).

The height of the image is $h_i = m \times h_o = -3 \times (2 \text{ cm}) = -6$ cm. The height is 6 cm, and the negative sign shows it is inverted.

Summary: The image is formed 60 cm in front of the mirror, is real, inverted, and magnified (3 times the size of the object).

(Note: Object is between F (-15 cm) and C (-30 cm). For concave mirror, object between F and C forms a real, inverted, magnified image beyond C. Our result $v = -60$ cm is beyond C=-30 cm and in front of the mirror, which is consistent with the ray diagram.)

Example 2. An object is placed at a distance of 30 cm in front of a convex mirror of focal length 20 cm. Find the position and nature of the image.

Answer:

Mirror type: Convex. Focal length $f = 20$ cm. By convention, convex mirror's focus is behind the mirror, so $f = +20$ cm in New Cartesian convention.

Object distance $u = 30$ cm. Object is in front of the mirror, to the left of the pole, so $u = -30$ cm.

Using the Mirror Formula: $ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} $

$ \frac{1}{20} = \frac{1}{-30} + \frac{1}{v} $

$ \frac{1}{v} = \frac{1}{20} - \frac{1}{-30} = \frac{1}{20} + \frac{1}{30} $

$ \frac{1}{v} = \frac{3}{60} + \frac{2}{60} = \frac{3+2}{60} = \frac{5}{60} = \frac{1}{12} $

$ v = 12 $ cm.

Position: The image is formed at a distance of 12 cm from the pole. Since $v$ is positive, the image is formed on the opposite side of the object, i.e., 12 cm behind the mirror.

Nature: Since the image is formed behind the mirror ($v$ is positive in this convention, corresponding to a virtual image), it is a virtual image. Virtual images formed by mirrors are always erect.

Size (Magnification): Calculate the magnification $m = -\frac{v}{u}$.

$ m = -\frac{(12 \text{ cm})}{(-30 \text{ cm})} = \frac{12}{30} = \frac{2}{5} = 0.4 $.

The magnification is 0.4. The positive sign confirms that the image is erect. The magnitude $|m| = 0.4 < 1$ indicates that the image is diminished.

Summary: The image is formed 12 cm behind the mirror, is virtual, erect, and diminished.

(Note: Convex mirrors always form virtual, erect, diminished images behind the mirror, between P and F. Here F is at +20 cm, image is at +12 cm, which is indeed behind the mirror and between P and F. This is consistent.)