### Reflection of Light at a Plane Surface:

We know that a ray of light is composed of packets of energy known as photons. The photons have the ability of colliding with any surface. Thus the photons transfer momentum & energy in the same way as the transfer of momentum & energy take place between any two particles during collision.

For elastic collision a particle gets rebounded with same speed where as the direction of its motion changes due to the impact (impulsive) force offered by the other colliding particle.

Similarly when a photon is incident on a fine plane surface, it gets rebounded. This is known as Reflection.

### What are the laws of reflection ?

There are two laws governing the physical phenomenon of reflection.

They are-

(i) The incident ray, normal to the interface and the reflected ray lie on the same plane.

(ii) The angle of incidence is equal to the angle of reflection. ∠i = ∠r

(iii) We can observe from the figure that two rays coming from a point object O are reflected at the points M & N. If the reflected rays are produced back they meet at a point P. This point P is known as image.

Since the image is formed at the back of the mirror, it is known as virtual image .

Geometrically we can see that object distance is equal to image distance from mirror surface p = q

(As ΔOMP & ΔONP are isosceles triangles)

__Properties of image formed by plane mirror__

Properties of image formed by plane mirror

(a) the image is virtual erect and laterally reversed.

(b) the image size = object size

(c) object distance = image distance

(d) when the plane mirror is rotated through an angle θ,

the reflected ray is rotated through an angle 2θ.

(e) the numbers of images formed by the combination of two mirrors inclined at an angle θ,

Number of image formed ,

(i) $ \displaystyle N = (\frac{360}{\theta}- 1) $ ; If 360/θ is an even integer

(i) $ \displaystyle N = (\frac{360}{\theta}) $ ; If 360/θ is an odd integer

e.g. When two plane mirrors are placed parallel to each other , then θ = 0

⇒ N = ∞

Therefore infinite number of images are formed.

Example : If an object moves towards a plane mirror with a speed V at an angle θ to the perpendicular to the plane of the mirror, find the relative velocity between the object & the image.

Solution :

Velocity of the object relative to the image is given as

$ \displaystyle \vec{V_{OI}} = \vec{V_O} – \vec{V_I} $

$ \displaystyle \vec{V_{OI}} = (Vcos\theta\hat{i}-Vsin\theta\hat{j}) – (- Vcos\theta\hat{i} – Vsin\theta\hat{j}) $

$ \displaystyle \vec{V_{OI}} = 2Vcos\theta\hat{i} $

$ \displaystyle V_{OI} = 2Vcos\theta $

Exercise : Prove that if the plane mirror rotates through an angle θ, the reflected ray for any incident ray is deviated through an angle 2θ.

Example : Find the minimum length of the mirror required to visualise the complete image of the body of a person of height H.

Solution: The person must have to obtain a reflected ray one from his head & another reflected ray from the feet.

Geometrically, AC = AB + BC where

$ \displaystyle AB = \frac{x}{2} , BC = \frac{H-x}{2} $

$ \displaystyle AC = \frac{x}{2} + \frac{H-x}{2} $

$ \displaystyle AC = \frac{H}{2}$

i.e. length of the mirror is just half the height of person.

Illustration: Show that if a mirror is rotated by angle θ the reflected ray rotates through an angle 2θ When the mirror rotates by θ , the angle between the normals is also equal to θ.

Solution :

Incidence angle for M’ = i + θ

Angle of reflection for M’ = i + θ

i.e. ∠N_{2}OR_{2} = i + θ

∠N_{1}OR_{2} = ∠N_{1}ON_{2} + ∠N_{2}OR_{2}

= θ + i + θ = i + 2 θ

or ∠R_{1}OR_{2} = ∠N_{2}OR_{2} -∠N_{2}OR_{1}

= i + 2 θ – i = 2 θ

Hence the reflected ray rotates by an angle 2θ

Exercise : Find the number of images formed when two plane mirrors are placed at right angle to each other & locate the images by drawing ray diagram.

Exercise : Draw the ray diagram for the images formed by an object placed in between two plane parallel mirrors due to both the reflecting surfaces

The reflection takes place in such a way that,

angle of incidence = angle of reflection

i = r

F → principal focus; where the parallel light rays meet (concave) or seems to be coming from (convex) on the optical axis

C → Center of curvature; center of the sphere, a part of which is the given mirror.

P → Pole; Geometrical center of the spherical reflecting surface (mirror)

f → Focal length; the distance between the focal point F & the pole P.

### Cartesian Sign Convention

(i) All distances are measured from the pole of the mirror / lens / refracting surface which should be placed at the origin.

(ii) Distances measured along the direction of incident rays are taken as positive.

(iii) Distances measured along a direction opposite to the incident rays are taken as negative.

(iv) Distances above the principal axis are positive.

(v) Distances below the principal axis are negative.

(vi) Angles when measured from the normal, in anti-clockwise direction are positive, while in clockwise direction are negative.

(vii)The focal length of the concave mirror / lens is negative & that of convex mirror / lens is positive; focal length of plane mirror is infinity, because its radius of curvature is infinitely large.

## How can the formation of images be visualized ?

Images can be visualised by drawing ray diagram. Some of the rays to be used to draw a ray diagram are listed:

(i) A ray parallel to the principal axis passes through the principal focus (or appears to diverge from the principal focus) after reflection.

(ii) A ray passing through the principal focus becomes parallel to the principal axis after reflection.

(iii) A ray passing through the centre of curvature of the mirror retraces its own path.

(iv) A ray incident at the pole is reflected with equal inclination to the principal axis.

## Mirror formula :

Consider the shown figure where O is a point object and I is corresponding image.

CB is normal to the mirror at B.

By laws of reflection, ∠OBC = ∠CBI = θ

α + θ = γ , γ + θ = β

⇒ α + β = 2 γ

For small aperture of the mirror, α , β , γ → 0

⇒ α ≈ tan α , β ≈ tan β , γ ≈ tan γ & M → P (for paraxial rays)

⇒ tan α + tan β = 2 tan γ

$ \displaystyle \frac{BM}{MO} + \frac{BM}{MI} = 2\frac{BM}{MC}$

$ \displaystyle \frac{1}{MO} + \frac{1}{MI} = \frac{2}{MC}$

As M is close to P

$ \displaystyle \frac{1}{PO} + \frac{1}{PI} = \frac{2}{PC}$

Applying Sign convention,

$ \displaystyle -\frac{1}{u} + \frac{1}{-v} = -\frac{2}{R}$

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

u = ∞

$ \displaystyle \frac{1}{v} = \frac{2}{R}$

If u = ∞ , v = f

Hence , f = R/2 and

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

**Alternate Method: **

As triangle ABC & A’B’C are similar

$\displaystyle \frac{AB}{A’B’} = \frac{CB}{CB’}$ …(i)

As triangle ABP & A’B’P are similar

$\displaystyle \frac{AB}{A’B’} = \frac{PB}{PB’}$ …(ii)

From (i) & (ii)

$\displaystyle \frac{PB}{PB’} = \frac{CB}{CB’}$

$\displaystyle \frac{PB}{PB’} = \frac{PB – PC}{PC -PB’}$

$\displaystyle \frac{-u}{-v} = \frac{-u-(-R)}{-R -(-v)}$

$\displaystyle \frac{u}{v} = \frac{-u+R}{-R +v)}$

-uv + v R = -u R + u v

u R + v R = 2 u v

Dividing by uvR ,

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

$ \displaystyle \frac{1}{v} + \frac{1}{u} = \frac{2}{2 f} $

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

__Power of mirror__

Power of mirror $ \displaystyle P = -\frac{1}{f} $

Where f should be taken in meter with proper sign

i.e. −ve for concave and +ve for convex.

Question: A concave mirror form on a screen a real image of twice the linear dimension of the object. The object and the screen are then moved until the image is 3 times the size of the object. If the shift of the screen in 25 cm, determine the shift of the object and the focal length of the mirror.

Solution:

We have $ \displaystyle \frac{1}{v} + \frac{1}{u} = \frac{1}{f} $ ….(1)

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

From the given data:-

$ \displaystyle 2 = \frac{v}{f}-1 $

$ \displaystyle 3 = \frac{v+25}{f}-1 $

Solving the above equations for v and f, we get

V = 75cms. and f = 25cms.

From equation (1) we can write

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

Substituting the data, we get

Before shifting: $ \displaystyle u = \frac{25\times 75}{75-25} = 37.5 cm $

After shifting: $ \displaystyle u = \frac{25\times 100}{100-25} = 33.3 cm $

Shift of the object = 33.3 – 37.5 = – 4.2 cm