# Reflection at Spherical Surface , Mirror Formula

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