**Reflection at Spherical Surface :**

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.

**Solved Example : ** 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

### Linear Magnification :

Linear magnification of a Spherical mirror is the ratio of the size of the image to the size of object .

$\displaystyle m = \frac{Size \;of \;image (h_2) }{size \;of \; object (h_1)} = \frac{A’B’}{AB}$

Magnification (m) will be +ve when h_{2} is +ve i.e image is erect and m will be -ve when h_{2} is -ve i.e. image is inverted . (Assuming h_{1} is always +ve)

In new cartesian sign convention, we define magnification in such a way that a negative sign (of m) implies inverted image and vice-versa. A real image is always inverted one and a virtual one is always erect.

Keeping these points in mind and that the real object and its real image would lie on the same sides in case of mirror and on opposite sides in case of lenses, we define m as in case of reflection by spherical mirror as :

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

Problem : An erect image is three times the size of object is obtained with a concave mirror of radius of curvature 36 cm . What is the position of the object ?

Solution : Since image is erect , m = + 3

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

$ \displaystyle 3 = -\frac{v}{u} $

v = – 3 u

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

$ \displaystyle \frac{1}{- 3 u} + \frac{1}{u} = \frac{1}{f} = \frac{2}{-36}$

$ \displaystyle -\frac{1}{3 u} + \frac{1}{u} = – \frac{1}{18}$

$ \displaystyle \frac{-1 + 3}{3 u} = – \frac{1}{18}$

$ \displaystyle \frac{2}{3 u} = – \frac{1}{18}$

3 u = -36

u = -12 cm

**Solved Example:** A short linear object of length L lies on the axis of a spherical mirror of focal length f at a distance b from the mirror. Find the size of the image ?

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

By differentiating both sides

$ \displaystyle \frac{dv}{v^2} = – \frac{du}{u^2} $

du : size of object = L

dv : size of image

u : object distance

v : image distance.

$ \displaystyle dv = – \frac{v^2}{u^2}(du) $

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

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

$ \displaystyle dv = – \frac{u^2 f^2}{(u-f)^2 u^2}du $

$ \displaystyle dv = -\frac{f^2}{(b-f)^2}L $

Negative sign implies that object is lying between u and u + du and the image will lie between v and v − dv

Solved Example : A point source S is placed midway between two converging mirrors having equal focal length f as shown in the figure. Find the values of d for which only one image is formed.

Solution : If the image is formed of the object, it must be formed at the center of curvature C.

since only one image is formed by two mirrors, the object should be placed at their common center of curvature as shown in the figure; the mirrors are the part of a sphere of center

⇒ 2f + 2f = d

⇒ d = 4f

Note : If the distance between the mirrors is 2f then also image will be found at the same position. So second value of d = 2f

Exercise : A concave mirror forms a real image on a screen of thrice the linear dimension of the object. Object and screen are moved until image is twice the size of the object. If the shift of the object is 6 cm, find the shift of the screen and the focal length of the mirror.