Mirror Formula

The Mirror Formula is an important principle in optics that explains how light behaves when it interacts with spherical mirrors, whether concave or convex. It connects three main factors: the object distance (u), image distance (v), and the focal length (f). This formula is key to figuring out where an image will form, how big it will be, and whether it will be real or virtual. It plays a major role in the design of optical devices like telescopes, microscopes, and cameras, helping us understand and predict how light reflects off mirrors.

1.0Reflection from Spherical Surface(Spherical Mirror)

• A spherical mirror is a segment of a spherical surface. If reflection occurs on the inner surface, it is concave, while reflection from the outer surface makes it convex.    

2.0Terms Related to Spherical Mirror    

C = Centre of curvature, R = Radius of curvature ,P = Pole (geometric centre of mirror)

1. Pole: The point on the mirror's reflecting surface, typically taken as the central point (P) of the mirror.
2. Principal Section: A section of the mirror, such as MM', passing through the pole.
3. Centre of Curvature: The center (C) of the sphere from which the mirror is a part.
4. Radius of Curvature: The radius (R) of the sphere to which the mirror belongs.
5. Principal Axis: The line (CP) connecting the pole to the center of curvature.

3.0Paraxial And Marginal Rays

Paraxial Rays: Rays that make a very small angle with the normal at the point of incidence are called paraxial rays.

Marginal Rays: Those rays which make a large angle with normal at a point of incidence are called marginal rays.

4.0Focal Length of Spherical Mirror

1. When a parallel beam of light strikes a concave mirror, the paraxial rays converge at a point F on the principal axis.
2. When the same beam strikes a convex mirror, the paraxial rays appear to diverge from a point F on the principal axis.
3. The spot F is called the principal focus of the mirror.    

$CNF\sim NFM$

$CN=NM=\frac{R}{2}$

$\cos i=\frac{CN}{CF}=\frac{\left(\frac{R}{2}\right)}{CF}$

$CF=\frac{R}{2\cos i}$

$FP=CP-CF=R-\frac{R}{2\cos i}$

If i is very small i.e. rays are paraxial, then $\cos i\approx 1$

$FP=R-\frac{R}{2}\Rightarrow FP=\frac{R}{2}=f$(Here, f is called focal length of mirror).

Conclusion:           

For marginal rays,

$FP=R-\frac{R}{2\cos i}$

If rays are paraxial then,

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

• When paraxial light rays parallel to the principal axis are incident on a mirror then the point where they meet (concave mirror) or appear to meet (convex mirror) after reflection is known as focus (F).

5.0Focal Plane

• When a parallel paraxial light beam strikes the mirror at an angle, the reflected rays converge (or seem to diverge) from a point in a plane through F, perpendicular to the principal axis. This is called the focal plane.
• If the parallel paraxial beam of light strikes the mirror at an angle to a plane passing over the focus and perpendicular to the principal axis.      

$\tan \alpha =\frac{h}{FP}$ ,Rays are paraxial so,

$\tan \alpha \approx \alpha$

$\alpha =\frac{h}{f}$

$h=f\alpha$

6.0Sign Convention For Spherical Mirrors

1. All separation is calculated from the pole of mirror or the optical center of the lens, with the pole being the origin.
2. Distances along the principal axis are measured from the pole (taken as the origin).
3. Distances in the direction of incident light are considered positive, while those in the opposite direction are negative.
4. Distances above the principal axis are considered positive, while distances below the axis are negative.
5. Incident light is assumed to travel from left to right whenever possible.                 

7.0Mirror Formula Derivation

• It is a mathematical relationship between object distance(u),image distance(v) and the focal length (f) of a spherical mirror.

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

$\frac{1}{ObjectDistance}+\frac{1}{ImageDistance}=\frac{1}{FocalLength}$

By using Sign Conventions

BP=-u,B'P=-v ,FP=-f,CP=-R=-2f

$△A^{\prime }{B}^{\prime }C\sim △ABC$

$\frac{A^{\prime }{B}^{\prime }}{AB}=\frac{C{B}^{\prime }}{BC}=\frac{CP-BP}{BP-CP}=\frac{-R+v}{-u+R}$………….(1)

As $\angle A^{\prime }{P}^{\prime }{B}^{\prime }=\angle APB$ therefore

$△A^{\prime }{A}^{\prime }{P}^{\prime }\sim △ABP$ consequently

$\frac{A^{\prime }{B}^{\prime }}{AB}=\frac{B{P}^{\prime }}{BP}=\frac{-v}{-u}=\frac{v}{u}$………………….(2)

From equation(1) and (2)

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

-uR+uv=-uv+vR

vR+uR=2uv

Dividing both sides by uvR, we get

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

R=2f

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

8.0Transverse or Lateral Magnification

• If a one dimensional object is placed perpendicular to the principal axis then ratio of image height and object height is called transverse or lateral magnification.

$m_t=\frac{h_t}{h_o}$

• If $m_t$ is positive means erect image is formed

• If $m_t$ is negative means inverted image is formed                   

$|m_t|>1$
$|m_t|<1$
$|m_t|=1$

Derivation of Transverse Magnification         

$\frac{h_t}{h_o}=\frac{v}{u}$

Putting the values with sign

$u\to -u,v\to -v,h_t\to -h_t,h_o\to +h_o$

$\frac{-h_t}{h_o}=\frac{-v}{-u}$

$\frac{h_t}{h_o}=\frac{v}{u}$              

$m_t=\frac{h_t}{h_o}=-\frac{v}{u}$

Example-1.

Find out position of an object placed in front of concave mirror of focal length 30cm so that 2 times magnified image is formed.

Solution:

Case-1.

m=-2

$-2=\frac{f}{f-u}$

$-2=\frac{-30}{-30-u}$

60+2u=-30

2u=-90u=-45 cm

Case-2

m=+2

$+2=\frac{f}{f-u}$

$-2=\frac{-30}{-30-u}$

-60-2u=-30

2u=-30u=-15 cm

9.0Longitudinal Magnification

• If an object is placed along the principal axis then the ratio of length of image and length of object is called longitudinal or axial magnification.

$m_L=\frac{\text{Length of Image}}{\text{Length of Object}}=\frac{I_1}{I_0}$

$m_L=\left|\frac{\text{Length of Image}}{\text{Length of Object}}\right|=\left|\frac{v-v_1}{u-u_1}\right|$

Longitudinal magnification for small objects (Lo << f):

object length=du, Image length=dv

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

Differentiating w.r.t u

$\frac{1}{v^2}\frac{dv}{du}-\frac{1}{u^2}=0\Rightarrow \frac{dv}{du}=-\frac{v^2}{u^2}$

If we use only magnitude then,

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

$M_L=\frac{I_1}{I_0}=\frac{v^2}{u^2}=m_t^2$

Example-2

A point object is placed at a separation of 15 cm from an inward curved mirror of radius of curvature 20 cm. If the object is made to oscillate along the principal axis with amplitude 2 mm. Find out the amplitude of its image.

Solution:

Amplitude of object du=2 mm

$m=\frac{f}{f-u}=\frac{-10}{-10-(-15)}=-2$

Amplitude of image

$|dv|=m^2|du|=(-2{)}^2\times 2=8\text{ mm}$

10.0Superficial Magnification

• When a two-dimensional object is positioned with its plane perpendicular to the principal axis, the resulting magnification is called superficial magnification.

$m_s=\frac{A_i}{A_0}$

$m_s=\frac{A_i}{A_0}$

$A_O=w_o{h}_{o}and{A}_i=w_i{h}_i$

$m_t=\frac{h_i}{h_0}=\frac{w_i}{w_0}$

$\frac{A_i}{A_0}=\frac{w_i{h}_i}{w_0{h}_0}=\frac{w_i}{w_0}$

$\frac{A_i}{A_0}=m_t^2=m_s$

Example-3.

A square with a side of 4 mm is positioned 30 cm away from a Inward-curved mirror having focal length of 10 cm. The center of the square lies along the mirror's axis, and the square's plane is perpendicular to this axis. Calculate the area of the image formed by the mirror.

Solution:

$m_t=\frac{f}{f-u}=\frac{-10}{-10-(-30)}=-\frac{1}{2}$

$\frac{A_i}{A_0}=m_t^2\Rightarrow \frac{A_i}{(4\text{ mm}{)}^2}={\left(-\frac{1}{2}\right)}^2\Rightarrow A_i=4{\text{ mm}}^2$

11.0Solved Examples

1.Find out position and nature of image?

Solution:

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

$\frac{1}{v}+\frac{1}{-25}=\frac{1}{-20}\Rightarrow v=-100\text{ cm, Real Image}$