🌟 Introduction to Optics

Optics is the branch of physics that deals with the study of light — its behavior, properties, and interactions with matter. Our vision, one of our most vital senses, is entirely dependent on the light that enters our eyes.

In physics, the study of light is divided into two broad branches:

• Ray Optics (Geometrical Optics): Treats light as rays that travel in straight lines. It explains phenomena such as reflection, refraction, and image formation using mirrors and lenses. (This is the focus of this chapter.)

• Wave Optics (Physical Optics): Considers the wave nature of light and is used to explain interference, diffraction, and polarization. (This will be covered in the next chapter.)

🔦 The Concept of a Light Ray

The fundamental assumption in ray optics is rectilinear propagation — light travels in straight lines.

• A ray is a straight line that indicates the direction of light's propagation.

• A beam is a collection or bundle of rays.

Ray optics is valid when the dimensions of the interacting objects are much larger than the wavelength of light.

📚 Phenomena Studied in Ray Optics

In this chapter, we will explore:

• Reflection: The bouncing back of light into the same medium after striking a surface.

• Refraction: The bending of light when it passes from one transparent medium to another.

These fundamental principles help us understand devices like the human eye, microscopes, and telescopes.

🔁 Reflection of Light

Reflection refers to the return of light into the same medium after hitting a surface.

🧾 Laws of Reflection

1. The angle of incidence is equal to the angle of reflection: $ \angle i = \angle r $

2. The incident ray, the reflected ray, and the normal to the surface all lie in the same plane.

🪞 Reflection by a Plane Mirror

Key characteristics of an image formed by a plane mirror:

• Virtual and erect

• Same size as the object

• Formed as far behind the mirror as the object is in front

• Laterally inverted

🪩 Reflection by Spherical Mirrors

A spherical mirror has a reflecting surface that is part of a sphere.

Types of Spherical Mirrors

• Concave Mirror: Reflecting surface is curved inwards (converging mirror).

• Convex Mirror: Reflecting surface is curved outwards (diverging mirror).

📌 Terminology for Spherical Mirrors

• Pole (P): Center of the mirror’s surface.

• Center of Curvature (C): Center of the sphere from which the mirror is cut.

• Radius of Curvature (R): Distance between pole and center of curvature ($ R = PC $).

• Principal Axis: Straight line passing through $ P $ and $ C $.

• Aperture: Diameter of the reflecting surface.

• Principal Focus (F):

  • Concave mirror: Rays parallel to the principal axis converge at $ F $ (real focus).

  • Convex mirror: Rays appear to diverge from $ F $ (virtual focus).

• Focal Length (f): Distance between $ P $ and $ F $ ($ f = PF $)

🔗 Relationship between Focal Length and Radius of Curvature

For a mirror with a small aperture:
$ f = \dfrac{R}{2} $

📏 Sign Convention (New Cartesian Convention)

Used to apply formulas for spherical mirrors:

• The pole (P) is the origin.

• The principal axis is the x-axis.

• The object is always placed to the left of the mirror.

• Distances measured:

  • Along the direction of incident light (left to right): positive (+)

  • Against the direction (right to left): negative (–)

• Heights:

  • Measured upward from principal axis: positive (+)

  • Measured downward: negative (–)

✅ Applying the Convention

📐 Derivation: $f = \dfrac{R}{2}$

Setup:

• Consider a concave mirror with a paraxial ray $AB$ parallel to the principal axis, striking the mirror at $B$ and reflecting through the focus $F$.

• Line $CB$ is the normal since $C$ is the center of curvature.

Geometry:

1. By law of reflection: $ \angle ABC = \angle CBF $

2. $ \angle ABC = \angle BCF $ (alternate interior angles)

3. Hence, $ \angle CBF = \angle BCF $, making triangle $ \triangle BCF $ isosceles

4. Thus, $ CF = FB $

5. If aperture is small, $ FB \approx FP $

6. So, $ CF \approx FP $

Now,

$PC=CF+FP$ $ \approx FP+FP$ $=2\times FP.PC $ $= CF + FP \approx FP + FP $

$=2\times FP.PC = CF + FP \approx FP + FP =2 \times FP$

Since $ PC = R $ and $ FP = f $,
we get:

$R=2f $ Or $ f = \dfrac{R}{2} $​

Mirror Formula

The mirror formula gives the relationship between the object distance $u$, image distance $v$, and focal length $f$ of a spherical mirror:

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

This formula is valid for both concave and convex mirrors. All distances are measured from the pole of the mirror, and the New Cartesian Sign Convention is used.

📐Derivation of Mirror Formula (for Concave Mirror)

Let an object $AB$ be placed in front of a concave mirror.
Let:

$u$ = object distance (distance of object from pole, always negative)

$v$ = image distance (distance of image from pole)

$f$ = focal length of the mirror (for concave mirror, $f < 0$)

$P$ = pole of the mirror

$F$ = focus of the mirror

$C$ = center of curvature of the mirror

We consider two rays:

A ray parallel to the principal axis reflects through the focus $F$

A ray passing through the center of curvature $C$ reflects back on the same path

These rays intersect at the image point $A'$. The image is formed at a distance $v$ from the pole.

Now, from geometry:

Triangles $ \triangle ABP $ and $ \triangle A'B'P $ are similar. So,

$ \frac{A'B'}{AB} = \frac{v}{u} \quad \text{(1)} $

Also, triangles $ \triangle A'B'F $ and $ \triangle CB'F $ are similar. So,

$ \frac{A'B'}{CB'} = \frac{v - f}{f} \quad \text{(2)} $

From the ray diagram, $CB' = AB$. So equation (2) becomes:

$ \frac{A'B'}{AB} = \frac{v - f}{f} \quad \text{(3)} $

Equating equation (1) and (3):

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

Cross-multiplying:

$ vf = u(v - f) $

Expanding the right-hand side:

$ vf = uv - uf $

Bringing all terms to one side:

$ uv - vf - uf = 0 $

Dividing every term by $uvf$:

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

This is the required mirror formula:

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

Example 1
Q. An object is placed at a distance of $20\ \text{cm}$ in front of a concave mirror of focal length $-10\ \text{cm}$. Find the position of the image and its nature.

Given:
$ u = -20\ \text{cm} $
$ f = -10\ \text{cm} $
$ v = ? $

Using mirror formula:
$ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} $

Substitute values:

$ \frac{1}{v} + \frac{1}{-20} = \frac{1}{-10} $

$ \frac{1}{v} - \frac{1}{20} = -\frac{1}{10} $

Take LCM and solve:

$ \frac{1}{v} = -\frac{1}{10} + \frac{1}{20} = -\frac{2 - 1}{20} = -\frac{1}{20} $

So,

$ v = -20\ \text{cm} $

Interpretation:
The image is formed $20\ \text{cm}$ in front of the mirror

Since $v$ is negative, the image is real and inverted

Example 2
Q. An object is placed $15\ \text{cm}$ in front of a convex mirror. Its focal length is $+10\ \text{cm}$. Find the position of the image and its nature.

Given:
$ u = -15\ \text{cm} $
$ f = +10\ \text{cm} $
$ v = ? $

Using mirror formula:
$ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} $

Substitute values:

$ \frac{1}{v} + \frac{1}{-15} = \frac{1}{10} $

$ \frac{1}{v} - \frac{1}{15} = \frac{1}{10} $

Solve:

$ \frac{1}{v} = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6} $

So,

$ v = 6\ \text{cm} $

Interpretation:
The image is formed $6\ \text{cm}$ behind the mirror

Since $v$ is positive, the image is virtual and erect.

Magnification by Spherical Mirrors

Magnification is the ratio of the height of the image to the height of the object. It tells us how much larger or smaller the image is compared to the object.

Formula:
$ m = \frac{h'}{h} = \frac{-v}{u} $

Where:

$m$ = magnification

$h$ = height of the object

$h'$ = height of the image

$v$ = image distance from the mirror

$u$ = object distance from the mirror (always negative)

The negative sign in the formula accounts for the inversion of real images.

Interpretation of $m$:
If $m > 0$, the image is virtual and erect

If $m < 0$, the image is real and inverted

If $|m| > 1$, the image is magnified

If $|m| < 1$, the image is diminished

If $|m| = 1$, the image is same size as the object

Example 3
Q. An object is placed $20\ \text{cm}$ in front of a concave mirror of focal length $-10\ \text{cm}$. Find the magnification.

Given:
$ u = -20\ \text{cm} $
$ f = -10\ \text{cm} $
Use mirror formula:

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

Substitute:

$ \frac{1}{v} + \frac{1}{-20} = \frac{1}{-10} $

$ \frac{1}{v} = -\frac{1}{10} + \frac{1}{20} = -\frac{1}{20} $

So,

$ v = -20\ \text{cm} $

Now use magnification formula:

$ m = \frac{-v}{u} = \frac{-(-20)}{-20} = \frac{20}{-20} = -1 $

Answer:
Magnification is $-1$

The image is real, inverted, and of the same size as the object

Example 4
Q. An object is placed $15\ \text{cm}$ in front of a convex mirror of focal length $+10\ \text{cm}$. Find the magnification.

Given:
$ u = -15\ \text{cm} $
$ f = +10\ \text{cm} $

Use mirror formula:

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

$ \frac{1}{v} + \frac{1}{-15} = \frac{1}{10} $

$ \frac{1}{v} = \frac{1}{10} + \frac{1}{15} = \frac{5}{30} = \frac{1}{6} $

So,

$ v = 6\ \text{cm} $

Now magnification:

$ m = \frac{-v}{u} = \frac{-6}{-15} = \frac{2}{5} = 0.4 $

Answer:
Magnification is $+0.4$

The image is virtual, erect, and diminished