Chapter 10: Wave Optics
1. Introduction: The Nature of Light
In the previous chapter (Ray Optics), we treated light as a stream of particles that travel in straight lines. However, this "Corpuscular Theory" (proposed by Newton) failed to explain phenomena like Interference, Diffraction, and Polarization.
To explain these, Christiaan Huygens (in 1678) proposed the Wave Theory of Light. According to this theory, light travels in the form of waves. This chapter is dedicated to understanding light as a wave.
1.1 The Concept of Wavefront
This is the most fundamental concept of Wave Optics. Imagine dropping a stone in a calm pond. Circular ripples spread out. Every point on a particular ripple oscillates together.
Types of Wavefronts:
- Spherical Wavefront: Produced by a point source of light (like a bulb). The energy spreads out in spheres.
- Cylindrical Wavefront: Produced by a linear source (like a tube light or a slit).
- Plane Wavefront: When a spherical or cylindrical wavefront travels a large distance, a small portion of it appears flat. (e.g., Light from the Sun).
2. Huygens' Principle
Huygens gave a geometric construction to determine the shape and position of a wavefront at any future time if its present position is known.
The Two Postulates
1. Secondary Wavelets: Every point on a given wavefront acts as a fresh source of new disturbance, called secondary wavelets. These wavelets travel in all directions with the speed of light in that medium.
2. New Wavefront: A surface touching these secondary wavelets tangentially in the forward direction at any instant gives the new wavefront at that instant.
3. Proof of Laws using Wave Theory
This section is extremely important for Board Exams (often asked as a 3-5 mark derivation).
A. Refraction of a Plane Wave (Deriving Snell's Law)
Setup: Consider a plane wavefront $AB$ incident on a surface $PP'$ separating medium 1 (velocity $v_1$) and medium 2 (velocity $v_2$). Let $v_1 > v_2$ (Rarer to Denser).
Step-by-Step Proof:
1. The wavefront strikes the surface at $A$ at $t=0$. Point $B$ is yet to hit the surface.
2. Let $t$ be the time taken by the disturbance to travel from $B$ to $C$.
$\Rightarrow BC = v_1 \tau$
3. In the same time $\tau$, the secondary wavelet from $A$ travels into medium 2. It covers a distance $AD = v_2 \tau$.
4. Geometry: In $\Delta ABC$ (Incident Triangle):
$\sin i = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{v_1 \tau}{AC}$
5. In $\Delta ADC$ (Refracted Triangle):
$\sin r = \frac{AD}{AC} = \frac{v_2 \tau}{AC}$
6. Dividing the two equations:
$\frac{\sin i}{\sin r} = \frac{v_1 \tau / AC}{v_2 \tau / AC} = \frac{v_1}{v_2}$
By definition, refractive index $n = c/v$, so $\frac{v_1}{v_2} = \frac{n_2}{n_1}$.
$\therefore \frac{\sin i}{\sin r} = \frac{n_2}{n_1}$ (Snell's Law Verified)
B. Reflection of a Plane Wave
Setup: Consider a plane wavefront $AB$ incident on a reflecting surface $MN$.
Step-by-Step Proof:
1. Incident wavefront $AB$ touches surface at $A$.
2. Time taken for $B$ to reach $C$ is $\tau$. So, $BC = v\tau$.
3. In same time, wavelet from $A$ reflects and forms a sphere of radius $AD = v\tau$.
4. Triangle Congruency: Consider $\Delta ABC$ and $\Delta ADC$.
$\quad$ • Side $AC$ is common (Hypotenuse).
$\quad$ • Side $BC = AD = v\tau$.
$\quad$ • Angle $\angle B = \angle D = 90^\circ$.
5. By RHS rule, $\Delta ABC \cong \Delta ADC$.
6. Therefore, corresponding angles are equal: $\angle BAC = \angle DCA$.
$\therefore \angle i = \angle r$ (Law of Reflection Verified)
Chapter 10: Wave Optics
1. Coherent & Incoherent Addition
Superposition Principle: When two or more waves overlap, the resultant displacement at any point is the vector sum of the individual displacements ($y = y_1 + y_2$).
Why don't two bulbs show interference?
To observe a stable interference pattern (bright and dark bands that don't flicker), the phase difference between the two sources must remain constant with time.
Coherent Sources: Sources that emit light of the same frequency and have a constant phase difference.
Incoherent Sources: Sources where phase difference changes randomly (like two independent bulbs). No interference is seen; intensities just add up ($I = I_1 + I_2$).
2. Theory of Interference
Let two waves be $y_1 = a \cos \omega t$ and $y_2 = a \cos(\omega t + \phi)$.
Resultant Amplitude ($A$) depends on phase difference ($\phi$):
$$I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$$
2.1 Constructive Interference (Maxima)
When crest meets crest, intensity is maximum.
- Phase Difference ($\phi$): $0, 2\pi, 4\pi \dots = 2n\pi$
- Path Difference ($\Delta x$): $0, \lambda, 2\lambda \dots = n\lambda$
- Resultant Intensity: $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 \approx 4I_0$
2.2 Destructive Interference (Minima)
When crest meets trough, intensity is minimum (Darkness).
- Phase Difference ($\phi$): $\pi, 3\pi, 5\pi \dots = (2n-1)\pi$
- Path Difference ($\Delta x$): $\lambda/2, 3\lambda/2 \dots = (2n-1)\frac{\lambda}{2}$
- Resultant Intensity: $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 \approx 0$
3. Young's Double Slit Experiment (YDSE)
Thomas Young (1801) demonstrated the wave nature of light using this historic experiment.
The Setup:
• Monochromatic light source ($S$).
• Two narrow slits ($S_1, S_2$) separated by distance $d$.
• A screen placed at distance $D$.
• Light from $S_1$ and $S_2$ overlaps to form alternate Bright and Dark bands called Fringes.
3.1 Position of Fringes (Formulas)
Let $x$ be the distance of a fringe from the central bright spot.
For Bright Fringes (Maxima):
Path difference $\frac{xd}{D} = n\lambda$
$$x_n = \frac{n \lambda D}{d}$$
($n=0$ is Central Bright, $n=1$ is First Bright...)
For Dark Fringes (Minima):
Path difference $\frac{xd}{D} = (2n-1)\frac{\lambda}{2}$
$$x_n = (2n-1) \frac{\lambda D}{2d}$$
3.2 Fringe Width ($\beta$)
The distance between two consecutive bright or two consecutive dark fringes.
Key Observations:
• $\beta \propto \lambda$: Red light fringes are wider than Violet.
• $\beta \propto 1/d$: If slits are brought closer, fringes become wider.
• If apparatus is dipped in water, $\lambda$ decreases, so Fringe Width decreases.
Chapter 10: Wave Optics
1. Diffraction of Light
Definition: The phenomenon of bending of light around the sharp corners of an obstacle or aperture and spreading into the region of the geometrical shadow is called Diffraction.
($a \approx \lambda$)
1.1 The Single Slit Experiment
When monochromatic light passes through a single narrow slit of width '$a$', a diffraction pattern is observed on the screen.
[attachment_0](attachment)Features of the Pattern:
1. Central Maximum: A broad, bright central band. It is widest and brightest.
2. Secondary Minima (Dark Fringes): Occur where path difference is integral multiples of $\lambda$.
3. Secondary Maxima (Bright Fringes): Occur between dark fringes but their intensity decreases rapidly.
Mathematical Conditions:
1. Condition for Minima (Dark):
$$a \sin \theta = n\lambda$$
Where $n = 1, 2, 3 \dots$
2. Condition for Secondary Maxima (Bright):
$$a \sin \theta = (n + \frac{1}{2})\lambda$$
3. Width of Central Maximum ($W$):
The angular width is $2\theta = 2\lambda/a$.
The linear width on screen (at distance D):
$$W = \frac{2\lambda D}{a}$$
⚠️ Crucial Comparison: Interference vs Diffraction
| Interference (YDSE) | Diffraction (Single Slit) |
|---|---|
| Caused by superposition of two separate wavefronts from two coherent sources. | Caused by superposition of wavelets from different parts of the same wavefront. |
| All bright fringes are of equal intensity. | Intensity of secondary maxima decreases rapidly. Central max is brightest. |
| Fringe width is generally constant ($\beta = \lambda D/d$). | Central fringe is twice as wide as secondary fringes. |
2. Polarization of Light
Light waves are Transverse electromagnetic waves. The electric field vector oscillates perpendicular to the direction of propagation.
Unpolarized Light: Vectors vibrate in all possible planes perpendicular to propagation (e.g., Sunlight, Bulb).
Polarized Light: Vibrations are restricted to a single plane.
2.1 Malus' Law
When an unpolarized beam of intensity $I_0$ passes through a Polarizer, intensity becomes $I_0/2$.
When this polarized light passes through an Analyzer whose axis makes an angle $\theta$ with the Polarizer:
(Where $I_0$ is intensity of polarized light incident on analyzer)
Chapter 10: Wave Optics
Section A: Multiple Choice Questions (1 Mark)
Q1. Which of the following properties shows that light is a transverse wave?
(a) Reflection (b) Interference (c) Diffraction (d) Polarization
Ans: (d) Polarization
Q2. In Young’s double slit experiment, if the separation between slits is halved and the distance of the screen is doubled, the fringe width will:
(a) Remain same (b) Become 4 times (c) Become half (d) Become double
Ans: (b) Become 4 times
Reason: $\beta = \frac{\lambda D}{d}$. If $D \to 2D$ and $d \to d/2$, then $\beta' = \frac{\lambda (2D)}{(d/2)} = 4 \frac{\lambda D}{d}$.
Q3. The phase difference between two points on the same wavefront is:
(a) $2\pi$ (b) $\pi$ (c) Zero (d) $\pi/2$
Ans: (c) Zero
Section B: Very Short Answer (1 Mark)
Q4. Define Wavefront.
Ans: It is the continuous locus of all particles of a medium vibrating in the same phase at any instant.
Q5. State the condition for constructive interference in terms of path difference.
Ans: Path difference $\Delta x = n\lambda$, where $n = 0, 1, 2 \dots$
Q6. Two independent light sources cannot produce interference. Why?
Ans: Because they are Incoherent. Their phase difference changes randomly with time, averaging the intensity to sum ($I_1+I_2$) rather than interfering.
Section C: Numericals (3 Marks)
Q7. In a YDSE, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is 1.2 cm. Determine the wavelength of light used.
Solution:
Given: $d = 0.28 \text{ mm} = 2.8 \times 10^{-4} \text{ m}$
$D = 1.4 \text{ m}$
$x_4 = 1.2 \text{ cm} = 1.2 \times 10^{-2} \text{ m}$
$n = 4$
Formula: $x_n = \frac{n \lambda D}{d}$
$\lambda = \frac{x_n d}{n D} = \frac{1.2 \times 10^{-2} \times 2.8 \times 10^{-4}}{4 \times 1.4}$
$\lambda = \frac{3.36 \times 10^{-6}}{5.6} = 0.6 \times 10^{-6} \text{ m}$
Ans: $\lambda = 600 \text{ nm}$
Q8. Two Polaroids are placed 90° to each other. A third Polaroid is placed between them at 45° to the first. If intensity of unpolarized light is $I_0$, find final intensity.
Solution:
1. After 1st Polaroid: $I_1 = I_0/2$.
2. After 2nd Polaroid (at $45^\circ$): $I_2 = I_1 \cos^2(45^\circ) = (I_0/2)(1/\sqrt{2})^2 = I_0/4$.
3. After 3rd Polaroid (at $45^\circ$ to 2nd): $I_3 = I_2 \cos^2(45^\circ) = (I_0/4)(1/2) = I_0/8$.
Ans: $I_0/8$
Section D: Long Answer / Derivations (5 Marks)
Q9. Define wavefront. Use Huygens' principle to verify the laws of refraction.
(Refer to Vol 1 of these notes for the complete step-by-step diagrammatic proof).
Q10. Describe Young's Double Slit Experiment to produce interference pattern due to monochromatic light. Deduce the expression for the fringe width.
(Refer to Vol 2 for theory and formula derivation).
Q11. Draw the intensity pattern for single slit diffraction. Obtain the condition for the first minimum. How does the width of central maximum change if: (i) slit width is decreased, (ii) distance between slit and screen is increased?
Ans Hint: Width $W = 2\lambda D / a$.
(i) If '$a$' decreases, Width increases.
(ii) If '$D$' increases, Width increases.
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