The study of light has historically transitioned between two competing frameworks: the corpuscular model and the wave model . In 1637, Descartes proposed the corpuscular model, which was later championed by Isaac Newton in his treatise OPTICKS . This model predicted that light travels faster in denser media. In 1678, Christiaan Huygens proposed the wave theory of light, which predicted the opposite: that the speed of light is slower in denser media. This was experimentally verified by Foucault in 1850, establishing the superiority of the wave model.
Historical Chronicle: Christiaan Huygens (1629–1695)
Historical Chronicle: Thomas Young (1773–1829)
| Model of Light | Key Proponents | Speed in Denser Medium | Primary Phenomena Explained |
|---|---|---|---|
| Corpuscular Model | Descartes, Newton | Greater than in vacuum/air ($v_{denser} > v_{rarer}$) | Rectilinear propagation, reflection, refraction |
| Wave Model | Huygens, Young, Fresnel, Maxwell | Less than in vacuum/air ($v_{denser} < v_{rarer}$) | Reflection, refraction, interference, diffraction, polarisation |
Huygens' principle is a geometrical construction used to determine the shape and position of a wavefront at a later time, given its shape at an initial instant.
A wavefront is defined as the locus of all points in a medium that oscillate in the same phase. It is a surface of constant phase.
Using Huygens' geometrical construction, we can rigorously derive the fundamental laws of reflection and refraction.
Let a plane wavefront $AB$ be incident at an angle $i$ on a plane interface separating medium 1 (speed $v_1$) and medium 2 (speed $v_2$). Let $\tau$ be the time taken by the wavefront to travel from $B$ to $C$. Thus, $BC = v_1 \tau$.
To find the refracted wavefront, we draw a sphere of radius $v_2 \tau$ from point $A$ in medium 2. The tangent plane $CE$ drawn from $C$ to this sphere represents the refracted wavefront, where $AE = v_2 \tau$.
From the geometry of the triangles $\triangle ABC$ and $\triangle AEC$:
$$\sin i = \frac{BC}{AC} = \frac{v_1 \tau}{AC}$$
$$\sin r = \frac{AE}{AC} = \frac{v_2 \tau}{AC}$$
Dividing these equations yields:
$$\frac{\sin i}{\sin r} = \frac{v_1}{v_2}$$
Defining the refractive indices $n_1 = c/v_1$ and $n_2 = c/v_2$, we obtain Snell's law of refraction :
$$n_1 \sin i = n_2 \sin r$$
During refraction, the wavelength changes such that $\frac{\lambda_1}{\lambda_2} = \frac{v_1}{v_2}$, but the frequency $\nu = \frac{v}{\lambda}$ remains constant.
Let a plane wavefront $AB$ be incident at an angle $i$ on a reflecting surface $MN$. Let $v$ be the speed of the wave. The time taken for the wavefront to advance from $B$ to $C$ is $\tau$, so $BC = v\tau$.
To construct the reflected wavefront, we draw a sphere of radius $v\tau$ from $A$. The tangent plane $CE$ from $C$ to this sphere represents the reflected wavefront, where $AE = v\tau$.
Since $BC = AE = v\tau$ and $AC$ is common, the right-angled triangles $\triangle EAC$ and $\triangle BAC$ are congruent. Therefore, the angle of incidence equals the angle of reflection:
$$i = r$$
Mathematical Problem Statement: Monochromatic light of wavelength 589 nm is incident from air onto a water surface of refractive index 1.33. Calculate the wavelength, frequency, and speed of the refracted light. (Speed of light in vacuum $c = 3.0 \times 10^8\text{ m s}^{-1}$)
Step-by-Step Algorithmic Solution Trace:
- Identify the given parameters: $\lambda_0 = 589\text{ nm} = 589 \times 10^{-9}\text{ m}$, $n = 1.33$, $c = 3.0 \times 10^8\text{ m s}^{-1}$.
- Calculate the frequency of light in air (which remains unchanged in water):
$$\nu = \frac{c}{\lambda_0} = \frac{3.0 \times 10^8}{589 \times 10^{-9}} \approx 5.09 \times 10^{14}\text{ Hz}$$- Calculate the speed of light in water:
$$v = \frac{c}{n} = \frac{3.0 \times 10^8}{1.33} \approx 2.26 \times 10^8\text{ m s}^{-1}$$- Calculate the wavelength of light in water:
$$\lambda = \frac{\lambda_0}{n} = \frac{589\text{ nm}}{1.33} \approx 443\text{ nm}$$Final Evaluation Output: Frequency: $5.09 \times 10^{14}\text{ Hz}$, Speed: $2.26 \times 10^8\text{ m s}^{-1}$, Wavelength: $443\text{ nm}$
The phenomenon of interference is based on the superposition principle , which states that the resultant displacement produced by multiple waves at a point is the vector sum of the individual displacements.
Consider two waves from sources $S_1$ and $S_2$ arriving at a point $P$ with displacements:
$$y_1 = a_1 \cos(\omega t)$$
$$y_2 = a_2 \cos(\omega t + \phi)$$
where $\phi$ is the constant phase difference between them. The resultant displacement is:
$$y = y_1 + y_2 = a_1 \cos(\omega t) + a_2 \cos(\omega t + \phi)$$
Using trigonometric identities, the amplitude $A$ of the resultant wave is given by:
$$A^2 = a_1^2 + a_2^2 + 2a_1 a_2 \cos\phi$$
Since intensity $I$ is proportional to the square of the amplitude ($I \propto A^2$), the resultant intensity is:
$$I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\phi$$
For identical sources where $I_1 = I_2 = I_0$:
$$I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$$
| Type of Interference | Phase Difference ($\phi$) | Path Difference ($\Delta$) | Resultant Intensity ($I$) |
|---|---|---|---|
| Constructive | $\phi = 2n\pi$ ($n = 0, 1, 2, \dots$) | $\Delta = n\lambda$ | $I_{max} = 4I_0$ |
| Destructive | $\phi = (2n+1)\pi$ ($n = 0, 1, 2, \dots$) | $\Delta = (n + \frac{1}{2})\lambda$ | $I_{min} = 0$ |
Thomas Young experimentally demonstrated the interference of light waves using two coherent pinholes $S_1$ and $S_2$ derived from a single primary source $S$.
Let $d$ be the distance between the two coherent sources $S_1$ and $S_2$, and $D$ be the distance from the slits to the screen. The path difference between the waves reaching an arbitrary point $P$ at a distance $x$ from the central maximum is:
$$\Delta = S_2P - S_1P \approx \frac{xd}{D}$$
Diffraction is the bending of light waves around the edges of obstacles or apertures into the region of geometrical shadow.
When a parallel beam of monochromatic light falls normally on a single narrow slit of width $a$, a diffraction pattern is formed on a screen placed at a distance $D$.
In both interference and diffraction, light energy is redistributed in accordance with the law of conservation of energy ; no energy is created or destroyed.
Polarisation is a phenomenon unique to transverse waves, where the vibrations of the wave are restricted to a single plane perpendicular to the direction of propagation.
When completely plane-polarised light of intensity $I_0$ is incident on an analyzer, the intensity $I$ of the light transmitted through the analyzer varies as the square of the cosine of the angle $\theta$ between the transmission axes of the polariser and the analyzer:
$$I = I_0 \cos^2\theta$$
Mathematical Problem Statement: Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids.
Step-by-Step Algorithmic Solution Trace:
- Let the first polariser be $P_1$ and the third be $P_3$. Since they are crossed, the angle between their pass-axes is $\pi/2$.
- Let the intermediate polaroid $P_2$ make an angle $\theta$ with the pass-axis of $P_1$.
- The intensity of light transmitted through $P_2$ is:
$$I_1 = I_0 \cos^2\theta$$- The angle between the pass-axes of $P_2$ and $P_3$ is $(\pi/2 - \theta)$. The intensity of light emerging from $P_3$ is:
$$I = I_1 \cos^2\left(\frac{\pi}{2} - \theta\right) = I_0 \cos^2\theta \sin^2\theta$$- Simplify using the double-angle identity $\sin(2\theta) = 2\sin\theta\cos\theta$:
$$I = \frac{I_0}{4} \sin^2(2\theta)$$- The transmitted intensity is maximum when $\sin^2(2\theta) = 1$, which occurs at $2\theta = \pi/2 \implies \theta = \pi/4$.
Final Evaluation Output: The transmitted intensity is given by $I = \frac{I_0}{4} \sin^2(2\theta)$, which reaches a maximum value of $\frac{I_0}{4}$ at $\theta = 45^\circ$.