Wave Optics

A wavefront is the locus of all points in a medium that are in the same phase. Types include spherical (from a point source), cylindrical (from a linear source), and plane (at very large distances). Huygens' principle states that every point on a wavefront acts as a secondary source of wavelets; the new wavefront is the forward envelope of these wavelets. This principle elegantly derives Snell's law: when a plane wavefront hits a boundary, the change in wave speed causes the wavefront to tilt, producing refraction.
The ratio sin $\theta_{1}$/sin $\theta_{2}$ = $\frac{v_{1}}{v_{2}}$ = $\frac{n_{2}}{n_{1}}$.

Young's double slit experiment (YDSE) demonstrates interference of coherent light. Two slits $S_{1}$ and $S_{2}$, separated by distance d, illuminate a screen at distance D (where D >> d).
The path difference at a point P at distance y from the center is $\Delta$ = d sin $\theta$ approximately equal to dy/D for small angles.

Bright fringes (constructive interference) occur when $\Delta$ = n $\lambda$, giving positions $y_{n}$ = n $\lambda$ D/d (n = 0, 1, 2, ...).
Dark fringes (destructive interference) occur when $\Delta$ = (2n - 1) $\frac{\lambda}{2}$, giving positions $y_{n}$ = (2n - 1) $\lambda$ D/(2d) (n = 1, 2, ...).
The fringe width $\beta$ = $\lambda$ D/d; [$\beta$] = [L], SI unit: m. The central fringe is always bright, and all fringes (except possibly the central one with white light) have equal width. Increasing D or $\lambda$ widens the fringes; increasing d narrows them.

Intensity distribution: For identical slits, I = $4I_{0} \; cos^{2}$($\frac{\phi}{2}$), where $\phi$ = (2 $\frac{\pi}{\lambda}$) $\Delta$ is the phase difference.
$I_{max}$ = $4I_{0}$ (constructive, $\phi$ = 0, $2\pi$, ...), $I_{min}$ = 0 (destructive, $\phi$ = $\pi$, $3\pi$, ...).
For unequal intensities: I = $I_{1}$ + $I_{2}$ + 2 $\sqrt{{I}_{1} {I}_{2}}$ cos $\phi$.

Coherent sources are essential for sustained interference — they must have the same frequency and a constant phase difference. Two independent sources (e.g., two light bulbs) are NOT coherent because their phases fluctuate randomly.

Single slit diffraction produces a central maximum of width 2 $\lambda$ D/a (where a is the slit width), which is the brightest and widest feature.
Secondary minima occur at a sin $\theta$ = n $\lambda$ (n = 1, 2, ...), and secondary maxima at a sin $\theta$ = (2n + 1) $\frac{\lambda}{2}$. Each secondary maximum has width $\lambda$ D/a — half the central maximum width. Intensity drops rapidly for higher-order maxima. Resolving power improves with larger aperture and shorter wavelength (Rayleigh's criterion).

Polarization proves that light is a transverse wave.
Brewster's law: When light hits a surface at the Brewster angle $\theta_{p}$, the reflected light is completely plane-polarized; tan $\theta_{p}$ = n (refractive index).
At Brewster's angle, reflected and refracted rays are perpendicular: $\theta_{p}$ + $\theta_{r}$ = 90 deg.
Malus's law: When polarized light of intensity $I_{0}$ passes through an analyzer at angle $\theta$ to the polarization direction, the transmitted intensity is I = $I_{0} \; cos^{2} \; \theta$.
Through crossed polaroids ($\theta$ = 90 deg), I = 0.
Inserting a third polaroid at 45 deg between two crossed polaroids allows light through: I = $\frac{I_{0}}{8}$.

The key testable concept is the YDSE fringe width formula $\beta$ = $\lambda$ D/d and its variations (effect of changing medium, slit separation, or screen distance), which is among the most frequently asked formulae in NEET Physics.

Solved Numericals

N₁. In YDSE, slit separation d = 0.5 mm, screen distance D = 1 m, wavelength $\lambda$ = 600 nm. Find fringe width, position of the 3rd bright fringe, and position of the 2nd dark fringe.

Given: d = 0.5 mm = 5 x $10^{-4}$ m, D = 1 m, $\lambda$ = 600 nm = 6 x $10^{-7}$ m.

Fringe width: $\beta$ = $\lambda$ D/d = (6 x $10^{-7}$ m x 1 m) / (5 x $10^{-4}$ m) = 6 x $10^{-7}$ / 5 x $10^{-4}$ m = 1.2 x $10^{-3}$ m = 1.2 mm.

3rd bright fringe (n = 3): $y_{3}$ = n $\lambda$ D/d = 3 x $\beta$ = 3 x 1.2 mm = 3.6 mm from center.

2nd dark fringe (n = 2): $y_{2}$ = (2n - 1) $\lambda$ D/(2d) = (2 x 2 - 1) x $\frac{\beta}{2}$ = 3 x 0.6 mm = 1.8 mm. Alternatively: $y_{2}$ = (2 x 2 - 1) x 600 x $10^{-9}$ x 1 / (2 x 5 x $10^{-4}$) = 3 x 6 x $10^{-7}$ / ($10^{-3}$) = 1.8 x $10^{-3}$ m = 1.8 mm.

Note: The 2nd dark fringe lies between the 1st and 2nd bright fringes (at 1.5 $\beta$ from center). The nth dark fringe position = (n - 0.5) $\beta$.

N₂. Unpolarized light of intensity $I_{0}$ passes through three polaroids. The first and third are crossed (90 deg apart). The second is at 30 deg to the first. Find the intensity after each polaroid.

After Polaroid 1 (polarizer): Unpolarized light becomes plane-polarized. By definition, the transmitted intensity is $I_{1}$ = $\frac{I_{0}}{2}$ (half the unpolarized intensity is transmitted).

After Polaroid 2 (at 30 deg to Polaroid 1): Apply Malus's law with $\theta$ = 30 deg: $I_{2}$ = $I_{1} \; cos^{2}$(30 deg) = ($\frac{I_{0}}{2}$)($\frac{\sqrt{3}}{2}$)² = ($\frac{I_{0}}{2}$)($\frac{3}{4}$) = $\frac{3I_{0}}{8}$.

After Polaroid 3 (at 90 deg to Polaroid 1, so at 90 - 30 = 60 deg to Polaroid 2): Apply Malus's law with $\theta$ = 60 deg: $I_{3}$ = $I_{2} \; cos^{2}$(60 deg) = ($\frac{3I_{0}}{8}$)($\frac{1}{2}$)² = ($\frac{3I_{0}}{8}$)($\frac{1}{4}$) = $\frac{3I_{0}}{32}$.

Final intensity: I = $\frac{3I_{0}}{32}$.

If Polaroid 2 were at 45 deg instead: $I_{3}$ = ($\frac{I_{0}}{2}$)($cos^{2}$ 45)($cos^{2}$ 45) = ($\frac{I_{0}}{2}$)($\frac{1}{2}$)($\frac{1}{2}$) = $\frac{I_{0}}{8}$.

Key insight: Without the middle polaroid, no light passes through crossed polaroids. Inserting a polaroid between them at any angle other than 0 deg or 90 deg allows some light through.

N₃. Find Brewster's angle for a glass surface with refractive index n = 1.732. Verify that reflected and refracted rays are perpendicular.

Brewster's law: tan $\theta_{p}$ = n = 1.732. $\theta_{p}$ = arctan(1.732) = 60 deg.

Verification: At Brewster's angle, the refracted angle $\theta_{r}$ satisfies: By Snell's law: sin $\theta_{p}$ = n sin $\theta_{r}$ sin 60 deg = 1.732 x sin $\theta_{r}$ 0.866 = 1.732 x sin $\theta_{r}$ sin $\theta_{r}$ = 0.$\frac{866}{1}$.732 = 0.5 $\theta_{r}$ = 30 deg.

Check perpendicularity: $\theta_{p}$ + $\theta_{r}$ = 60 + 30 = 90 deg. Verified.

The reflected and refracted rays are perpendicular at the Brewster angle. The reflected light is completely plane-polarized with the electric field vector perpendicular to the plane of incidence.