Class 12 — Wave Optics

Chapter Overview & Weightage

Wave Optics in Class 12 Physics carries 4–6 marks every year and is one of the most predictable scoring chapters. The questions revolve around Young’s double-slit experiment, single-slit diffraction, and Huygens’ principle.

CBSE Class 12 Board — Wave Optics Weightage

Year	Marks	Question Type
2024	5	5-mark long: YDSE derivation + numerical
2023	4	3-mark numerical + 1-mark MCQ
2022	5	5-mark: single-slit diffraction graph + width
2021	6	1+5: polarisation + Huygens construction
2020	4	4-mark: fringe-shift due to glass plate

Roughly 6% of the Physics paper. Combined with Ray Optics (also high-weightage), the optics unit alone gives ~10 marks.

Key Concepts You Must Know

Ranked by board frequency:

Huygens’ principle — every point on a wavefront is a source of secondary wavelets. Used to derive laws of reflection, refraction.
Young’s Double-Slit Experiment (YDSE) — formation of interference, fringe width $\beta = \lambda D/d$ , conditions for max/min.
Conditions for sustained interference — coherent sources, equal intensities, monochromatic, narrow slits.
Single-slit diffraction — central maximum, minima at $a\sin\theta = n\lambda$ .
Polarisation — Malus’s law $I = I_0 \cos^2\theta$ , polarisation by reflection, Brewster’s angle $\tan i_B = \mu$ .
Difference between interference and diffraction — comparison table is asked almost every year.

Important Formulas

YDSE fringe width: $\beta = \dfrac{\lambda D}{d}$ .

Position of bright fringe: $y_n = \dfrac{n\lambda D}{d}$ .

Position of dark fringe: $y_n = \dfrac{(2n-1)\lambda D}{2d}$ .

Path difference for max: $\Delta = n\lambda$ .

Path difference for min: $\Delta = (n + \tfrac{1}{2})\lambda$ .

Single-slit minima: $a\sin\theta = n\lambda$ ( $n = \pm1, \pm2, \ldots$ ).

Width of central diffraction max: $W = 2\lambda D/a$ (twice the spacing of side fringes).

Brewster’s law: $\tan i_B = n$ . At this angle, reflected light is fully polarised.

Malus’s law: $I = I_0 \cos^2\theta$ .

Solved Previous Year Questions

PYQ 1 — Fringe width (CBSE 2023, 3 marks)

Q. In a YDSE setup, slits are 1 mm apart, screen 1 m away, light wavelength 600 nm. Find the fringe width.

Solution. $\beta = \lambda D/d = (6 \times 10^{-7})(1)/(10^{-3}) = 6 \times 10^{-4}$ m $= 0.6$ mm. Always convert wavelength to metres before plugging in.

PYQ 2 — Single-slit width (CBSE 2022, 3 marks)

Q. Light of wavelength 500 nm passes through a slit of width 0.2 mm and falls on a screen 1 m away. Find the width of the central maximum.

Solution.

W = \frac{2\lambda D}{a} = \frac{2 \times 5 \times 10^{-7} \times 1}{2 \times 10^{-4}} = 5 \times 10^{-3} \text{ m} = 5 \text{ mm}

PYQ 3 — Interference vs Diffraction (CBSE 2024, 4 marks)

Q. State three differences between interference and diffraction.

Solution.

Interference	Diffraction
From two (or more) coherent sources	From a single source via slit/edge
Fringes of equal width	Central max wider; side fringes diminish
All maxima have ~same intensity	Central max much brighter than side maxima

Difficulty Distribution

Easy (40%): Fringe-width plug-in, identifying conditions for interference, Malus’s law numerical.
Medium (45%): YDSE derivation, single-slit diffraction graph, Brewster’s angle.
Hard (15%): Glass-plate fringe shift, intensity ratio in YDSE, YDSE with mixed wavelengths.

Expert Strategy

Toppers’ approach:

Master the YDSE derivation. Path difference $= d\sin\theta \approx d \cdot y/D$ gives $\beta = \lambda D/d$ in 4 lines. This is a guaranteed 3 marks.
Remember the central-max-is-double rule. Width of central diffraction max = $2\lambda D/a$ . Side maxima are half that wide.
Polarisation = easy 2 marks. Memorise Brewster’s law and Malus’s law. Practise 5 numericals.
Diagrams matter. YDSE setup, single-slit pattern with intensity profile, Huygens’ construction — neat labelled diagrams earn 1 mark each.

Common Traps

Trap 1 — Confusing $a$ (slit width) with $d$ (slit separation). YDSE uses $d$ , single-slit uses $a$ . Mix them and the answer is off by orders of magnitude.

Trap 2 — Forgetting $(2n-1)$ for dark fringes. Bright: $y_n = n\beta$ . Dark: $y_n = (2n-1)\beta/2$ . Some students confuse the formulas.

Trap 3 — Brewster angle requires $\tan$ , not $\sin$ . $\tan i_B = n$ . Many write $\sin i_B = n$ which is total internal reflection — different concept.

Trap 4 — Polariser pairs at 90° give zero, not minimum. $I = I_0\cos^2 90° = 0$ , exactly zero in the ideal case.