Board Previous Year Questions

How to Use This Section

CBSE Board-pattern questions on Wave Optics by mark value, with examiner-rewarded model answers — wavefront and Huygens definitions, the wavefront derivations of Snell's and reflection laws, interference conditions, the Young's fringe-width derivation, the interference-vs-diffraction contrast, and Malus' law. Concept-dense and high-yield: expect a definition, a derivation and a numerical each year.

1-Mark Questions (Definitions & Direct)

Q1. Define a wavefront. Answer: A wavefront is a surface of constant phase — the locus of all points of a wave oscillating in phase. Energy travels perpendicular to it.

Q2. State Huygens' principle. Answer: Every point on a wavefront is a source of secondary wavelets spreading with the wave's speed; the new wavefront at a later time is the forward envelope (common tangent) of these wavelets.

Q3. What is the effect on the wavelength of light when it enters a denser medium? Answer: The wavelength decreases (to $\lambda/n$); the frequency stays the same and the speed decreases.

Q4. Write the condition for the dark fringe in Young's double-slit experiment. Answer: Path difference $= (n+\tfrac12)\lambda$, i.e. $x = (n+\tfrac12)\frac{\lambda D}{d}$.

Q5. On what does the fringe width in Young's experiment depend? Answer: $\beta = \lambda D/d$ — directly on wavelength and screen distance, inversely on slit separation.

Q6. State Malus' law. Answer: When polarised light of intensity $I_0$ passes through a polaroid whose axis makes angle $\theta$ with the polarisation, the transmitted intensity is $I = I_0\cos^2\theta$.

2-Mark Questions (Short Answer)

Q7. Using Huygens' principle, why is the frequency of light unchanged on refraction while the wavelength changes? Answer: The wavefronts in the two media stay continuous at the interface (a crest meeting the boundary produces a crest beyond it), so the number of wavefronts arriving per second equals the number leaving per second — the frequency is conserved. Since $v = \nu\lambda$ and the speed changes by 1/n, the wavelength must change by the same factor while $\nu$ holds fixed.

Q8. State two conditions for sustained (observable) interference of light. Answer: (i) The two sources must be coherent — same frequency and a constant phase difference. (ii) They should have equal (or comparable) amplitudes and the same plane of polarisation, so the bright/dark contrast is good. (Practically, both are derived from a single source, as in Young's experiment.)

Q9. Distinguish between interference and diffraction. Answer: Interference arises from the superposition of waves from a few discrete coherent sources (e.g. two slits), giving equally spaced fringes of equal brightness; diffraction arises from the superposition of secondary wavelets from a continuous wavefront (e.g. one slit), giving a broad central maximum with weaker, unequal secondary maxima.

Q10. Two independent sodium lamps cannot produce interference fringes. Why? Answer: Independent sources emit light in random bursts with abrupt phase changes (~$10^{-10}$ s), so the phase difference between them is not constant — they are incoherent, and the intensities simply add with no stable fringe pattern.

Q11. Unpolarised light of intensity $I_0$ passes through two polaroids whose axes are at 60 degrees. Find the transmitted intensity. Answer: After the first polaroid: $I_0/2$. After the second (Malus): $\frac{I_0}{2}\cos^2 60^{\circ} = \frac{I_0}{2}\times\frac14 = \frac{I_0}{8}$.

3-Mark Questions (Derivations & Numericals)

Q12. Using Huygens' principle, derive Snell's law of refraction. Answer: A plane wavefront AB meets the interface at angle i. While B travels to C in time $\tau$ ($BC = v_1\tau$), the wavelet from A grows to $AE = v_2\tau$ in medium 2; CE is the refracted wavefront. From triangles ABC and AEC sharing AC: $\sin i = BC/AC = v_1\tau/AC$, $\sin r = AE/AC = v_2\tau/AC$. Dividing, $\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{n_2}{n_1}$, i.e. $n_1\sin i = n_2\sin r$.

Q13. In Young's experiment, derive the expression for fringe width. Answer: For slits separated by d, screen at distance D, a point at distance x has path difference $\Delta = xd/D$. Bright fringes: $\Delta = n\lambda \Rightarrow x_n = n\lambda D/d$. The spacing between consecutive bright fringes is $\beta = x_{n+1} - x_n = \lambda D/d$. (Dark fringes are equally spaced and interleave the bright ones.)

Q14. In a double-slit experiment, d = 1.0 mm, D = 1.0 m and $\lambda$ = 500 nm. Find the fringe width and the distance of the 3rd bright fringe from the centre. Answer: $\beta = \lambda D/d = (5\times10^{-7})(1)/(10^{-3}) = 5\times10^{-4}$ m = 0.5 mm. The 3rd bright fringe: $x_3 = 3\beta = 1.5$ mm.

Q15. Light of 600 nm falls on a single slit of width 0.1 mm. Find the angular width of the central maximum. Answer: First minima at $\sin\theta = \lambda/a = (6\times10^{-7})/(10^{-4}) = 6\times10^{-3}$ rad. Angular width of the central maximum $= 2\lambda/a = 1.2\times10^{-2}$ rad.

5-Mark Questions (Long Answer)

Q16. (a) State Huygens' principle and use it to verify the law of reflection. (b) Draw the reflected wavefront for a plane wave incident on a concave mirror. (c) What is the relation between the angle of incidence and reflection? Answer:

1. (a) Each point of a wavefront is a source of secondary wavelets advancing at the wave's speed; the new wavefront is their forward envelope. For reflection, a plane wave AB hits the surface; while B reaches C ($BC = v\tau$), the wavelet from A grows to $AE = v\tau$ in the same medium. Triangles BAC and EAC share AC, have equal legs ($AE = BC$) and are right-angled — hence congruent, so the wavefront (and ray) angles are equal.
2. (b) A plane wave reflecting off a concave mirror emerges as a spherical wavefront converging to the focus F.
3. (c) The angle of incidence equals the angle of reflection ($i = r$).

Q17. (a) Describe Young's double-slit experiment and derive the fringe-width formula. (b) How does the pattern change if (i) the whole apparatus is immersed in water, (ii) the slit separation is increased, (iii) white light is used? Answer:

1. (a) A single source S illuminates two close, coherent slits S1, S2; their overlapping waves produce equally spaced bright/dark fringes. Path difference $\Delta = xd/D$; bright at $\Delta = n\lambda$ gives $x_n = n\lambda D/d$, so $\beta = \lambda D/d$.
2. (b)(i) In water $\lambda \to \lambda/n$, so $\beta \to \beta/n$ — fringes narrow. (ii) Larger d means smaller $\beta$ — fringes crowd together. (iii) With white light the central fringe is white (zero path difference for all colours); the neighbouring fringes are coloured (each wavelength's fringe falls at a slightly different place), and they soon overlap into white beyond a few orders.