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CBSE Class 12 Physics · 11 questions · 26 marks
Ray optics explains reflection and refraction but goes silent on why light spreads into fringes at a narrow slit or why two beams can cancel each other into darkness. Treating light as a wave, following Huygens' construction, answers both. This chapter builds up to Young's double-slit experiment, the single clearest classroom proof that light truly interferes like a wave.
In Young's double-slit experiment, if the distance between the slits is doubled while all other quantities remain unchanged, the fringe width will:
Answer
Halve — fringe width β = λD/d is inversely proportional to the slit separation d. Doubling d halves β, packing the fringes closer together on the screen while λ and D stay fixed.
Two coherent sources of equal intensity I₀ superpose at a point where the path difference is λ/2. The resultant intensity at that point is:
Answer
Zero — a path difference of λ/2 corresponds to a phase difference of π (destructive interference). Using I = I₁ + I₂ + 2√(I₁I₂) cosφ with I₁ = I₂ = I₀ and cosπ = -1: I = I₀ + I₀ - 2I₀ = 0. The waves arrive exactly out of step and cancel completely.
Unpolarised light of intensity I₀ is incident on a polaroid, and the transmitted light then passes through a second polaroid whose axis is at 60° to the first. The final intensity is:
Answer
I₀/8 — the first polaroid always halves the intensity of unpolarised light, giving I₀/2 regardless of its own orientation. The second polaroid then applies Malus's law: I = (I₀/2) cos²60° = (I₀/2) × (0.5)² = (I₀/2) × 0.25 = I₀/8.
In a single-slit diffraction experiment, if the slit width is decreased, the width of the central maximum:
Answer
Increases — the first minimum occurs at sinθ = λ/a, so a narrower slit (smaller a) pushes the first minima out to larger angles θ, spreading the central maximum wider. This inverse relationship is why diffraction effects become more noticeable as an opening is made narrower relative to the wavelength.
Assertion (A): Two independent sodium lamps cannot produce a sustained interference pattern on a screen. Reason (R): The light emitted by two separate sources has a rapidly and randomly changing phase relationship, so any interference pattern shifts faster than the eye or a detector can register.
Answer
Both A and R are true and R is the correct explanation of A — independent sources emit light in uncorrelated bursts from countless atoms, so the phase difference between them fluctuates randomly on a timescale of nanoseconds. The pattern of bright and dark fringes shifts too fast to observe, averaging out to uniform illumination, which is why coherent sources derived from a single source (as in Young's experiment) are essential.
In a Young's double-slit experiment, slits separated by 0.30 mm are illuminated by light of wavelength 600 nm, and the screen is 1.5 m away. Calculate the fringe width.
Answer
Fringe width, β = λD/d. λ = 600×10⁻⁹ m, D = 1.5 m, d = 0.30×10⁻³ m. β = (600×10⁻⁹ × 1.5)/(0.30×10⁻³) β = 9.0×10⁻⁷/3.0×10⁻⁴ β = 3.0×10⁻³ m = 3.0 mm.
State Huygens' principle and use it to explain, with the help of the construction of wavefronts, why the speed of light is smaller in a denser medium such as glass compared with air (a qualitative explanation is expected, not a full derivation).
Answer
Huygens' principle states that every point on an existing wavefront becomes the source of secondary spherical wavelets that spread out with the speed of the wave in that medium, and the new wavefront at any later instant is the surface tangent to all these secondary wavelets (their envelope). When a plane wavefront in air meets a denser medium like glass obliquely, the part of the wavefront that enters the glass first starts producing secondary wavelets that travel more slowly than the part still in air, since the wave speed is lower in a denser medium. Because the wavelets in glass expand more slowly while those still in air keep expanding at the higher air-speed, the new envelope (wavefront) inside the glass is tilted compared to the incident wavefront, bending the ray toward the normal. This same geometric argument, applied with Snell's law, shows that n = c/v, so a higher refractive index directly corresponds to a lower speed of light in that medium.
Light of wavelength 500 nm falls normally on a single slit of width 0.50 mm. Find the angular position (in terms of sinθ) of the first minimum, and the linear distance of that minimum from the centre on a screen 1.0 m away.
Answer
For the first minimum in single-slit diffraction, sinθ = λ/a. sinθ = 500×10⁻⁹/0.50×10⁻³ = 1.0×10⁻³. For such a small angle, sinθ ≈ θ (in radians) ≈ 1.0×10⁻³ rad. Distance from the centre, y = Dθ = 1.0 × 1.0×10⁻³ = 1.0×10⁻³ m = 1.0 mm.
Describe Young's double-slit experiment and derive the expression for the fringe width of the interference pattern observed on the screen. Two slits 1.0 mm apart are illuminated by monochromatic light and the fringe width observed on a screen 1.0 m away is 0.60 mm. Find the wavelength of light used.
Answer
Description: two narrow, closely spaced slits S₁ and S₂, illuminated by a single monochromatic source (or a common source through a narrow slit before them so as to be coherent), act as two coherent sources. Light spreading out from them overlaps on a distant screen, producing alternating bright and dark fringes. Derivation: Let the slits be separated by distance d, and the screen be at distance D from the slits (D >> d). Consider a point P on the screen at distance y from the centre O (the point equidistant from both slits). Path difference between the two waves reaching P: Δ = S₂P - S₁P ≈ yd/D, using the small-angle approximation valid since D >> d and y is small compared to D (derived by drawing the perpendicular from S₁ to S₂P and treating the triangle formed). For a bright fringe (constructive interference): Δ = nλ, so yd/D = nλ, giving y_n = nλD/d. For the next bright fringe, y_(n+1) = (n+1)λD/d. Fringe width, β = y_(n+1) - y_n = λD/d. This is the same for every consecutive pair of bright (or dark) fringes, so the fringes are equally spaced. Numerical: β = 0.60×10⁻³ m, D = 1.0 m, d = 1.0×10⁻³ m. λ = βd/D = (0.60×10⁻³ × 1.0×10⁻³)/1.0 = 6.0×10⁻⁷ m = 600 nm.
(a) Explain what is meant by plane polarised light and describe how a polaroid produces it from unpolarised light. Derive Malus's law. (b) Unpolarised light of intensity 100 W/m² is incident on a polaroid. It then passes through a second polaroid whose axis makes 30° with the first. Find the intensity of light emerging from the second polaroid.
Answer
(a) In unpolarised light, the electric field vibrates randomly in all directions perpendicular to the direction of propagation, changing orientation many times per second. Plane polarised light has its electric field vibrating in one fixed plane containing the direction of propagation. A polaroid contains long-chain molecules aligned in one direction; it absorbs the component of the electric field parallel to these chains and transmits only the component perpendicular to them (along its transmission axis). Passed through a single polaroid, unpolarised light emerges plane polarised along that axis, with half its original intensity, since on average half the random field components lie along the absorption direction. Derivation of Malus's law: let plane polarised light of amplitude E₀ (intensity I₀ ∝ E₀²) fall on a second polaroid (analyser) whose transmission axis makes angle θ with the direction of E₀. Only the component of E₀ along the analyser's axis is transmitted: E = E₀ cosθ. Since intensity is proportional to the square of amplitude: I = I₀ cos²θ, which is Malus's law: I is maximum (= I₀) when θ = 0° and zero when θ = 90°. (b) First polaroid: I₁ = 100/2 = 50 W/m² (unpolarised light is always halved). Second polaroid at θ = 30°: I₂ = I₁ cos²30° = 50 × (0.866)² = 50 × 0.75 = 37.5 W/m².
Read the passage and answer the questions that follow: A student sets up Young's double-slit experiment using a laser of wavelength 632.8 nm (a common He-Ne laser line), with slits separated by 0.20 mm and a screen placed 2.0 m away. She records the position of the fringes and then repeats the experiment after covering one slit, observing that the sharp fringe pattern disappears and only a smooth diffraction-like glow remains. (a) Calculate the fringe width observed in the original two-slit setup. (b) Find the distance of the 4th bright fringe from the central maximum. (c) Explain, in one or two sentences, why covering one slit destroys the interference pattern. (d) If the student replaces the laser with white light, describe what she would observe at the centre and away from the centre of the pattern.
Answer
(a) β = λD/d = (632.8×10⁻⁹ × 2.0)/(0.20×10⁻³) = 1.2656×10⁻⁶/2.0×10⁻⁴ = 6.328×10⁻³ m ≈ 6.3 mm. (b) Position of the nth bright fringe, y_n = nλD/d = nβ. y₄ = 4 × 6.328×10⁻³ = 2.53×10⁻² m ≈ 25.3 mm from the centre. (c) Interference requires two coherent sources whose waves overlap and combine; with only one slit open there is just a single source of light, so there is nothing for it to interfere with, and only that slit's own single-slit diffraction pattern (a broad central glow) remains. (d) With white light, the central fringe (zero path difference for every wavelength) remains white, since all colours reinforce there together. Moving away from the centre, different wavelengths satisfy the bright-fringe condition at different positions (since y_n = nλD/d depends on λ), so the fringes spread out into overlapping coloured bands, and beyond the first few fringes the pattern becomes a blur of overlapping colours rather than distinct fringes.
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