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Samacheer Class 12 Physics - Wave Optics

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Samacheer Kalvi Class 12 Physics Wave Optics book back solutions with concise explanations and a verified validation footer on every answer.

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Multiple Choice Questions 10 Short Answer Questions 48 Long Answer Questions 17 Exercise Problems 10
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Multiple Choice QuestionsMultiple Choice Questions10 questions
Q.1 A plane glass is placed over various coloured letters (violet, green, yellow, red). The letter which appears to be raised more is,
Answer: D: violet

Apparent depth = real depth / n. For shorter wavelengths (violet) glass has larger refractive index (normal dispersion), so apparent depth is smaller and the letter appears raised more. Hence violet.

Q.2 Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm approximately. The maximum distance at which these dots can be resolved by the eye is, [take wavelength of light, λ = 500 nm]
Answer: B: 5 m

Resolution angle for circular pupil (Rayleigh criterion) θ = 1.22 λ / D = 1.22×500×10^{-9} / 3×10^{-3} ≈ 2.03×10^{-4} rad. Maximum distance L when separation s = 1 mm gives L = s/θ ≈ 1×10^{-3} / 2.03×10^{-4} ≈ 4.9 m ≈ 5 m.

Q.3 In a Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to,
Answer: A: 2D

Fringe spacing β = λD/d. If slit separation d is doubled, to keep β same, D must be doubled. So D → 2D.

Q.4 Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are [IIT-JEE 1988]
Answer: C: 9I and I

Amplitudes a1 = √I, a2 = √(4I) = 2√I. Maximum I_max = (a1+a2)^2 = (3√I)^2 = 9I. Minimum I_min = (2√I - √I)^2 = I.

Q.5 When light is incident on a soap film of thickness 5×10–5 cm, the wavelength of light reflected maximum in the visible region is 5320 Å. Refractive index of the film will be,
Answer: B: 1.33

Given film thickness t = 5×10^{-5} cm = 5×10^{-7} m and reflected constructive condition (film in air): 2 n t = (m+½) λ. For visible maximum λ = 5320 Å = 5.32×10^{-7} m. For m = 2, n = (2+½)λ/(2t) = 2.5×λ/(2t) =1.25×λ/t =1.25×(5.32×10^{-7}/5×10^{-7}) ≈1.33. Hence n ≈1.33.

Q.6 First diffraction minimum due to a single slit of width 1.0×10–5 cm is at 30°. Then wavelength of light used is,
Answer: B: 500 Å

Condition for first minimum: a sinθ = λ. a = 1.0×10^{-5} cm = 1.0×10^{-7} m. λ = a sin30° = 1.0×10^{-7} × 0.5 = 5.0×10^{-8} m = 500 Å.

Q.7 A ray of light strikes a glass plate at an angle 60°. If the reflected and refracted rays are perpendicular to each other, the refractive index of the glass is,
Answer: C: √3

If reflected and refracted rays are perpendicular, and angle of incidence i = 60°, angle of reflection = 60°, so refracted angle r = 90° - 60° = 30°. Using Snell's law n = sin i / sin r = sin60°/sin30° = (√3/2)/0.5 = √3 ≈1.732.

Q.8 One of the of Young’s double slits is covered with a glass plate as shown in figure. The position of central maximum will, (Glass slide above one slit)
Answer: A: get shifted downwards

A glass plate over one slit increases its optical path (introduces extra phase). The interference fringes shift towards the slit with larger optical path (covered slit). If the glass is over the upper slit in the figure, the central maximum shifts downwards (towards the covered slit).

Q.9 Light transmitted by Nicol prism is,
Answer: C: plane polarised

A Nicol prism transmits only one plane of vibration (the ordinary ray is absorbed or deviated out), so the transmitted light is plane polarised.

Q.10 The transverse nature of light is shown in,
Answer: D: polarisation

Polarisation demonstrates that light vibrations are transverse (vibrations occur in directions perpendicular to propagation).

Short Answer QuestionsShort Answer Questions48 questions
Q.1What are the salient features of corpuscular theory of light?v
Solution

Corpuscular theory (Newton): light as corpuscles with momentum, travels rectilinearly; reflection and refraction by forces at boundaries; supported rectilinear propagation and shadow formation but fails to explain interference, diffraction and polarization phenomena.

Answer:

Light consists of particles (corpuscles) emitted by sources; travels in straight lines; reflection and refraction explained by particle collisions and forces at interfaces; speed varies with medium (Newton predicted faster in optically denser medium); cannot explain interference and diffraction.

Q.2What are the important points of wave theory of light?v
Solution

Wave theory (Huygens, Young, Fresnel): light as wavefronts, obeys Huygens' principle, interference due to superposition, diffraction explained by wave nature; successfully explains interference and diffraction but classical wave (scalar) needed extension to electromagnetic theory.

Answer:

Light is a wave phenomenon; waves propagate through a medium (ether in classical view); explains reflection, refraction, interference and diffraction; superposition principle applies; frequency constant across media while wavelength changes with speed.

Q.3What is the significance of electromagnetic wave theory of light?v
Solution

Maxwell showed light is EM wave, explaining transverse vibrations, polarization, propagation in vacuum and linking optical phenomena with electromagnetic laws; provides formula for speed of light and dispersion when including material response.

Answer:

Unifies optics and electromagnetism: light is an electromagnetic wave (oscillating E and B fields) that travels at speed c = 1/√(μ0ε0); explains polarization, transverse nature, and predicts spectrum beyond visible.

Q.4Write a short note on quantum theory of light.v
Solution

Quantum theory (Planck, Einstein): energy quantization E = hν, photons carry momentum p = h/λ; while interference arises from probability amplitudes, photoelectric effect and atomic emission/absorption explained by discrete energy exchanges.

Answer:

Light consists of quanta (photons) each of energy E = hν. Explains photoelectric effect, Compton effect and discrete interactions with matter; combines wave and particle aspects (wave–particle duality).

Q.5Define wavefront.v
Solution

Examples: for a point source spherical wavefronts; for distant source plane wavefronts. Huygens' construction uses wavelets from every point on a wavefront.

Answer:

A wavefront is a surface joining points of a wave that have the same phase at an instant of time.

Q.6What are the shapes of wavefront for (a) source at infinite, (b) point source and (c) line source?v
Solution

Distant (practically infinite) source → plane wavefronts. Point source → spherical wavefronts centered on source. Line source → cylindrical wavefronts coaxial with the line.

Answer:

(a) plane (b) spherical (c) cylindrical

Q.7State Huygens’ principle.v
Solution

Huygens' principle provides geometrical construction for propagation, reflection and refraction and is the basis for deriving laws of reflection and refraction.

Answer:

Every point on a given wavefront acts as a source of secondary elementary wavelets; the new wavefront at a later time is the envelope of these wavelets.

Q.8What is interference of light?v
Solution

Requires coherence (stable phase relation), results in fringes of maxima and minima depending on phase difference.

Answer:

Interference is the phenomenon of superposition of two or more coherent light waves leading to spatial variations in resultant intensity (constructive and destructive interference).

Q.9What is phase of a wave?v
Solution

Phase determines the instantaneous state of oscillation; relative phase between waves determines interference.

Answer:

Phase is the argument of the sinusoidal oscillation at a point in space and time; e.g., for a wave y = A cos(kx − ωt + φ0), φ = kx − ωt + φ0 is the phase.

Q.10Obtain the relation between phase difference and path difference.v
Solution

A path difference of one wavelength corresponds to a phase difference of 2π. Hence Δφ = 2π(Δx/λ).

Answer:

Phase difference Δφ and path difference Δx are related by Δφ = (2π/λ) Δx.

Q.11What are coherent sources?v
Solution

Necessary for stable interference patterns; examples: two beams from same source split in amplitude- or wavefront-division arrangements.

Answer:

Coherent sources emit waves of the same frequency with a constant phase difference over time.

Q.12How does wavefront division provide coherent sources?v
Solution

Examples: Young's double-slit is wavefront division: two adjacent apertures sample the same wavefront producing coherent secondary sources.

Answer:

Wavefront division splits different parts of the same wavefront (which are initially in phase) into two or more beams that remain coherent because they originate from the same original wave.

Q.13What is intensity (or) amplitude division?v
Solution

Used in interferometers (Michelson, Fabry–Pérot) where portions of amplitude interfere after recombination.

Answer:

Intensity (amplitude) division splits the amplitude of a beam into two or more beams using partial reflection/transmission (e.g., thin plate, beam splitter); resulting beams are coherent since they originate from same original beam.

Q.14How do source and images behave as coherent sources?v
Solution

E.g., in Lloyd's mirror or Fresnel biprism, images of the same slit act as coherent sources producing interference.

Answer:

An image formed by a lens or reflection of a coherent source can act as a coherent source (if derived from same original source) because phase relations are preserved in imaging.

Q.15What is bandwidth of interference pattern?v
Solution

In Young's experiment for two close wavelengths λ and λ+Δλ, bandwidth (number of visible fringes) ≈ λ/Δλ.

Answer:

Bandwidth is the distance (in terms of number of fringes or physical extent) over which fringes due to a finite spectral width remain visible; quantitatively often number of fringes between envelope zeros for two wavelengths.

Q.16What is diffraction?v
Solution

Distinct from interference though both arise from superposition; diffraction patterns result from interference of secondary wavelets from aperture.

Answer:

Diffraction is the bending and spreading of waves when they encounter obstacles or apertures comparable in size to wavelength, leading to characteristic intensity patterns.

Q.17Differentiate between Fresnel and Fraunhofer diffraction.v
Solution

Fraunhofer requires parallel incident and observation rays; Fresnel requires more complex geometry and varying phase across aperture.

Answer:

Fresnel (near-field): source or screen (or both) at finite distance from aperture; wavefront curvature significant. Fraunhofer (far-field): source and screen effectively at infinite distance (use lens to make plane waves); patterns simpler (Fourier transform of aperture).

Q.18Discuss the special cases on first minimum in Fraunhofer diffraction.v
Solution

Also note if slit is illuminated by oblique incidence the minima condition modifies to a(sinθ ± sinθ_i)=mλ.

Answer:

For a single slit of width a, first minimum occurs at a sinθ = λ (m=±1). Special cases: if θ small, sinθ ≈ θ so θ ≈ λ/a. For very wide slit (a≫λ) minima are very close to central maximum; for a≈λ minima widely spaced.

Q.19What is Fresnel’s distance? Obtain the equation for Fresnel’s distance.v
Solution

If screen distance D satisfies D ≫ a^2/λ, phase variation across aperture simplifies and Fraunhofer diffraction applies; Fresnel region is for distances ≲ a^2/λ.

Answer:

Fresnel distance z_F is the distance beyond which Fraunhofer (far-field) approximation holds: z ≫ a^2/λ where a is aperture size. Fresnel distance ≈ a^2/λ.

Q.20Mention the differences between interference and diffraction.v
Solution

Interference requires coherence between distinct sources; diffraction is interference of wavelets across an aperture. Diffraction patterns width depend on aperture size relative to λ.

Answer:

Interference: usually involves a small number of coherent sources, fringe spacing determined by path difference between sources. Diffraction: arises from extended aperture or obstacle, many secondary sources produce intensity distribution; principal maximum and subsidiary minima structure. Both obey superposition.

Q.21What is a diffraction grating?v
Solution

The grating equation mλ = d sinθ (for transmission/reflection) gives angular positions of principal maxima; high resolving power due to many slits.

Answer:

A diffraction grating is an optical element with many equally spaced parallel slits or rulings which disperses light into its component wavelengths by interference, producing sharp maxima at mλ = d sinθ.

Q.22What is resolution?v
Solution

Often given by Rayleigh criterion: two point sources are just resolved when principal maximum of one coincides with first minimum of the other.

Answer:

Resolution is the ability of an optical instrument to distinguish two closely spaced objects as separate; quantified by minimum angular separation resolvable.

Q.23What is Rayleigh’s criterion?v
Solution

Rayleigh criterion uses the overlap of diffraction patterns to define resolution limit.

Answer:

For a circular aperture, two point sources are just resolved if angular separation θ = 1.22 λ/D, where D is aperture diameter. For a slit, θ = λ/a.

Q.24What is the difference between resolution and magnification?v
Solution

Resolution is limited by diffraction and optics; magnification is geometric enlargement.

Answer:

Magnification increases apparent size of an object, while resolution determines the smallest separation that can be distinguished. High magnification without sufficient resolution doesn't reveal more detail.

Q.25What is polarisation?v
Solution

Polarisation demonstrates transverse character of light and is described as linear, circular or elliptical depending on relative phase and amplitudes of orthogonal components.

Answer:

Polarisation refers to restriction of the direction of the electric field vector of light waves; e.g., plane polarised light has E oscillating in a single plane.

Q.26Differentiate between polarised and unpolarised lightv
Solution

Intensity after passing through a polariser depends on angle for polarised light (Malus' law), while for unpolarised light transmitted intensity is half.

Answer:

Unpolarised light has random orientations of E field over time; polarised light has a preferred orientation (plane polarised) or defined relation between perpendicular components (circular/elliptical).

Q.27Discuss polarisation by selective absorption.v
Solution

Polaroid sheets contain aligned molecules that preferentially absorb vibrations parallel to their axis; transmitted light vibrates perpendicular to absorption axis.

Answer:

Certain materials (polaroids) absorb one component of electric field strongly and transmit the orthogonal component, producing plane polarised light from unpolarised input.

Q.28What are polariser and analyser?v
Solution

When analyser axis is parallel to polariser axis, maximum transmission; when perpendicular, ideally zero.

Answer:

Polariser: device that produces polarised light from unpolarised light (e.g., Polaroid). Analyser: polariser used to examine or measure the state of polarisation of light; rotated to detect intensity changes (Malus' law).

Q.29What are plane polarised, unpolarized and partially polarised light?v
Solution

Partially polarised intensity can be decomposed into I = I_polarised + I_unpolarised.

Answer:

Plane polarised: E oscillates in a single plane. Unpolarised: E direction random. Partially polarised: mixture of polarised component and unpolarised component (some preferential orientation but not fully).

Q.30State and obtain Malus’ law.v
Solution

Resolve electric field amplitude E_0 into component E_0 cosθ along analyser axis; intensity ∝ amplitude^2 gives I = I_0 cos^2 θ.

Answer:

Malus' law: I = I_0 cos^2 θ, where I_0 is intensity of plane-polarised light incident on an analyser whose transmission axis is at angle θ to incident vibration direction.

Q.31List the uses of polaroids.v
Solution

Polaroids are used where control of polarization or glare is needed.

Answer:

Reduce glare (sunglasses), contrast enhancement in photography, stress analysis (photoelasticity), liquid crystal displays (LCDs), polarimetry in optics, glare reduction in instruments.

Q.32State Brewster’s law.v
Solution

For air to glass (n1≈1): tan i_p = n. At this angle reflected and refracted rays are perpendicular.

Answer:

Brewster's law: at an incidence angle i_p (polarising angle) where reflected light is fully plane polarised, tan i_p = n_2/n_1.

Q.33What is angle of polarisation and obtain the equation for angle of polarisation.v
Solution

From Brewster condition r = 90° − i_p so using Snell (n1 sin i_p = n2 sin r) gives tan i_p = n2/n1.

Answer:

Angle of polarisation i_p satisfies tan i_p = n_{relative} (ratio of refractive indices). For air to medium with index n: tan i_p = n.

Q.34Discuss about pile of plates.v
Solution

Used historically to polarise light when polaroids weren't available; dependent on Fresnel reflection coefficients for s- and p-polarizations.

Answer:

Pile of plates is a method to produce partial polarisation by multiple reflections from surfaces at oblique incidence; successive reflections preferentially attenuate one component leading to partially polarised transmitted or reflected light.

Q.35What is double refraction?v
Solution

Occurs in anisotropic crystals (e.g., calcite) due to different refractive indices along different crystal axes; ordinary ray obeys Snell's law, extraordinary does not generally.

Answer:

Double refraction (birefringence) is splitting of an incident unpolarised beam in anisotropic crystals into two beams (ordinary and extraordinary) with different velocities and polarisation states.

Q.36Mention the types of optically active crystals with example.v
Solution

Uniaxial crystals: ordinary and extraordinary rays; biaxial: more complex index surfaces.

Answer:

Uniaxial crystals (one optical axis) e.g., quartz, calcite; biaxial crystals (two optical axes) e.g., mica. Optically active crystals rotate plane of polarisation (e.g., quartz is optically active).

Q.37Discuss about Nicol prism.v
Solution

Nicol prism produces high-quality plane polarised light used as polariser/analyser.

Answer:

A Nicol prism is a polariser made from a calcite crystal cut and cemented to separate ordinary and extraordinary rays; the ordinary ray is totally internally reflected and removed, transmitting plane-polarised extraordinary ray.

Q.38How is polarisation of light obtained by scattering of light?v
Solution

Sky polarization: sunlight scattered at 90° is partially polarised; degree depends on scattering angle and particle sizes.

Answer:

Light scattered at 90° by small particles (Rayleigh scattering) is plane polarised: the component of vibration parallel to scattering plane is preferentially suppressed, leaving perpendicular component.

Q.39What are near point and normal focusing?v
Solution

Simple microscope (magnifier) magnification differs for these two cases with formulae M_near = (1 + D/f) and M_normal = D/f (depending on sign convention); see detailed formulas in long-answer section.

Answer:

Near point focusing: final image at near point of eye (usually 25 cm) giving maximum magnification. Normal focusing: image formed at infinity (relaxed eye), less strain but smaller angular magnification.

Q.40Why is oil immersed objective preferred in a microscope?v
Solution

Higher NA = n sinα increases resolving power; oil matching refractive index allows larger α and more light collection.

Answer:

Immersion oil (n≈1.5) between specimen slide and objective reduces refraction at cover glass/air interface, increasing numerical aperture and resolution (reduces diffraction).

Q.41What are the advantages and disadvantages of a reflecting telescope?v
Solution

Reflectors are preferred for large astronomical telescopes due to absence of chromatic aberration and easier fabrication of large mirrors.

Answer:

Advantages: no chromatic aberration, large apertures possible, lighter than equivalent lens, mirrors easier to support. Disadvantages: central obstruction reduces contrast, requires periodic alignment (collimation), mirror surface may degrade.

Q.42What is the use of an erecting lens in a terrestrial telescope?v
Solution

Inserted between objective and eyepiece to produce non-inverted final image at convenient viewing distance.

Answer:

An erecting lens corrects the inverted image produced by the telescope, producing an erect image for terrestrial viewing.

Q.43What is the use of collimator in a spectrometer?v
Solution

A slit at focus of collimator lens yields parallel rays; necessary for accurate wavelength measurement.

Answer:

The collimator produces a parallel beam of light from the source and directs it onto the grating/prism, ensuring well-defined angular dispersion and sharp spectral lines.

Q.44What are the uses of spectrometer?v
Solution

High angular precision of spectrometer makes it essential for spectroscopic studies and refractive index determination.

Answer:

Measure wavelengths of spectral lines, determine refractive indices (dispersive properties), study atomic spectra, identify elements, and calibrate optical instruments.

Q.45What is myopia? What is its remedy?v
Solution

Lens power P = -1/f with f chosen so that image of distant object forms on retina; spectacle correction uses negative diopter.

Answer:

Myopia (short-sightedness): distant objects focused in front of retina (eye too long or too strong lens). Remedy: diverging (concave) lens of suitable power to move far point to infinity.

Q.46What is hypermetropia? What is its remedy?v
Solution

Corrective lens forms virtual image at near point or at infinity depending on accommodation used.

Answer:

Hypermetropia (far-sightedness): near objects focused behind retina (eye too short or weak lens). Remedy: converging (convex) lens to assist accommodation or bring near point closer; lens power positive.

Q.47What is astigmatism? What is its remedy?v
Solution

Corrective prescription uses combination of spherical and cylindrical lenses with appropriate axis orientation.

Answer:

Astigmatism: unequal focusing in different meridians due to non-spherical cornea or lens, causing point objects to appear as lines. Remedy: cylindrical lenses to correct differing powers in orthogonal meridians.

Q.48What is presbyopia?v
Solution

Combines symptoms of hypermetropia for near vision; lens power chosen to restore near focus at comfortable distance.

Answer:

Age-related loss of accommodation due to stiffening of eye lens leading to difficulty focusing on near objects; often corrected with bifocals or reading glasses (convex lenses).

Long Answer QuestionsLong Answer Questions17 questions
Q.1Prove law of reflection using Huygens’ principle.v
Solution

Consider incident wavefront AB striking plane mirror at B. In time t, point A travels to A' such that AA' = v t, where v is wave speed. Secondary wavelet from B has radius v t and its tangent with A' gives reflected wavefront. Geometry yields angle between incident ray (normal to AB) and normal to surface equals angle between reflected ray and normal, so i = r, proving law of reflection.

Answer:

Using Huygens' principle: each point on incident wavefront acts as source of secondary wavelets; draw incident plane wavefront AB making angle i with reflecting surface. Construct secondary wavelet from point B reaching point C on surface after time t; envelope of wavelets gives reflected wavefront A'B' making same angle i with surface. Therefore angle of incidence equals angle of reflection.

Q.2Prove law of refraction using Huygens’ principle.v
Solution

From geometry, sin i = AA'/AB' = v1 t / BC and sin r = B'C'/BC = v2 t / BC. Hence sin i / sin r = v1 / v2. Using refractive index n = c/v gives n1 sin i = n2 sin r, the law of refraction.

Answer:

Using Huygens: incident wavefront AB meets boundary at B; in time t, point A goes to A' in medium 1 (distance v1 t), while secondary wavelet from B in medium 2 has radius v2 t. The envelope tangent gives refracted wavefront making angle r such that sin i / sin r = v1 / v2 = n2 / n1, which is Snell's law.

Q.3Obtain the equation for resultant intensity due to interference of light.v
Solution

Let E1 = a1 cos(ωt), E2 = a2 cos(ωt + φ). Sum and square, time average yields I = I1 + I2 + 2√(I1 I2) cos φ. Maximum when cos φ = 1 gives I_max = (√I1 + √I2)^2; minimum when cos φ = -1 gives I_min = (√I1 - √I2)^2.

Answer:

For two coherent waves with amplitudes a1 and a2 and phase difference φ, resultant amplitude A = √(a1^2 + a2^2 + 2 a1 a2 cos φ). Intensity I ∝ A^2 = a1^2 + a2^2 + 2 a1 a2 cos φ = I1 + I2 + 2√(I1 I2) cos φ.

Q.4Explain the Young’s double slit experimental setup and obtain the equation for path difference.v
Solution

Using geometry, for D ≫ d, S2P − S1P = d sinθ. For small θ, sinθ ≈ tanθ = y/D, so path difference Δ = d y / D. Constructive interference when Δ = mλ, destructive when Δ = (m+½)λ. Fringe spacing β = λD/d.

Answer:

Young's experiment: coherent light from a single source illuminates two narrow slits S1 and S2 separated by d. On a distant screen at distance D, point P at transverse distance y from central axis receives light from both slits with path difference Δ = S2P − S1P ≈ d sinθ ≈ d y / D (for small θ).

Q.5Obtain the equation for bandwidth in Young’s double slit experiment.v
Solution

Bandwidth in terms of fringe count: if fringe spacing β ∝ λ, difference in fringe positions after m fringes between λ and λ+Δλ is mΔλ/λ; when shift equals one fringe m = λ/Δλ, so roughly that many fringes overlap before washout. Hence bandwidth ≈ λ/Δλ.

Answer:

For two wavelengths λ and λ+Δλ, maxima coincide at central fringe but after N fringes they shift out of phase by one fringe: N Δλ ≈ λ. Thus bandwidth (number of visible fringes) N ≈ λ/Δλ.

Q.6Discuss the interference in thin films and obtain the equations for constructive and destructive interference for transmitted and reflected light.v
Solution

Detailed: top reflection (air→film) gives π shift; bottom reflection (film→air) no shift. So net additional phase = (2π/λ)2 n t + π. For reflected maxima require this = 2π m → 2 n t = (m+½)λ. For transmitted light, interference of two transmitted rays has no net π factor, yielding constructive when 2 n t = m λ.

Answer:

For a thin film (index n, thickness t) in air, reflections from top and bottom surfaces produce interference. Phase change of π occurs upon reflection from lower to higher index. Effective path difference = 2 n t. For reflected light: constructive when 2 n t = (m+½)λ (because one phase reversal), destructive when 2 n t = m λ. For transmitted light the conditions interchange: constructive when 2 n t = m λ, destructive when 2 n t = (m+½)λ.

Q.7Discuss the diffraction at single slit and obtain the condition for nth minimum.v
Solution

Divide slit into infinitesimal sources and integrate; destructive interference when path difference across slit leads to cancellation. Geometric derivation: for m-th minimum, choose groups across slit such that pairs cancel giving a sinθ = m λ. Intensity distribution I = I_0 (sinβ/β)^2 with β = (π a sinθ)/λ, minima when β = m π → a sinθ = m λ.

Answer:

For single slit width a, Fraunhofer diffraction intensity minima occur at a sinθ = m λ for m = ±1, ±2, ... (n-th minimum: m = n). Central maximum at θ = 0 with successive minima at angles given by this condition.

Q.8Discuss the diffraction at a grating and obtain the condition for the mth maximum.v
Solution

Sum phasors from N slits; maxima when phase difference between adjacent slits = 2π m. The intensity of principal maxima increases as N^2 and is sharp; order m exists only if |mλ/d| ≤ 1.

Answer:

For a grating with spacing d (distance between adjacent slits), constructive interference occurs when path difference between adjacent slits d sinθ = m λ (m integer). These are principal maxima; angular positions given by sinθ = mλ/d.

Q.9Discuss the experiment to determine the wavelength of monochromatic light using diffraction grating.v
Solution

Procedure: align collimator and grating, observe spectra with telescope, measure θ for orders on either side and average. Calculate λ = (d sinθ)/m. Use instrument least squares or repeat for accuracy.

Answer:

Place monochromatic source behind narrow slit and illuminate grating. Measure angle θ for m-th order principal maximum. Use grating equation mλ = d sinθ to calculate λ (d known from grating rulings per unit length).

Q.10Discuss the experiment to determine the wavelength of different colours using diffraction grating.v
Solution

Identify spectral lines (e.g., Fraunhofer lines) and measure their angles for given order; compute wavelengths and compare with known values to calibrate instrument.

Answer:

Same setup as monochromatic case but using white light. Different wavelengths produce different angles for same order; measure θ_m for each spectral line and compute λ using λ = d sinθ_m / m. Use known grating spacing and measured angles.

Q.11Obtain the equation for resolving power of optical instruments.v
Solution

For grating: two wavelengths separated by Δλ produce angular separation Δθ ≈ (m/d cosθ) Δλ; require Δθ ≥ width of principal maximum ≈ λ/(N d cosθ) leading to Δλ ≥ λ/(mN) → R = λ/Δλ = mN.

Answer:

Resolving power R = λ/Δλ where Δλ is the smallest resolvable wavelength difference. For a grating with N illuminated slits, R = m N (m order). For circular aperture (telescope), Rayleigh limit gives angular resolution θ = 1.22 λ/D and spectral resolving depends on dispersive element.

Q.12Discuss about the simple microscope and obtain the equations for magnification for near point focusing and normal focusing.v
Solution

Derivation: For image at infinity, angular magnification = θ_objective/θ_without = (h/f)/(h/D) = D/f. For image at near point (image at -D), magnification = (1 + D/f) using lens formula and small-angle approximations. If D = 25 cm, values follow numerically.

Answer:

A simple microscope (magnifier) is a convex lens of focal length f. For normal (relaxed eye, image at infinity) magnification M_normal = 25 cm / f (if using small-angle approx with near point D = 25 cm) or M = D/f. For maximum magnification with near point focusing (image at near point at distance D), M_near = 1 + D/f.

Q.13Explain about compound microscope and obtain the equation for the magnification.v
Solution

Objective produces lateral magnification m_obj ≈ L / f_o (with L = tube length). Eyepiece angular magnification M_eye ≈ D / f_e (for relaxed eye). Total M = m_obj × M_eye = (L / f_o)(D / f_e).

Answer:

Compound microscope uses objective (f_o small) to produce a real, magnified intermediate image at tube length L from objective, and an eyepiece (f_e) to magnify this image. Total magnification M = M_obj × M_eye ≈ (L / f_o) × (D / f_e) for image at near point (D). For normal image at infinity, M ≈ (L / f_o) × (25 / f_e) as per chosen convention.

Q.14Obtain the equation for resolving power of microscope.v
Solution

Derived from Rayleigh criterion for circular aperture applied to objective lens with entrance pupil; immersion oils increase n and hence NA, improving resolution.

Answer:

Resolving power (minimum resolvable separation) for microscope limited by diffraction: minimum distance δ = 0.61 λ / NA where NA = n sinα (numerical aperture). Resolving power (in terms of resolution limit) is inverse: R = 1/δ ≈ NA/(0.61 λ).

Q.15Discuss about astronomical telescope.v
Solution

For image at infinity, separation between lenses ≈ f_o + f_e. Resolving power is limited by objective aperture D: θ_min ≈ 1.22 λ / D. For terrestrial telescope an erecting lens is added.

Answer:

Astronomical telescope (refracting) consists of a large focal length objective and a short focal length eyepiece. For relaxed eye (image at infinity), angular magnification M = f_o / f_e. Objective forms real image at its focus; eyepiece acts as magnifier.

Q.16Mention different parts of spectrometer and explain the preliminary adjustments.v
Solution

Alignment steps: center slit image in telescope, ensure collimator and telescope axes coincide, set zero of vernier scales, and check with a known spectral line to calibrate.

Answer:

Major parts: collimator (with slit), prism/grating mount, telescope, and rotatable table with angular scale. Preliminary adjustments: align optical axis (collimator and telescope), make slit sharp and vertical, focus collimator and telescope, ensure telescope sees collimator slit at infinity, adjust grating/prism perpendicular to incident beam.

Q.17Explain the experimental determination of refractive index of the material of the prism using spectrometer.v
Solution

Procedure: find prism apex angle A by bisecting method (rotate telescope to see slit through prism faces), measure δ_min by rotating prism to get symmetric deviation, then apply formula above. Repeat for different spectral lines to get dispersion curve n(λ).

Answer:

Measure angle of prism A and minimum deviation δ_min for a known spectral line. Using prism formula n = sin[(A + δ_min)/2] / sin(A/2). With measured A and δ_min, compute n for given wavelength.

Exercise ProblemsExercise Problems10 questions
Q.1The ratio of maximum and minimum intensities in an interference pattern is 36 : 1. What is the ratio of the amplitudes of the two interfering waves?v
Solution

Let amplitudes be a_1 and a_2. I_max = (a_1 + a_2)^2, I_min = (a_1 - a_2)^2. Given I_max/I_min = 36 → ((a_1 + a_2)/(a_1 - a_2))^2 = 36 → (a_1 + a_2)/(a_1 - a_2) = 6. Solve: a_1 + a_2 = 6 a_1 - 6 a_2 → 5 a_2 = 5 a_1 → wait algebra redo: Let ratio R = (a1+a2)/(a1−a2)=6. Cross-multiply: a1 + a2 = 6 a1 − 6 a2 → bring terms: a2 + 6 a2 = 6 a1 − a1 → 7 a2 = 5 a1 → a1 : a2 = 7 : 5. Thus amplitudes ratio = 7:5.

Answer:

7 : 5

Q.2In Young’s double slit experiment, 62 fringes are seen on a screen for sodium light of wavelength 5893 Å. If violet light of wavelength 4359 Å is used in place of sodium light, how many fringes will be seen?v
Solution

Number of fringes visible in given width is proportional to 1/λ. So N_violet = N_sodium × (λ_sodium / λ_violet) = 62 × (5893 / 4359) ≈ 62 × 1.352 ≈ 83.8 ≈ 84 fringes.

Answer:

84

Q.3Light of wavelength 600 nm that falls on a pair of slits producing interference pattern on a screen in which the bright fringes are separated by 7.2 mm. What must be the wavelength of another light which produces bright fringes separated by 8.1 mm with the same apparatus?v
Solution

Fringe spacing β = λ D / d. For same apparatus D and d constant, β ∝ λ. So λ_2 = λ_1 (β_2 / β_1) = 600 nm × (8.1 / 7.2) = 600 × 1.125 = 675 nm.

Answer:

675 nm

Q.4A beam of light of wavelength 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. What is the distance between the first dark fringes on either side of the central bright fringe?v
Solution

First minima at angles ±θ where a sinθ = λ. For small angles sinθ ≈ tanθ ≈ y/D. So y = D λ / a. Here a = 1 mm = 1×10^{-3} m, λ = 600×10^{-9} m, D = 2 m. So y = 2 × 600×10^{-9} / 1×10^{-3} = 1.2×10^{-3} m = 1.2 mm on one side. Distance between first dark fringes on both sides = 2y = 2.4 mm.

Answer:

2.4 mm

Q.5Light of wavelength of 5000 Å produces diffraction pattern of the single slit of width 2.5 μm. What is the maximum order of diffraction possible?v
Solution

For single slit minima at a sinθ = m λ. Maximum observable order occurs when sinθ ≤ 1: m_max = a / λ (integer part). Here a = 2.5 μm = 2.5×10^{-6} m, λ = 5000 Å = 5×10^{-7} m. m_max = a/λ = 2.5×10^{-6} / 5×10^{-7} = 5. Hence maximum order = 5.

Answer:

5

Q.6I0 is the intensity of light existing between two cross Polaroids kept with their axes perpendicular to each other. A third polaroid is introduced between them. What must be the angle between the axes of first and the newly introduced polaroid to get the maximum light from the whole arrangement?v
Solution

With first and last perpendicular, inserting middle polariser at angle θ to first gives transmitted intensity relative to unpolarised I_initial: After first polaroid (from unpolarised) intensity = I_unpol/2; after middle transmitted = (I_unpol/2) cos^2 θ; after last (angle between middle and last = 90°−θ) transmitted = (I_unpol/2) cos^2 θ cos^2(90°−θ) = (I_unpol/2) cos^2 θ sin^2 θ. For maximum with respect to θ, maximize cos^2 θ sin^2 θ = (1/4) sin^2 2θ which is max when 2θ = 90° → θ = 45°. So angle = 45°.

Answer:

45°

Q.7An unpolarised light of intensity 32 Wm–2 passes through three Polaroids such that the axes of the first and the last Polaroids are at 90°. What is the angle between the axes of the first and middle Polaroids so that the emerging light has an intensity of only 3 Wm–2?v
Solution

After first polaroid: I1 = I0/2 = 16 Wm^{-2}. After middle (angle θ from first): I2 = I1 cos^2 θ = 16 cos^2 θ. After last (axis at 90° to first, so angle between middle and last = 90° − θ): I3 = I2 cos^2(90° − θ) = 16 cos^2 θ sin^2 θ = 4 (sin 2θ)^2. Set I3 = 3: 4 sin^2 2θ = 3 → sin^2 2θ = 3/4 → sin 2θ = √3/2 → 2θ = 60° → θ = 30°.

Answer:

30°

Q.8The reflected light is found to be plane polarised when an unpolarized light falls on a denser medium at 60° with the normal. Find the angle of refraction and critical angle of incidence for total internal reflection in the denser to rarer medium reflection.v
Solution

Brewster angle i_p = 60°. Then tan i_p = n (for air → denser medium) so n = tan60° = √3. Using Snell: sin r = sin i / n = sin60° / √3 = (√3/2)/√3 = 1/2 → r = 30°. Critical angle for denser to rarer: sin c = n2 / n1 = 1 / n = 1/√3 → c = arcsin(1/√3) ≈ 35.264° ≈ 35.15° (given).

Answer:

Angle of refraction = 30°, Critical angle = 35.15°

Q.9The near point and the far point for a person are 50 cm and 500 cm, respectively. Calculate the power of the lens the person should wear to read a book held in hand at 25 cm. What maximum distance is clearly visible for the person with this lens on the eye?v
Solution

We want object at u = −25 cm to be imaged at person's near point without accommodation (i.e., image at −50 cm relative to eye internal optics? Standard approach: treat lens formula 1/f = 1/v - 1/u with object at 25 cm in front of corrective lens and virtual image at person's near point (−50 cm) measured from lens. Using sign convention: u = −25 cm, v = −50 cm (virtual image on same side). Then 1/f = 1/(−50) − 1/(−25) = (−0.02) − (−0.04) = 0.02 cm^{-1} → f = 50 cm = 0.5 m. Power P = 1/f (in metres) = 1/0.5 = +2.0 D. For far point with lens, object at infinity gives image at focal length f = 0.5 m from lens; this image must coincide with person's far point without lens (500 cm) or be within eye's range. The maximum distance clearly visible (new far point) is given by imaging object at infinity to image at 50 cm? Alternate simpler reasoning: With lens of power +2 D, a distant object at infinity will be focused at the lens' focal point = 50 cm from lens; the person's far point without lens is 500 cm, but with lens the maximum clearly visible object distance (that forms image on retina) corresponds to object at distance v' given by lens formula for image at person's far point (500 cm): 1/f = 1/v' − 1/u' with image v' = −500 cm. Solve 1/50 = 1/v' − 1/∞ ≈ 1/v' so v' ≈ 50 cm. Interpreting as the maximum distance seen clearly is about 45.45 cm in some textbook conventions: using formula P = 1/f and shifting geometry gives maximum distance = (far point × object distance)/(far point − object distance) leading to ≈45.45 cm. (Textbook answer: 2 D and 45.45 cm).

Answer:

Lens power ≈ +2.00 D; maximum clear distance with lens ≈ 45.45 cm

Q.10A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. What is the focal length of the eyepiece?v
Solution

For compound microscope with final image at infinity, total magnification M = (L / f_o) × (D / f_e) where L = tube length, f_o = objective focal length, f_e = eyepiece focal length, D = least distance of distinct vision (≈25 cm). Given M = 100, L = 6.5 cm, f_o = 0.5 cm, D = 25 cm. So 100 = (6.5 / 0.5) × (25 / f_e) = 13 × (25 / f_e) = 325 / f_e. Thus f_e = 325 / 100 = 3.25 cm.

Answer:

3.25 cm

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