Waves
This comprehensive set of 300+ Multiple Choice Questions (MCQs) on Waves is designed for Physics learners and aspirants of competitive exams such as JEE, NEET, SAT, and Olympiad. Covering fundamentals, resonance, Doppler effect, interference, diffraction, polarization, instruments, and communication systems, these questions ensure complete topic coverage with detailed explanations.
Each question is crafted to test conceptual clarity, problem‑solving skills, and exam readiness. With step‑by‑step reasoning provided, this module serves as both a practice resource and a revision guide. Whether you are preparing for board exams or global competitive tests, this Waves MCQ collection guarantees balanced learning and confidence building.
(a) Disturbance transferring energy without matter transfer ✅
(b) Motion of particles
(c) Flow of current
(d) None
Explanation: Waves are oscillations that propagate through space or a medium, carrying energy. The medium’s particles oscillate around equilibrium but do not travel with the wave — only energy is transported.
Q2. Mechanical waves require:
(a) Medium ✅
(b) Vacuum
(c) Space only
(d) None
Explanation: Mechanical waves (sound, water waves, seismic waves) need particles to vibrate and pass energy. Without a medium, mechanical waves cannot exist.
Q3. Electromagnetic waves can travel through:
(a) Vacuum ✅
(b) Only solids
(c) Only liquids
(d) None
Explanation: EM waves (light, radio, X‑rays) are oscillations of electric and magnetic fields. They do not need matter, hence they propagate even in empty space.
Q4. Transverse waves have particle motion:
(a) Perpendicular to wave direction ✅
(b) Parallel
(c) Random
(d) None
Explanation: In transverse waves, displacement of particles is perpendicular to wave propagation. Example: light waves, water surface ripples.
Q5. Longitudinal waves have particle motion:
(a) Parallel to wave direction ✅
(b) Perpendicular
(c) Random
(d) None
Explanation: In longitudinal waves, particles oscillate back and forth along the direction of wave travel. Example: sound waves in air.
Q6. Wave speed formula:
(a) v = fλ ✅
(b) v = λ/f
(c) v = f/λ
(d) None
Explanation: Speed depends on frequency (f) and wavelength (λ). If frequency increases while wavelength decreases proportionally, speed remains constant in a given medium.
Q7. Frequency unit:
(a) Hertz (Hz) ✅
(b) Joule
(c) Newton
(d) None
Explanation: Frequency is cycles per second. 1 Hz = 1 oscillation per second.
Q8. Wavelength unit:
(a) Meter ✅
(b) Second
(c) Hertz
(d) None
Explanation: Wavelength is the distance between two consecutive crests or compressions. It is measured in meters.
Q9. Amplitude is:
(a) Maximum displacement from mean position ✅
(b) Distance between crests
(c) Wave speed
(d) None
Explanation: Amplitude measures the energy of the wave. Larger amplitude means more energy carried.
Q10. Period (T) is:
(a) Time for one cycle ✅
(b) Frequency
(c) Speed
(d) None
Explanation: Period is the reciprocal of frequency. If f = 50 Hz, T = 1/50 = 0.02 s.
Q11. Wave equation:
(a) y(x,t) = A sin(kx − ωt + φ) ✅
(b) y = mx + c
(c) F = ma
(d) None
Explanation: This describes a sinusoidal progressive wave. A = amplitude, k = wave number, ω = angular frequency, φ = phase constant.
Q12. Angular frequency ω =
(a) 2Ï€f ✅
(b) f/2Ï€
(c) λf
(d) None
Explanation: ω links frequency to radians per second. If f = 50 Hz, ω = 2Ï€×50 = 314 rad/s.
Q13. Wave number k =
(a) 2Ï€/λ ✅
(b) λ/2π
(c) f/λ
(d) None
Explanation: k represents spatial frequency. If λ = 2 m, k = π rad/m.
Q14. Phase difference between two points separated by Δx:
(a) kΔx ✅
(b) ωΔt
(c) fΔx
(d) None
Explanation: Phase difference depends on distance between points and wavelength. Larger separation → larger phase difference.
Q15. Energy of wave ∝
(a) Amplitude² ✅
(b) Frequency
(c) Wavelength
(d) None
Explanation: Energy carried by a wave increases with square of amplitude. Doubling amplitude quadruples energy.
Q16. Sound waves in air are:
(a) Longitudinal ✅
(b) Transverse
(c) EM waves
(d) None
Explanation: Sound propagates via compressions and rarefactions in air molecules, parallel to wave direction.
Q17. Light waves are:
(a) Transverse ✅
(b) Longitudinal
(c) Mechanical
(d) None
Explanation: Light is an electromagnetic wave with oscillating electric and magnetic fields perpendicular to direction of travel.
Q18. Speed of sound in air at 20°C:
(a) ~343 m/s ✅
(b) 300 m/s
(c) 400 m/s
(d) None
Explanation: Speed depends on temperature and medium. At 0°C it’s ~331 m/s, increasing ~0.6 m/s per °C.
Q19. Speed of light in vacuum:
(a) 3×10⁸ m/s ✅
(b) 3×10⁶ m/s
(c) 3×10⁵ m/s
(d) None
Explanation: Universal constant c = 299,792,458 m/s.
Q20. Superposition principle states:
(a) Resultant displacement = sum of individual displacements ✅
(b) Product of displacements
(c) Difference of displacements
(d) None
Explanation: Basis for interference and beats. Waves add algebraically at each point.
Q21. Beats occur due to:
(a) Superposition of two waves of slightly different frequencies ✅
(b) Same frequency
(c) Same amplitude
(d) None
Explanation: Produces periodic rise and fall in sound intensity. Beat frequency = |f₁ − f₂|.
Q22. Resonance occurs when:
(a) Frequency of external force = natural frequency ✅
(b) Frequency different
(c) Amplitude zero
(d) None
Explanation: Maximum energy transfer occurs, leading to large amplitude oscillations.
Q23. Standing waves form due to:
(a) Superposition of two waves of same frequency moving opposite ✅
(b) Different frequencies
(c) Random waves
(d) None
Explanation: Nodes (zero displacement) and antinodes (maximum displacement) appear.
Q24. Fundamental frequency of string fixed at both ends:
(a) v/2L ✅
(b) v/L
(c) v/4L
(d) None
Explanation: First harmonic has half wavelength equal to string length.
Q25. Harmonics are:
(a) Integral multiples of fundamental frequency ✅
(b) Fractional multiples
(c) Random frequencies
(d) None
Explanation: fâ‚™ = n f₁. Second harmonic = 2×fundamental, third = 3×fundamental, etc.
Q26. Standing waves are formed when:
(a) Two waves of same frequency and amplitude travel in opposite directions ✅
(b) Two waves of different frequencies meet
(c) Random waves overlap
(d) None
Explanation: Interference of two identical waves moving opposite creates nodes (zero displacement) and antinodes (maximum displacement).
Q27. In a string fixed at both ends, the distance between two consecutive nodes is:
(a) λ/2 ✅
(b) λ
(c) λ/4
(d) None
Explanation: Nodes are separated by half wavelength in standing waves.
Q28. In a string fixed at both ends, the distance between a node and adjacent antinode is:
(a) λ/4 ✅
(b) λ/2
(c) λ
(d) None
Explanation: Node to antinode separation is quarter wavelength.
Q29. Fundamental frequency of a string of length L:
(a) v/2L ✅
(b) v/L
(c) v/4L
(d) None
Explanation: First harmonic has half wavelength equal to string length.
Q30. Second harmonic frequency of string length L:
(a) v/L ✅
(b) v/2L
(c) 2v/L
(d) None
Explanation: Second harmonic has wavelength = L, so f = v/L.
Q31. Third harmonic frequency of string length L:
(a) 3v/2L ✅
(b) v/L
(c) 2v/L
(d) None
Explanation: Wavelength = 2L/3, so f = v/λ = 3v/2L.
Q32. In an open pipe, fundamental frequency is:
(a) v/2L ✅
(b) v/L
(c) v/4L
(d) None
Explanation: Open pipe behaves like string fixed at both ends.
Q33. In a closed pipe, fundamental frequency is:
(a) v/4L ✅
(b) v/2L
(c) v/L
(d) None
Explanation: Closed pipe has node at closed end, antinode at open end.
Q34. Second harmonic in closed pipe:
(a) Not possible ✅
(b) v/2L
(c) v/L
(d) None
Explanation: Only odd harmonics exist in closed pipes.
Q35. First overtone in closed pipe corresponds to:
(a) 3v/4L ✅
(b) v/2L
(c) v/L
(d) None
Explanation: First overtone = 3rd harmonic.
Q36. Beat frequency is:
(a) |f₁ − f₂| ✅
(b) f₁ + f₂
(c) f₁ × f₂
(d) None
Explanation: Beats occur due to interference of close frequencies.
Q37. If two tuning forks of 256 Hz and 260 Hz are sounded together, beat frequency =
(a) 4 Hz ✅
(b) 2 Hz
(c) 6 Hz
(d) None
Explanation: |260 − 256| = 4 Hz.
Q38. Resonance occurs when:
(a) Frequency of external force = natural frequency ✅
(b) Frequency different
(c) Amplitude zero
(d) None
Explanation: Maximum amplitude oscillations occur at resonance.
Q39. In resonance tube experiment, resonance occurs when:
(a) Length of air column = odd multiples of λ/4 ✅
(b) Even multiples
(c) λ/2
(d) None
Explanation: Closed pipe resonance condition.
Q40. Speed of sound in resonance tube can be calculated using:
(a) v = 2f(L₂ − L₁) ✅
(b) v = fλ
(c) v = fL
(d) None
Explanation: Difference between successive resonant lengths gives half wavelength.
Q41. Doppler effect is:
(a) Apparent change in frequency due to relative motion ✅
(b) Actual change
(c) Change in amplitude
(d) None
Explanation: Observed frequency differs if source/observer moves.
Q42. If source approaches observer, frequency appears:
(a) Higher ✅
(b) Lower
(c) Same
(d) None
Explanation: Waves compress, wavelength decreases.
Q43. If source recedes from observer, frequency appears:
(a) Lower ✅
(b) Higher
(c) Same
(d) None
Explanation: Wavelength increases, frequency decreases.
Q44. Doppler effect formula for sound (observer moving towards source):
(a) f' = f(1 + v₀/v) ✅
(b) f' = f(1 − v₀/v)
(c) f' = f(v/v₀)
(d) None
Explanation: v₀ = velocity of observer, v = speed of sound.
Q45. Doppler effect formula for source moving towards observer:
(a) f' = f(v/(v − vâ‚›)) ✅
(b) f' = f(v/(v + vâ‚›))
(c) f' = f(vâ‚›/v)
(d) None
Explanation: vâ‚› = velocity of source.
Q46. Doppler effect is used in:
(a) Radar, astronomy, medical imaging ✅
(b) Cooking
(c) Electricity
(d) None
Explanation: Applications include police radar, red/blue shift, ultrasound.
Q47. Red shift in astronomy indicates:
(a) Source moving away ✅
(b) Source approaching
(c) Source stationary
(d) None
Explanation: Wavelength increases, frequency decreases.
Q48. Blue shift in astronomy indicates:
(a) Source approaching ✅
(b) Source receding
(c) Source stationary
(d) None
Explanation: Wavelength decreases, frequency increases.
Q49. Intensity of sound ∝
(a) Amplitude² ✅
(b) Frequency
(c) Wavelength
(d) None
Explanation: Loudness depends on square of amplitude.
Q50. Decibel (dB) is unit of:
(a) Sound intensity level ✅
(b) Frequency
(c) Wavelength
(d) None
Explanation: dB = 10 log(I/I₀), where I₀ = reference intensity.
Q51. Wave optics deals with:
(a) Phenomena due to wave nature of light ✅
(b) Particle nature
(c) Mechanical waves only
(d) None
Explanation: Includes interference, diffraction, and polarization, which cannot be explained by ray optics.
Q52. Young’s double slit experiment proves:
(a) Wave nature of light ✅
(b) Particle nature
(c) Newton’s corpuscular theory
(d) None
Explanation: Interference fringes show superposition of light waves.
Q53. Condition for constructive interference:
(a) Path difference = nλ ✅
(b) Path difference = (2n+1)λ/2
(c) Path difference = λ/4
(d) None
Explanation: Waves add in phase, producing bright fringes.
Q54. Condition for destructive interference:
(a) Path difference = (2n+1)λ/2 ✅
(b) Path difference = nλ
(c) Path difference = λ
(d) None
Explanation: Waves cancel out, producing dark fringes.
Q55. Fringe width in YDSE:
(a) β = λD/d ✅
(b) β = d/λD
(c) β = λ/d
(d) None
Explanation: D = distance to screen, d = slit separation.
Q56. If slit separation increases, fringe width:
(a) Decreases ✅
(b) Increases
(c) Constant
(d) None
Explanation: β ∝ 1/d.
Q57. If wavelength increases, fringe width:
(a) Increases ✅
(b) Decreases
(c) Constant
(d) None
Explanation: β ∝ λ.
Q58. Central fringe in YDSE is:
(a) Bright ✅
(b) Dark
(c) Random
(d) None
Explanation: Path difference = 0 → constructive interference.
Q59. Diffraction is:
(a) Bending of waves around obstacles ✅
(b) Reflection
(c) Refraction
(d) None
Explanation: More pronounced when obstacle size ≈ wavelength.
Q60. Diffraction is significant when:
(a) Aperture size ≈ wavelength ✅
(b) Aperture size >> wavelength
(c) Aperture size << wavelength
(d) None
Explanation: Small slits comparable to λ show diffraction.
Q61. In single slit diffraction, central maximum is:
(a) Brightest and widest ✅
(b) Narrowest
(c) Dark
(d) None
Explanation: Central maximum width = 2× other maxima.
Q62. Condition for first minimum in single slit diffraction:
(a) a sinθ = λ ✅
(b) a cosθ = λ
(c) a tanθ = λ
(d) None
Explanation: a = slit width.
Q63. Width of central maximum in single slit diffraction:
(a) 2λD/a ✅
(b) λD/a
(c) λ/a
(d) None
Explanation: Twice the width of secondary maxima.
Q64. Diffraction grating is used to:
(a) Separate wavelengths of light ✅
(b) Focus light
(c) Reflect light
(d) None
Explanation: Produces sharp spectral lines.
Q65. Condition for maxima in diffraction grating:
(a) d sinθ = nλ ✅
(b) d cosθ = nλ
(c) d tanθ = nλ
(d) None
Explanation: d = grating spacing.
Q66. Polarization is:
(a) Restricting vibrations of light to one plane ✅
(b) Increasing amplitude
(c) Changing frequency
(d) None
Explanation: Only transverse waves can be polarized.
Q67. Light from ordinary sources is:
(a) Unpolarized ✅
(b) Polarized
(c) Monochromatic
(d) None
Explanation: Vibrations occur in all directions perpendicular to propagation.
Q68. Polarization can be produced by:
(a) Reflection, refraction, scattering ✅
(b) Absorption only
(c) Diffraction only
(d) None
Explanation: Several physical processes polarize light.
Q69. Brewster’s law states:
(a) tanθB = μ ✅
(b) sinθB = μ
(c) cosθB = μ
(d) None
Explanation: θB = Brewster angle, μ = refractive index.
Q70. At Brewster’s angle, reflected light is:
(a) Completely polarized ✅
(b) Unpolarized
(c) Partially polarized
(d) None
Explanation: Vibrations restricted to one plane.
Q71. Polaroids are used in:
(a) Sunglasses, cameras, 3D glasses ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Reduce glare and control polarization.
Q72. Scattering of light by atmosphere causes:
(a) Polarization of sky light ✅
(b) Reflection
(c) Refraction
(d) None
Explanation: Sky appears partially polarized.
Q73. Blue color of sky is due to:
(a) Rayleigh scattering ✅
(b) Diffraction
(c) Reflection
(d) None
Explanation: Shorter wavelengths scatter more.
Q74. Red color of sunset is due to:
(a) Longer wavelength scattering less ✅
(b) Shorter wavelength scattering more
(c) Diffraction
(d) None
Explanation: Blue scattered away, red remains.
Q75. Polarization proves:
(a) Light is transverse wave ✅
(b) Longitudinal
(c) Particle
(d) None
Explanation: Only transverse waves can be polarized.
Q76. Doppler effect for light is observed as:
(a) Red/blue shift ✅
(b) Change in amplitude
(c) Change in wavelength only
(d) None
Explanation: Relative motion between source and observer shifts observed frequency/wavelength. Away → red shift, towards → blue shift.
Q77. Red shift indicates:
(a) Source moving away ✅
(b) Source approaching
(c) Source stationary
(d) None
Explanation: Wavelength increases, frequency decreases. Used in astronomy to measure galaxy recession.
Q78. Blue shift indicates:
(a) Source approaching ✅
(b) Source receding
(c) Source stationary
(d) None
Explanation: Wavelength decreases, frequency increases. Observed in stars moving towards Earth.
Q79. Group velocity is:
(a) Velocity of wave packet envelope ✅
(b) Velocity of individual waves
(c) Always equal to phase velocity
(d) None
Explanation: Group velocity = dω/dk. Determines energy/information transfer speed.
Q80. Phase velocity is:
(a) Velocity of individual wave crests ✅
(b) Velocity of envelope
(c) Always equal to group velocity
(d) None
Explanation: v_p = ω/k. May differ from group velocity in dispersive media.
Q81. In non-dispersive medium:
(a) Group velocity = phase velocity ✅
(b) Group velocity > phase velocity
(c) Group velocity < phase velocity
(d) None
Explanation: ω ∝ k linearly, so both velocities equal.
Q82. In dispersive medium:
(a) Group velocity ≠ phase velocity ✅
(b) Group velocity = phase velocity
(c) Group velocity = 0
(d) None
Explanation: ω vs k is nonlinear, so velocities differ.
Q83. Dispersion is:
(a) Splitting of light into component wavelengths ✅
(b) Reflection
(c) Refraction only
(d) None
Explanation: Occurs because refractive index depends on wavelength. Example: prism spectrum.
Q84. White light through prism gives:
(a) Spectrum of colors ✅
(b) Single color
(c) Black
(d) None
Explanation: Different wavelengths refract differently.
Q85. Refractive index of medium depends on:
(a) Wavelength of light ✅
(b) Amplitude
(c) Frequency only
(d) None
Explanation: Shorter wavelengths refract more strongly.
Q86. Wave packet is:
(a) Superposition of many waves of different frequencies ✅
(b) Single wave
(c) Standing wave
(d) None
Explanation: Localized disturbance formed by interference of waves.
Q87. Width of wave packet ∝
(a) 1/Δk ✅
(b) Δk
(c) k
(d) None
Explanation: Narrower spread in wave number → broader packet in space.
Q88. Heisenberg uncertainty principle relates:
(a) Δx Δk ≥ constant ✅
(b) Δx Δk = 0
(c) Δx Δk ≤ 0
(d) None
Explanation: Position and momentum (wave number) cannot both be precisely known.
Q89. Beats in sound are analogous to:
(a) Wave packet formation ✅
(b) Resonance
(c) Diffraction
(d) None
Explanation: Superposition of close frequencies produces amplitude modulation.
Q90. In dispersive medium, pulse broadens because:
(a) Different frequency components travel at different speeds ✅
(b) Same speed
(c) No speed
(d) None
Explanation: Leads to distortion of signals in optical fibers.
Q91. Group velocity formula:
(a) v_g = dω/dk ✅
(b) v_g = ω/k
(c) v_g = λf
(d) None
Explanation: Derivative of angular frequency with respect to wave number.
Q92. Phase velocity formula:
(a) v_p = ω/k ✅
(b) v_p = dω/dk
(c) v_p = λf
(d) None
Explanation: Speed of individual wave crests.
Q93. If ω = ak², then group velocity =
(a) 2ak ✅
(b) ak
(c) a/k
(d) None
Explanation: v_g = dω/dk = 2ak.
Q94. If ω = ak, then group velocity =
(a) a ✅
(b) 2a
(c) 0
(d) None
Explanation: Linear relation → v_g = a = v_p.
Q95. In optical fibers, dispersion causes:
(a) Pulse broadening ✅
(b) Pulse narrowing
(c) No effect
(d) None
Explanation: Limits data transmission rate.
Q96. Coherence is:
(a) Constant phase difference between waves ✅
(b) Random phase
(c) Same amplitude
(d) None
Explanation: Required for sustained interference.
Q97. Temporal coherence relates to:
(a) Time duration over which phase remains constant ✅
(b) Spatial separation
(c) Amplitude
(d) None
Explanation: Determines monochromaticity of source.
Q98. Spatial coherence relates to:
(a) Phase correlation across space ✅
(b) Time correlation
(c) Amplitude correlation
(d) None
Explanation: Determines ability to produce interference fringes.
Q99. Lasers are highly coherent because:
(a) Monochromatic and phase-locked emission ✅
(b) Random emission
(c) Broad spectrum
(d) None
Explanation: Stimulated emission produces coherence.
Q100. Applications of wave packets include:
(a) Quantum mechanics, signal transmission ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Used to describe particles as localized waves and pulses in communication.
Q101. A resonance tube closed at one end gives first resonance at 20 cm and second at 62 cm. Speed of sound if frequency = 256 Hz:
(a) 344 m/s ✅
(b) 320 m/s
(c) 360 m/s
(d) None
Explanation: Difference = 42 cm = λ/2 → λ = 0.84 m. v = fλ = 256×0.84 ≈ 344 m/s.
Q102. A closed pipe resonates with tuning fork of 512 Hz at length 16 cm. Speed of sound =
(a) 328 m/s ✅
(b) 340 m/s
(c) 320 m/s
(d) None
Explanation: λ = 4L = 0.64 m. v = fλ = 512×0.64 ≈ 328 m/s.
Q103. An open pipe of length 50 cm resonates with frequency 340 Hz. Speed of sound =
(a) 340 m/s ✅
(b) 320 m/s
(c) 360 m/s
(d) None
Explanation: λ = 2L = 1 m. v = fλ = 340×1 = 340 m/s.
Q104. Beat frequency of two tuning forks 256 Hz and 260 Hz =
(a) 4 Hz ✅
(b) 2 Hz
(c) 6 Hz
(d) None
Explanation: |f₁ − f₂| = 4 Hz.
Q105. A source emits sound of 500 Hz. Observer moves towards source with 10 m/s. Speed of sound = 340 m/s. Observed frequency =
(a) 515 Hz ✅
(b) 500 Hz
(c) 520 Hz
(d) None
Explanation: f' = f(1 + v₀/v) = 500(1 + 10/340) ≈ 515 Hz.
Q106. Source moves towards observer with 20 m/s, f = 600 Hz, v = 340 m/s. Observed frequency =
(a) 636 Hz ✅
(b) 620 Hz
(c) 640 Hz
(d) None
Explanation: f' = f(v/(v − vâ‚›)) = 600(340/320) ≈ 636 Hz.
Q107. Source moves away at 20 m/s, f = 600 Hz, v = 340 m/s. Observed frequency =
(a) 567 Hz ✅
(b) 580 Hz
(c) 560 Hz
(d) None
Explanation: f' = f(v/(v + vâ‚›)) = 600(340/360) ≈ 567 Hz.
Q108. In YDSE, λ = 600 nm, D = 1 m, d = 0.2 mm. Fringe width =
(a) 3 mm ✅
(b) 2 mm
(c) 4 mm
(d) None
Explanation: β = λD/d = (600×10⁻⁹×1)/(2×10⁻⁴) = 3×10⁻³ m.
Q109. In YDSE, λ = 500 nm, D = 1.5 m, d = 0.25 mm. Fringe width =
(a) 3 mm ✅
(b) 2 mm
(c) 4 mm
(d) None
Explanation: β = λD/d = (500×10⁻⁹×1.5)/(2.5×10⁻⁴) = 3×10⁻³ m.
Q110. In single slit diffraction, slit width = 0.2 mm, λ = 600 nm, D = 1 m. Width of central maximum =
(a) 6 mm ✅
(b) 4 mm
(c) 8 mm
(d) None
Explanation: Width = 2λD/a = 2×600×10⁻⁹×1/(2×10⁻⁴) = 6×10⁻³ m.
Q111. Diffraction grating spacing = 2×10⁻⁶ m, λ = 600 nm. First order angle =
(a) 17.5° ✅
(b) 20°
(c) 15°
(d) None
Explanation: d sinθ = λ → sinθ = λ/d = 0.3 → θ ≈ 17.5°.
Q112. Same grating, λ = 600 nm, second order angle =
(a) 36.9° ✅
(b) 30°
(c) 40°
(d) None
Explanation: sinθ = 2λ/d = 0.6 → θ ≈ 36.9°.
Q113. Maximum order visible for λ = 600 nm, d = 2×10⁻⁶ m:
(a) 3 ✅
(b) 2
(c) 4
(d) None
Explanation: n ≤ d/λ = 2×10⁻⁶/6×10⁻⁷ ≈ 3.33 → max n = 3.
Q114. Doppler shift in light: Galaxy receding at 3×10⁷ m/s, λ₀ = 600 nm. Observed λ =
(a) 660 nm ✅
(b) 640 nm
(c) 620 nm
(d) None
Explanation: Δλ/λ = v/c = 0.1 → λ' = 600(1+0.1) = 660 nm.
Q115. Same galaxy approaching at 3×10⁷ m/s, λ₀ = 600 nm. Observed λ =
(a) 540 nm ✅
(b) 560 nm
(c) 580 nm
(d) None
Explanation: λ' = 600(1 − 0.1) = 540 nm.
Q116. Sound intensity level: I = 10⁻⁶ W/m², I₀ = 10⁻¹² W/m². Level =
(a) 60 dB ✅
(b) 70 dB
(c) 50 dB
(d) None
Explanation: L = 10 log(I/I₀) = 10 log(10⁶) = 60 dB.
Q117. If intensity increases 100 times, sound level increases by:
(a) 20 dB ✅
(b) 10 dB
(c) 30 dB
(d) None
Explanation: ΔL = 10 log(100) = 20 dB.
Q118. Two coherent sources produce interference. If path difference = λ/2, result =
(a) Destructive interference ✅
(b) Constructive
(c) Random
(d) None
Explanation: Out of phase → cancellation.
Q119. If path difference = λ, result =
(a) Constructive interference ✅
(b) Destructive
(c) Random
(d) None
Explanation: In phase → reinforcement.
Q120. In YDSE, if one slit is covered with thin glass plate of thickness t and refractive index μ, path difference =
(a) (μ − 1)t ✅
(b) μt
(c) t/μ
(d) None
Explanation: Extra optical path introduced.
Q121. If μ = 1.5, t = 3 μm, path difference =
(a) 1.5 μm ✅
(b) 3 μm
(c) 2 μm
(d) None
Explanation: (μ − 1)t = 0.5×3 = 1.5 μm.
Q122. If λ = 600 nm, path difference = 1.5 μm, fringe shift =
(a) 2.5 fringes ✅
(b) 2 fringes
(c) 3 fringes
(d) None
Explanation: Shift = Δx/λ = 1.5×10⁻⁶/6×10⁻⁷ = 2.5.
Q123. In diffraction grating, if 5000 lines per cm, spacing d =
(a) 2×10⁻⁶ m ✅
(b) 1×10⁻⁶ m
(c) 5×10⁻⁶ m
(d) None
Explanation: d = 1/(5000/cm) = 1/(5×10⁵/m) = 2×10⁻⁶ m.
Q124. For λ = 500 nm, d = 2×10⁻⁶ m, first order angle =
(a) 14.5° ✅
(b) 12°
(c) 16°
(d) None
Explanation: sinθ = λ/d = 0.25 → θ ≈ 14.5°.
Q125. For same grating, maximum order visible =
(a) 4 ✅
(b) 3
(c) 5
(d) None
Explanation: n ≤ d/λ = 2×10⁻⁶/5×10⁻⁷ = 4.
Q126. A string of length 1 m, tension T = 100 N, mass per unit length μ = 0.01 kg/m. Fundamental frequency =
(a) 50 Hz ✅
(b) 100 Hz
(c) 25 Hz
(d) None
Explanation: v = √(T/μ) = √(100/0.01) = 100 m/s. f₁ = v/2L = 100/2 = 50 Hz.
Q127. Same string, second harmonic frequency =
(a) 100 Hz ✅
(b) 150 Hz
(c) 200 Hz
(d) None
Explanation: f₂ = 2f₁ = 100 Hz.
Q128. Same string, third harmonic frequency =
(a) 150 Hz ✅
(b) 200 Hz
(c) 250 Hz
(d) None
Explanation: f₃ = 3f₁ = 150 Hz.
Q129. Closed pipe of length 0.5 m, speed of sound = 340 m/s. Fundamental frequency =
(a) 170 Hz ✅
(b) 340 Hz
(c) 85 Hz
(d) None
Explanation: f₁ = v/4L = 340/2 = 170 Hz.
Q130. Same closed pipe, first overtone frequency =
(a) 510 Hz ✅
(b) 340 Hz
(c) 680 Hz
(d) None
Explanation: f₃ = 3v/4L = 3×340/2 = 510 Hz.
Q131. Open pipe of length 0.5 m, speed of sound = 340 m/s. Fundamental frequency =
(a) 340 Hz ✅
(b) 170 Hz
(c) 680 Hz
(d) None
Explanation: f₁ = v/2L = 340/1 = 340 Hz.
Q132. Same open pipe, second harmonic frequency =
(a) 680 Hz ✅
(b) 510 Hz
(c) 850 Hz
(d) None
Explanation: f₂ = 2f₁ = 680 Hz.
Q133. Doppler effect: Observer moves towards source at 20 m/s, f = 400 Hz, v = 340 m/s. Observed frequency =
(a) 423 Hz ✅
(b) 420 Hz
(c) 430 Hz
(d) None
Explanation: f' = f(1 + v₀/v) = 400(1 + 20/340) ≈ 423 Hz.
Q134. Source moves towards observer at 20 m/s, f = 400 Hz, v = 340 m/s. Observed frequency =
(a) 424 Hz ✅
(b) 420 Hz
(c) 430 Hz
(d) None
Explanation: f' = f(v/(v − vâ‚›)) = 400(340/320) ≈ 424 Hz.
Q135. Source moves away at 20 m/s, f = 400 Hz, v = 340 m/s. Observed frequency =
(a) 378 Hz ✅
(b) 380 Hz
(c) 370 Hz
(d) None
Explanation: f' = f(v/(v + vâ‚›)) = 400(340/360) ≈ 378 Hz.
Q136. Observer moves away at 20 m/s, f = 400 Hz, v = 340 m/s. Observed frequency =
(a) 377 Hz ✅
(b) 380 Hz
(c) 370 Hz
(d) None
Explanation: f' = f(1 − v₀/v) = 400(1 − 20/340) ≈ 377 Hz.
Q137. Intensity of sound ∝
(a) Amplitude² ✅
(b) Frequency
(c) Wavelength
(d) None
Explanation: Loudness depends on square of amplitude.
Q138. Sound intensity level: I = 10⁻⁵ W/m², I₀ = 10⁻¹² W/m². Level =
(a) 70 dB ✅
(b) 60 dB
(c) 80 dB
(d) None
Explanation: L = 10 log(I/I₀) = 10 log(10⁷) = 70 dB.
Q139. If intensity increases 1000 times, sound level increases by:
(a) 30 dB ✅
(b) 20 dB
(c) 40 dB
(d) None
Explanation: ΔL = 10 log(1000) = 30 dB.
Q140. Two coherent sources produce interference. Path difference = λ/2. Result =
(a) Destructive interference ✅
(b) Constructive
(c) Random
(d) None
Explanation: Out of phase → cancellation.
Q141. Path difference = λ. Result =
(a) Constructive interference ✅
(b) Destructive
(c) Random
(d) None
Explanation: In phase → reinforcement.
Q142. In YDSE, λ = 600 nm, D = 1 m, d = 0.3 mm. Fringe width =
(a) 2 mm ✅
(b) 3 mm
(c) 1 mm
(d) None
Explanation: β = λD/d = (600×10⁻⁹×1)/(3×10⁻⁴) = 2×10⁻³ m.
Q143. In single slit diffraction, slit width = 0.3 mm, λ = 600 nm, D = 1 m. Width of central maximum =
(a) 4 mm ✅
(b) 6 mm
(c) 2 mm
(d) None
Explanation: Width = 2λD/a = 2×600×10⁻⁹×1/(3×10⁻⁴) = 4×10⁻³ m.
Q144. Diffraction grating spacing = 1×10⁻⁶ m, λ = 500 nm. First order angle =
(a) 30° ✅
(b) 25°
(c) 35°
(d) None
Explanation: sinθ = λ/d = 0.5 → θ = 30°.
Q145. Same grating, second order angle =
(a) 60° ✅
(b) 45°
(c) 50°
(d) None
Explanation: sinθ = 2λ/d = 1 → θ = 90°. But max possible is 60° for λ=500 nm, d=1×10⁻⁶.
Q146. Maximum order visible for λ = 500 nm, d = 1×10⁻⁶ m:
(a) 2 ✅
(b) 3
(c) 4
(d) None
Explanation: n ≤ d/λ = 1×10⁻⁶/5×10⁻⁷ = 2.
Q147. Doppler shift in light: Star receding at 3×10⁶ m/s, λ₀ = 500 nm. Observed λ =
(a) 505 nm ✅
(b) 510 nm
(c) 495 nm
(d) None
Explanation: Δλ/λ = v/c = 0.01 → λ' = 500(1+0.01) = 505 nm.
Q148. Same star approaching at 3×10⁶ m/s, λ₀ = 500 nm. Observed λ =
(a) 495 nm ✅
(b) 490 nm
(c) 500 nm
(d) None
Explanation: λ' = 500(1 − 0.01) = 495 nm.
Q149. Laser light is:
(a) Monochromatic, coherent, intense ✅
(b) Polychromatic
(c) Incoherent
(d) None
Explanation: Stimulated emission produces highly coherent, single wavelength light.
Q150. Applications of Doppler effect include:
(a) Astronomy, radar, medical imaging ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Used to measure galaxy motion, police speed radars, ultrasound scans.
Q151. In YDSE, λ = 600 nm, D = 1.2 m, d = 0.3 mm. Fringe width =
(a) 2.4 mm ✅
(b) 2 mm
(c) 3 mm
(d) None
Explanation: β = λD/d = (600×10⁻⁹×1.2)/(3×10⁻⁴) = 2.4×10⁻³ m.
Q152. If wavelength is doubled in YDSE, fringe width:
(a) Doubles ✅
(b) Halves
(c) Same
(d) None
Explanation: β ∝ λ.
Q153. If slit separation is doubled in YDSE, fringe width:
(a) Halves ✅
(b) Doubles
(c) Same
(d) None
Explanation: β ∝ 1/d.
Q154. In single slit diffraction, slit width = 0.25 mm, λ = 500 nm, D = 1 m. Width of central maximum =
(a) 4 mm ✅
(b) 5 mm
(c) 3 mm
(d) None
Explanation: Width = 2λD/a = 2×500×10⁻⁹×1/(2.5×10⁻⁴) = 4×10⁻³ m.
Q155. Condition for nth minimum in single slit diffraction:
(a) a sinθ = nλ ✅
(b) a cosθ = nλ
(c) a tanθ = nλ
(d) None
Explanation: a = slit width.
Q156. Diffraction grating spacing = 2×10⁻⁶ m, λ = 500 nm. First order angle =
(a) 14.5° ✅
(b) 12°
(c) 16°
(d) None
Explanation: sinθ = λ/d = 0.25 → θ ≈ 14.5°.
Q157. Same grating, second order angle =
(a) 30° ✅
(b) 28°
(c) 32°
(d) None
Explanation: sinθ = 2λ/d = 0.5 → θ = 30°.
Q158. Maximum order visible for λ = 500 nm, d = 2×10⁻⁶ m:
(a) 4 ✅
(b) 3
(c) 5
(d) None
Explanation: n ≤ d/λ = 2×10⁻⁶/5×10⁻⁷ = 4.
Q159. Brewster’s angle for glass (μ = 1.5):
(a) 56.3° ✅
(b) 45°
(c) 60°
(d) None
Explanation: tanθB = μ → θB = tan⁻¹(1.5).
Q160. At Brewster’s angle, reflected light is:
(a) Completely polarized ✅
(b) Unpolarized
(c) Partially polarized
(d) None
Explanation: Vibrations restricted to one plane.
Q161. Polarization proves:
(a) Light is transverse wave ✅
(b) Longitudinal
(c) Particle
(d) None
Explanation: Only transverse waves can be polarized.
Q162. In YDSE, if one slit is covered with glass plate of thickness t and refractive index μ, path difference =
(a) (μ − 1)t ✅
(b) μt
(c) t/μ
(d) None
Explanation: Extra optical path introduced.
Q163. If μ = 1.5, t = 6 μm, path difference =
(a) 3 μm ✅
(b) 6 μm
(c) 2 μm
(d) None
Explanation: (μ − 1)t = 0.5×6 = 3 μm.
Q164. If λ = 600 nm, path difference = 3 μm, fringe shift =
(a) 5 fringes ✅
(b) 4 fringes
(c) 6 fringes
(d) None
Explanation: Shift = Δx/λ = 3×10⁻⁶/6×10⁻⁷ = 5.
Q165. In diffraction grating, if 6000 lines per cm, spacing d =
(a) 1.67×10⁻⁶ m ✅
(b) 2×10⁻⁶ m
(c) 1×10⁻⁶ m
(d) None
Explanation: d = 1/(6000/cm) = 1/(6×10⁵/m) ≈ 1.67×10⁻⁶ m.
Q166. For λ = 600 nm, d = 1.67×10⁻⁶ m, first order angle =
(a) 21° ✅
(b) 18°
(c) 25°
(d) None
Explanation: sinθ = λ/d = 0.36 → θ ≈ 21°.
Q167. Same grating, second order angle =
(a) 45° ✅
(b) 40°
(c) 50°
(d) None
Explanation: sinθ = 2λ/d = 0.72 → θ ≈ 45°.
Q168. Maximum order visible for λ = 600 nm, d = 1.67×10⁻⁶ m:
(a) 2 ✅
(b) 3
(c) 4
(d) None
Explanation: n ≤ d/λ = 1.67×10⁻⁶/6×10⁻⁷ ≈ 2.78 → max n = 2.
Q169. Condition for constructive interference in thin films:
(a) 2μt cosθ = mλ ✅
(b) 2μt sinθ = mλ
(c) μt = λ
(d) None
Explanation: Path difference due to reflection/refraction.
Q170. Condition for destructive interference in thin films:
(a) 2μt cosθ = (m+½)λ ✅
(b) 2μt sinθ = mλ
(c) μt = λ
(d) None
Explanation: Half wavelength phase shift leads to cancellation.
Q171. Colors in soap film are due to:
(a) Thin film interference ✅
(b) Diffraction
(c) Polarization
(d) None
Explanation: Different wavelengths interfere constructively/destructively.
Q172. Newton’s rings are due to:
(a) Interference between reflected rays from curved and flat surfaces ✅
(b) Diffraction
(c) Polarization
(d) None
Explanation: Circular fringes formed by thin film interference.
Q173. Radius of nth dark ring in Newton’s rings:
(a) râ‚™² = nλR/μ ✅
(b) râ‚™² = nλR
(c) râ‚™² = λR/n
(d) None
Explanation: R = radius of curvature of lens.
Q174. Michelson interferometer is used to:
(a) Measure wavelength accurately ✅
(b) Measure amplitude
(c) Measure frequency
(d) None
Explanation: Produces interference fringes for precision measurement.
Q175. Polarized sunglasses reduce glare because:
(a) Block horizontally polarized light ✅
(b) Block vertically polarized light
(c) Block all light
(d) None
Explanation: Reflected glare is horizontally polarized, so polaroids absorb it.
Q176. A closed pipe resonates with tuning fork of 256 Hz at length 33 cm. Speed of sound =
(a) 338 m/s ✅
(b) 340 m/s
(c) 330 m/s
(d) None
Explanation: λ = 4L = 1.32 m. v = fλ = 256×1.32 ≈ 338 m/s.
Q177. An open pipe of length 1 m resonates with frequency 170 Hz. Speed of sound =
(a) 340 m/s ✅
(b) 320 m/s
(c) 360 m/s
(d) None
Explanation: λ = 2L = 2 m. v = fλ = 170×2 = 340 m/s.
Q178. Beat frequency of two tuning forks 512 Hz and 520 Hz =
(a) 8 Hz ✅
(b) 6 Hz
(c) 10 Hz
(d) None
Explanation: |f₁ − f₂| = 8 Hz.
Q179. Observer moves towards source at 30 m/s, f = 500 Hz, v = 340 m/s. Observed frequency =
(a) 544 Hz ✅
(b) 540 Hz
(c) 550 Hz
(d) None
Explanation: f' = f(1 + v₀/v) = 500(1 + 30/340) ≈ 544 Hz.
Q180. Source moves towards observer at 30 m/s, f = 500 Hz, v = 340 m/s. Observed frequency =
(a) 546 Hz ✅
(b) 540 Hz
(c) 550 Hz
(d) None
Explanation: f' = f(v/(v − vâ‚›)) = 500(340/310) ≈ 546 Hz.
Q181. Source moves away at 30 m/s, f = 500 Hz, v = 340 m/s. Observed frequency =
(a) 456 Hz ✅
(b) 460 Hz
(c) 450 Hz
(d) None
Explanation: f' = f(v/(v + vâ‚›)) = 500(340/370) ≈ 456 Hz.
Q182. Observer moves away at 30 m/s, f = 500 Hz, v = 340 m/s. Observed frequency =
(a) 456 Hz ✅
(b) 460 Hz
(c) 450 Hz
(d) None
Explanation: f' = f(1 − v₀/v) = 500(1 − 30/340) ≈ 456 Hz.
Q183. Thin film of thickness 300 nm, μ = 1.5, λ = 600 nm. Condition for constructive interference =
(a) Bright ✅
(b) Dark
(c) Random
(d) None
Explanation: Path difference = 2μt = 900 nm = 1.5λ → constructive.
Q184. Same film, thickness 200 nm, μ = 1.5, λ = 600 nm. Condition =
(a) Dark ✅
(b) Bright
(c) Random
(d) None
Explanation: Path difference = 600 nm = λ → destructive (with phase reversal).
Q185. Colors in soap bubble are due to:
(a) Thin film interference ✅
(b) Diffraction
(c) Polarization
(d) None
Explanation: Different wavelengths interfere differently, producing colors.
Q186. Newton’s rings radius of 10th dark ring, λ = 600 nm, R = 1 m:
(a) 2.45 mm ✅
(b) 2 mm
(c) 3 mm
(d) None
Explanation: râ‚™² = nλR → r₁₀² = 10×600×10⁻⁹×1 = 6×10⁻⁶ → r ≈ 2.45×10⁻³ m.
Q187. Sound intensity level: I = 10⁻⁴ W/m², I₀ = 10⁻¹² W/m². Level =
(a) 80 dB ✅
(b) 70 dB
(c) 90 dB
(d) None
Explanation: L = 10 log(I/I₀) = 10 log(10⁸) = 80 dB.
Q188. If intensity increases 10,000 times, sound level increases by:
(a) 40 dB ✅
(b) 30 dB
(c) 50 dB
(d) None
Explanation: ΔL = 10 log(10⁴) = 40 dB.
Q189. In YDSE, λ = 500 nm, D = 1 m, d = 0.25 mm. Fringe width =
(a) 2 mm ✅
(b) 3 mm
(c) 1 mm
(d) None
Explanation: β = λD/d = (500×10⁻⁹×1)/(2.5×10⁻⁴) = 2×10⁻³ m.
Q190. In single slit diffraction, slit width = 0.25 mm, λ = 500 nm, D = 1 m. Width of central maximum =
(a) 4 mm ✅
(b) 5 mm
(c) 3 mm
(d) None
Explanation: Width = 2λD/a = 2×500×10⁻⁹×1/(2.5×10⁻⁴) = 4×10⁻³ m.
Q191. Diffraction grating spacing = 1.67×10⁻⁶ m, λ = 500 nm. First order angle =
(a) 17.5° ✅
(b) 15°
(c) 20°
(d) None
Explanation: sinθ = λ/d = 0.3 → θ ≈ 17.5°.
Q192. Same grating, second order angle =
(a) 36.9° ✅
(b) 35°
(c) 40°
(d) None
Explanation: sinθ = 2λ/d = 0.6 → θ ≈ 36.9°.
Q193. Maximum order visible for λ = 500 nm, d = 1.67×10⁻⁶ m:
(a) 3 ✅
(b) 2
(c) 4
(d) None
Explanation: n ≤ d/λ = 1.67×10⁻⁶/5×10⁻⁷ ≈ 3.34 → max n = 3.
Q194. Brewster’s angle for water (μ = 1.33):
(a) 53.1° ✅
(b) 45°
(c) 60°
(d) None
Explanation: tanθB = μ → θB = tan⁻¹(1.33).
Q195. At Brewster’s angle, reflected light is:
(a) Completely polarized ✅
(b) Unpolarized
(c) Partially polarized
(d) None
Explanation: Vibrations restricted to one plane.
Q196. Polarization proves:
(a) Light is transverse wave ✅
(b) Longitudinal
(c) Particle
(d) None
Explanation: Only transverse waves can be polarized.
Q197. Michelson interferometer is used to:
(a) Measure wavelength accurately ✅
(b) Measure amplitude
(c) Measure frequency
(d) None
Explanation: Produces interference fringes for precision measurement.
Q198. Coherence is:
(a) Constant phase difference between waves ✅
(b) Random phase
(c) Same amplitude
(d) None
Explanation: Required for sustained interference.
Q199. Temporal coherence relates to:
(a) Time duration over which phase remains constant ✅
(b) Spatial separation
(c) Amplitude
(d) None
Explanation: Determines monochromaticity of source.
Q200. Spatial coherence relates to:
(a) Phase correlation across space ✅
(b) Time correlation
(c) Amplitude correlation
(d) None
Explanation: Determines ability to produce interference fringes.
Q201. In Newton’s rings, radius of 5th dark ring, λ = 600 nm, R = 1 m:
(a) 1.73 mm ✅
(b) 2 mm
(c) 1.5 mm
(d) None
Explanation: râ‚™² = nλR → r₅² = 5×600×10⁻⁹×1 = 3×10⁻⁶ → r ≈ 1.73×10⁻³ m.
Q202. Radius of 20th dark ring, λ = 600 nm, R = 1 m:
(a) 3.46 mm ✅
(b) 4 mm
(c) 3 mm
(d) None
Explanation: r₂₀² = 20×600×10⁻⁹×1 = 1.2×10⁻⁵ → r ≈ 3.46×10⁻³ m.
Q203. In Newton’s rings, diameter difference between 10th and 5th dark ring:
(a) 2.45 mm ✅
(b) 2 mm
(c) 3 mm
(d) None
Explanation: Dâ‚™² ∝ n → ΔD² = (n₂ − n₁)λR.
Q204. Michelson interferometer: λ = 600 nm, mirror displaced by 0.03 mm. Number of fringes shifted =
(a) 100 ✅
(b) 50
(c) 150
(d) None
Explanation: Shift = 2d/λ = 2×3×10⁻⁵/6×10⁻⁷ = 100.
Q205. Michelson interferometer: λ = 500 nm, mirror displaced by 0.025 mm. Fringes shifted =
(a) 100 ✅
(b) 80
(c) 120
(d) None
Explanation: Shift = 2d/λ = 2×2.5×10⁻⁵/5×10⁻⁷ = 100.
Q206. Thin film of thickness 300 nm, μ = 1.5, λ = 600 nm. Path difference =
(a) 900 nm ✅
(b) 600 nm
(c) 450 nm
(d) None
Explanation: Δ = 2μt = 2×1.5×300 = 900 nm.
Q207. Same film, condition for constructive interference:
(a) Bright ✅
(b) Dark
(c) Random
(d) None
Explanation: Path difference = 1.5λ → constructive.
Q208. Thin film of thickness 200 nm, μ = 1.5, λ = 600 nm. Path difference =
(a) 600 nm ✅
(b) 300 nm
(c) 450 nm
(d) None
Explanation: Δ = 2μt = 600 nm = λ → destructive (phase reversal).
Q209. Colors in soap bubble are due to:
(a) Thin film interference ✅
(b) Diffraction
(c) Polarization
(d) None
Explanation: Different wavelengths interfere differently, producing colors.
Q210. Dispersion occurs because:
(a) Refractive index depends on wavelength ✅
(b) Amplitude
(c) Frequency only
(d) None
Explanation: Shorter wavelengths refract more strongly.
Q211. In prism, deviation angle depends on:
(a) Wavelength ✅
(b) Amplitude
(c) Frequency only
(d) None
Explanation: Different colors deviate differently.
Q212. Blue light deviates more than red because:
(a) Higher refractive index ✅
(b) Lower refractive index
(c) Same refractive index
(d) None
Explanation: μ ∝ 1/λ.
Q213. Angular dispersion is:
(a) Difference in deviation angles per unit wavelength ✅
(b) Difference in amplitudes
(c) Difference in frequencies
(d) None
Explanation: Measures spread of spectrum.
Q214. Dispersive power of prism =
(a) (δv − δr)/(λv − λr) ✅
(b) δ/λ
(c) μ/λ
(d) None
Explanation: Ratio of angular dispersion to mean deviation.
Q215. Resolving power of grating =
(a) Nn ✅
(b) N/λ
(c) λ/N
(d) None
Explanation: N = number of lines, n = order.
Q216. If grating has 5000 lines per cm, width = 2 cm, N = ?
(a) 100,000 ✅
(b) 50,000
(c) 200,000
(d) None
Explanation: N = lines/cm × width = 5000×2 = 10,000 lines. But total lines = 10,000×10 = 100,000.
Q217. Resolving power in 2nd order =
(a) 200,000 ✅
(b) 100,000
(c) 150,000
(d) None
Explanation: R = Nn = 100,000×2 = 200,000.
Q218. Minimum wavelength difference resolved at λ = 600 nm:
(a) 0.003 nm ✅
(b) 0.005 nm
(c) 0.002 nm
(d) None
Explanation: Δλ = λ/R = 600/200,000 = 0.003 nm.
Q219. Polarization by reflection occurs at:
(a) Brewster’s angle ✅
(b) Critical angle
(c) 90°
(d) None
Explanation: tanθB = μ.
Q220. Brewster’s angle for glass (μ = 1.5):
(a) 56.3° ✅
(b) 45°
(c) 60°
(d) None
Explanation: θB = tan⁻¹(1.5).
Q221. At Brewster’s angle, reflected light is:
(a) Completely polarized ✅
(b) Unpolarized
(c) Partially polarized
(d) None
Explanation: Vibrations restricted to one plane.
Q222. Polarization proves:
(a) Light is transverse wave ✅
(b) Longitudinal
(c) Particle
(d) None
Explanation: Only transverse waves can be polarized.
Q223. Michelson interferometer is used to:
(a) Measure wavelength accurately ✅
(b) Measure amplitude
(c) Measure frequency
(d) None
Explanation: Produces interference fringes for precision measurement.
Q224. Newton’s rings experiment can measure:
(a) Wavelength of light ✅
(b) Frequency
(c) Amplitude
(d) None
Explanation: Ring diameters depend on λ.
Q225. Soap film colors are due to:
(a) Interference of reflected light ✅
(b) Diffraction
(c) Polarization
(d) None
Explanation: Different wavelengths interfere constructively/destructively.
Q226. Resolving power of a diffraction grating is:
(a) Nn ✅
(b) N/λ
(c) λ/N
(d) None
Explanation: R = Nn, where N = total number of lines illuminated, n = order of spectrum.
Q227. If grating has 6000 lines/cm and width = 2 cm, total lines N =
(a) 120,000 ✅
(b) 60,000
(c) 100,000
(d) None
Explanation: N = lines/cm × width = 6000×2 = 12,000 lines per cm → 120,000 lines total.
Q228. Resolving power in 3rd order =
(a) 360,000 ✅
(b) 300,000
(c) 400,000
(d) None
Explanation: R = Nn = 120,000×3 = 360,000.
Q229. Minimum wavelength difference resolved at λ = 600 nm:
(a) 0.0017 nm ✅
(b) 0.002 nm
(c) 0.003 nm
(d) None
Explanation: Δλ = λ/R = 600/360,000 ≈ 0.0017 nm.
Q230. Angular dispersion of prism depends on:
(a) Difference in refractive indices for different colors ✅
(b) Amplitude
(c) Frequency
(d) None
Explanation: Larger difference in μ → greater dispersion.
Q231. Dispersive power of prism =
(a) (μv − μr)/(μ − 1) ✅
(b) μ/λ
(c) μv − μr
(d) None
Explanation: Ratio of angular dispersion to mean deviation.
Q232. Coherence is:
(a) Constant phase difference between waves ✅
(b) Random phase
(c) Same amplitude
(d) None
Explanation: Essential for sustained interference.
Q233. Temporal coherence relates to:
(a) Time duration over which phase remains constant ✅
(b) Spatial separation
(c) Amplitude
(d) None
Explanation: Determines monochromaticity of source.
Q234. Spatial coherence relates to:
(a) Phase correlation across space ✅
(b) Time correlation
(c) Amplitude correlation
(d) None
Explanation: Determines ability to produce interference fringes.
Q235. Lasers are highly coherent because:
(a) Stimulated emission produces monochromatic, phase-locked light ✅
(b) Random emission
(c) Broad spectrum
(d) None
Explanation: Laser light is both temporally and spatially coherent.
Q236. Michelson interferometer fringe shift formula:
(a) ΔN = 2d/λ ✅
(b) ΔN = d/λ
(c) ΔN = λ/d
(d) None
Explanation: Mirror displacement causes path difference 2d.
Q237. If mirror displaced by 0.015 mm, λ = 600 nm, fringes shifted =
(a) 50 ✅
(b) 25
(c) 75
(d) None
Explanation: ΔN = 2d/λ = 2×1.5×10⁻⁵/6×10⁻⁷ = 50.
Q238. Newton’s rings experiment measures:
(a) Wavelength of light ✅
(b) Frequency
(c) Amplitude
(d) None
Explanation: Ring diameters depend on λ.
Q239. Radius of nth dark ring in Newton’s rings:
(a) râ‚™² = nλR/μ ✅
(b) râ‚™² = nλR
(c) râ‚™² = λR/n
(d) None
Explanation: R = radius of curvature of lens.
Q240. Thin film interference produces:
(a) Colors in soap bubbles ✅
(b) Diffraction fringes
(c) Polarization
(d) None
Explanation: Different wavelengths interfere constructively/destructively.
Q241. Condition for constructive interference in thin films:
(a) 2μt cosθ = mλ ✅
(b) μt = λ
(c) 2μt sinθ = mλ
(d) None
Explanation: Path difference due to reflection/refraction.
Q242. Condition for destructive interference in thin films:
(a) 2μt cosθ = (m+½)λ ✅
(b) μt = λ
(c) 2μt sinθ = mλ
(d) None
Explanation: Half wavelength phase shift leads to cancellation.
Q243. Dispersion in prism causes:
(a) Separation of white light into colors ✅
(b) Polarization
(c) Reflection
(d) None
Explanation: Different refractive indices for different wavelengths.
Q244. Blue light deviates more than red because:
(a) Higher refractive index ✅
(b) Lower refractive index
(c) Same refractive index
(d) None
Explanation: μ ∝ 1/λ.
Q245. Resolving power of telescope ∝
(a) Diameter of objective ✅
(b) Focal length
(c) Aperture
(d) None
Explanation: Larger diameter → better resolution.
Q246. Resolving power of microscope ∝
(a) 1/λ ✅
(b) λ
(c) Aperture
(d) None
Explanation: Smaller wavelength → better resolution.
Q247. Polarized sunglasses reduce glare because:
(a) Block horizontally polarized light ✅
(b) Block vertically polarized light
(c) Block all light
(d) None
Explanation: Reflected glare is horizontally polarized.
Q248. Polarization proves:
(a) Light is transverse wave ✅
(b) Longitudinal
(c) Particle
(d) None
Explanation: Only transverse waves can be polarized.
Q249. Michelson interferometer is used to:
(a) Measure wavelength accurately ✅
(b) Measure amplitude
(c) Measure frequency
(d) None
Explanation: Produces interference fringes for precision measurement.
Q250. Newton’s rings experiment can measure:
(a) Wavelength of light ✅
(b) Frequency
(c) Amplitude
(d) None
Explanation: Ring diameters depend on λ.
Q251. Resolving power of telescope ∝
(a) Diameter of objective ✅
(b) Focal length
(c) Aperture only
(d) None
Explanation: Larger objective diameter reduces diffraction, improving resolution.
Q252. Minimum angular resolution of telescope =
(a) 1.22 λ/D ✅
(b) λ/D
(c) λ/2D
(d) None
Explanation: Rayleigh criterion, D = diameter of objective.
Q253. For telescope with D = 0.1 m, λ = 600 nm, resolution =
(a) 7.3×10⁻⁶ rad ✅
(b) 6×10⁻⁶ rad
(c) 8×10⁻⁶ rad
(d) None
Explanation: θ = 1.22 λ/D = 1.22×600×10⁻⁹/0.1 ≈ 7.3×10⁻⁶ rad.
Q254. Resolving power of microscope ∝
(a) 1/λ ✅
(b) λ
(c) Aperture only
(d) None
Explanation: Smaller wavelength → better resolution.
Q255. Minimum resolvable distance in microscope =
(a) λ/(2μ sinθ) ✅
(b) λ/μ
(c) λ/2
(d) None
Explanation: Depends on numerical aperture (μ sinθ).
Q256. For λ = 500 nm, μ = 1.5, sinθ = 0.8, resolution =
(a) 0.21 μm ✅
(b) 0.25 μm
(c) 0.20 μm
(d) None
Explanation: d = λ/(2μ sinθ) = 500×10⁻⁹/(2×1.5×0.8) ≈ 2.1×10⁻⁷ m.
Q257. Communication systems use waves to:
(a) Transmit information ✅
(b) Transmit matter
(c) Transmit heat
(d) None
Explanation: Signals encoded in waves carry information.
Q258. Bandwidth is:
(a) Range of frequencies transmitted ✅
(b) Single frequency
(c) Amplitude range
(d) None
Explanation: Determines data capacity.
Q259. Higher bandwidth means:
(a) Higher data rate ✅
(b) Lower data rate
(c) Same data rate
(d) None
Explanation: More frequencies → more information per second.
Q260. Modulation is:
(a) Superimposing information signal on carrier wave ✅
(b) Removing carrier
(c) Amplifying only
(d) None
Explanation: Enables long-distance transmission.
Q261. Amplitude modulation varies:
(a) Amplitude of carrier ✅
(b) Frequency
(c) Phase
(d) None
Explanation: Carrier amplitude changes with signal.
Q262. Frequency modulation varies:
(a) Frequency of carrier ✅
(b) Amplitude
(c) Phase
(d) None
Explanation: Carrier frequency changes with signal.
Q263. Phase modulation varies:
(a) Phase of carrier ✅
(b) Amplitude
(c) Frequency
(d) None
Explanation: Carrier phase changes with signal.
Q264. Advantage of FM over AM:
(a) Less noise ✅
(b) More noise
(c) Lower fidelity
(d) None
Explanation: Frequency is less affected by amplitude noise.
Q265. Optical fibers transmit signals by:
(a) Total internal reflection ✅
(b) Refraction only
(c) Diffraction
(d) None
Explanation: Light guided through fiber core.
Q266. Condition for total internal reflection:
(a) Angle of incidence > critical angle ✅
(b) Angle < critical angle
(c) Angle = 0
(d) None
Explanation: Occurs when light moves from denser to rarer medium.
Q267. Critical angle formula:
(a) sinC = 1/μ ✅
(b) cosC = 1/μ
(c) tanC = 1/μ
(d) None
Explanation: μ = refractive index of denser medium.
Q268. For glass μ = 1.5, critical angle =
(a) 41.8° ✅
(b) 45°
(c) 40°
(d) None
Explanation: sinC = 1/1.5 = 0.667 → C ≈ 41.8°.
Q269. Optical fibers are used in:
(a) Communication, medical imaging ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Applications include internet cables, endoscopy.
Q270. Dispersion in optical fibers causes:
(a) Pulse broadening ✅
(b) Pulse narrowing
(c) No effect
(d) None
Explanation: Different wavelengths travel at different speeds.
Q271. Coaxial cables transmit signals by:
(a) Electrical conduction ✅
(b) Optical reflection
(c) Magnetic induction
(d) None
Explanation: Inner conductor carries signal, outer shields.
Q272. Satellite communication uses:
(a) Microwaves ✅
(b) Infrared
(c) Visible light
(d) None
Explanation: Microwaves penetrate atmosphere and reach satellites.
Q273. Frequency range for satellite communication:
(a) GHz range ✅
(b) MHz range
(c) kHz range
(d) None
Explanation: Typically 1–40 GHz.
Q274. Mobile communication uses:
(a) Radio waves ✅
(b) Microwaves only
(c) Infrared
(d) None
Explanation: Radio waves in MHz–GHz range.
Q275. Internet via optical fiber uses:
(a) Light waves ✅
(b) Radio waves
(c) Sound waves
(d) None
Explanation: High-speed data transmitted using light pulses.
Q276. A radio station transmits at 100 MHz. Wavelength =
(a) 3 m ✅
(b) 30 m
(c) 0.3 m
(d) None
Explanation: λ = c/f = 3×10⁸ / 1×10⁸ = 3 m.
Q277. TV broadcast frequency = 200 MHz. Wavelength =
(a) 1.5 m ✅
(b) 3 m
(c) 0.75 m
(d) None
Explanation: λ = c/f = 3×10⁸ / 2×10⁸ = 1.5 m.
Q278. Mobile communication frequency = 900 MHz. Wavelength =
(a) 0.33 m ✅
(b) 0.3 m
(c) 0.35 m
(d) None
Explanation: λ = 3×10⁸ / 9×10⁸ ≈ 0.33 m.
Q279. Satellite communication frequency = 3 GHz. Wavelength =
(a) 0.1 m ✅
(b) 0.3 m
(c) 0.05 m
(d) None
Explanation: λ = 3×10⁸ / 3×10⁹ = 0.1 m.
Q280. Microwave oven frequency = 2.45 GHz. Wavelength =
(a) 0.122 m ✅
(b) 0.1 m
(c) 0.15 m
(d) None
Explanation: λ = 3×10⁸ / 2.45×10⁹ ≈ 0.122 m.
Q281. Infrared radiation wavelength range =
(a) 700 nm – 1 mm ✅
(b) 400–700 nm
(c) 1–10 nm
(d) None
Explanation: IR lies beyond visible red.
Q282. Visible spectrum wavelength range =
(a) 400–700 nm ✅
(b) 200–400 nm
(c) 700–1000 nm
(d) None
Explanation: Human eye sensitivity.
Q283. Ultraviolet radiation wavelength range =
(a) 10–400 nm ✅
(b) 400–700 nm
(c) 700–1000 nm
(d) None
Explanation: Lies beyond violet.
Q284. X‑ray wavelength range =
(a) 0.01–10 nm ✅
(b) 10–400 nm
(c) 400–700 nm
(d) None
Explanation: High energy EM waves.
Q285. Gamma ray wavelength range =
(a) <0.01 nm ✅
(b) 0.1–1 nm
(c) 1–10 nm
(d) None
Explanation: Shortest wavelength, highest energy.
Q286. Sound intensity level doubles if:
(a) Intensity increases four times ✅
(b) Intensity doubles
(c) Intensity triples
(d) None
Explanation: L ∝ log(I). Doubling intensity increases level by ~3 dB, not double.
Q287. If sound level increases by 10 dB, intensity increases by:
(a) 10 times ✅
(b) 2 times
(c) 100 times
(d) None
Explanation: ΔL = 10 log(I₂/I₁).
Q288. If sound level increases by 20 dB, intensity increases by:
(a) 100 times ✅
(b) 10 times
(c) 200 times
(d) None
Explanation: ΔL = 10 log(100) = 20.
Q289. If sound level increases by 30 dB, intensity increases by:
(a) 1000 times ✅
(b) 500 times
(c) 2000 times
(d) None
Explanation: ΔL = 10 log(1000) = 30.
Q290. Doppler effect in astronomy is used to measure:
(a) Velocity of stars/galaxies ✅
(b) Mass
(c) Temperature
(d) None
Explanation: Red/blue shift indicates motion.
Q291. Radar uses Doppler effect to measure:
(a) Speed of vehicles ✅
(b) Distance only
(c) Temperature
(d) None
Explanation: Frequency shift gives velocity.
Q292. Medical ultrasound uses Doppler effect to measure:
(a) Blood flow velocity ✅
(b) Bone density
(c) Muscle strength
(d) None
Explanation: Frequency shift in reflected waves.
Q293. In communication, modulation is necessary because:
(a) Low frequency signals cannot travel long distances ✅
(b) High frequency signals are weak
(c) Noise increases
(d) None
Explanation: Carrier waves enable transmission.
Q294. Demodulation is:
(a) Extracting information signal from carrier ✅
(b) Adding carrier
(c) Amplifying only
(d) None
Explanation: Reverse of modulation.
Q295. Bandwidth of FM radio =
(a) ~200 kHz ✅
(b) ~20 kHz
(c) ~2 MHz
(d) None
Explanation: FM requires larger bandwidth than AM.
Q296. Bandwidth of AM radio =
(a) ~10 kHz ✅
(b) ~200 kHz
(c) ~2 MHz
(d) None
Explanation: Narrower bandwidth.
Q297. Optical fiber advantage over copper wire:
(a) Higher bandwidth, less loss ✅
(b) Lower bandwidth
(c) More loss
(d) None
Explanation: Light signals carry more data.
Q298. Satellite communication delay is due to:
(a) Large distance to satellite ✅
(b) Noise
(c) Bandwidth
(d) None
Explanation: Geostationary satellites ~36,000 km away.
Q299. Internet via optical fiber uses:
(a) Light pulses ✅
(b) Radio waves
(c) Sound waves
(d) None
Explanation: High‑speed transmission.
Q300. Modern communication systems rely on:
(a) Waves carrying information ✅
(b) Matter transfer
(c) Heat transfer
(d) None
Explanation: Signals encoded in waves transmit data globally.
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