Wave OpticsNEET MCQs with solutions
Wave Optics covers Huygens' principle, Young's double-slit experiment (YDSE), diffraction, polarisation and interference. NEET tests YDSE fringe width and position, single-slit diffraction, Brewster's angle and conditions for constructive/destructive interference.
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Tap an option to check it. Questions from every NCERT topic in this chapter, from easy to hard.
Q1PYQ Mixed Concepts
An unpolarised light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster's angle. Which statement is correct?
Not quite — the answer is A.
tanθB = μ = √3 gives θB = 60°. Reflected beam is completely plane polarised; transmitted beam is only partially polarised. Angle of reflection equals angle of incidence = 60°. Trap: option B incorrectly claims transmitted light is completely polarised.
Q2PYQ Mixed Concepts
In the phenomenon of interference, light energy at destructive interference points is:
Not quite — the answer is B.
Energy is never destroyed in interference. Dark regions have zero or near-zero intensity because energy is redistributed to bright regions where intensity exceeds the average. Total energy over the entire pattern is conserved. Trap: confusing redistribution with absorption or permanent loss.
Q3Grand Test
Which of the following correctly defines a wavefront?
Not quite — the answer is A.
A wavefront is the locus of all points in a wave oscillating in the same phase at a given instant. Option D describes a consequence in isotropic homogeneous media only — perpendicularity is derived from the equal-phase property, not the definition. Option C describes a ray.
Q4Grand Test
A plane wavefront enters a denser medium where the speed of light decreases. Which of the following correctly describes the change?
Not quite — the answer is C.
Frequency is fixed by the source and never changes at an interface. From v = fλ, since f is constant and v decreases, λ decreases in the same ratio. Speed and wavelength change together; frequency never does. Trap: option A reverses the wavelength change — wavelength decreases, not increases, in a denser medium.
Q5Mixed Revision
Which of the following is the correct decreasing order of wavelength?
Not quite — the answer is A.
The electromagnetic spectrum in decreasing wavelength order is: Radio > Microwave > IR > Visible > UV > X-ray > Gamma. Option C is the trap: it places Visible before IR, reversing their correct order (IR has longer wavelength than visible).
Q6Mixed Revision
Unpolarised light of intensity I₀ passes through a polariser P₁, then through a polariser P₂ whose axis is at 30° to P₁. The final transmitted intensity is:
Not quite — the answer is B.
After P₁: I₁ = I₀/2. After P₂ at 30°: I₂ = I₁cos²30° = (I₀/2)(3/4) = 3I₀/8. Option C (I₀/4) is the trap: students use cos²45° = 1/2 by reflex rather than cos²30° = 3/4 — a direct substitution error with the wrong standard angle.
Q7Reflection and Refraction of Wavefront
A plane wavefront is incident obliquely on a plane mirror. Which statement about the reflected wavefront obtained by Huygens' construction is correct?
Not quite — the answer is A.
Huygens' equal-time construction gives secondary wavelets of equal radius on the reflecting surface; their envelope geometry directly yields i = r. Reflection from a plane surface preserves the plane shape of the wavefront and does not alter frequency.
Q8Reflection and Refraction of Wavefront
A plane wavefront passes normally from air into glass of refractive index 1.5. Which pair correctly describes the changes in the wave?
Not quite — the answer is B.
Speed: v = c/n = 3×10⁸/1.5 = 2×10⁸ m/s. Frequency is unchanged (set by source). Wavelength: λ = v/f = λ₀/n = 2λ₀/3. Option C multiplies c by n — the classic corpuscular-theory error. Option D is wrong because frequency never changes at an interface.
Q9Interference of Light
For two coherent waves to produce a sustained interference pattern, which condition is essential?
Not quite — the answer is C.
Sustained interference requires coherence: same frequency and constant phase difference. Equal amplitudes are not mandatory — unequal amplitudes still give a stable pattern with reduced visibility. A randomly varying phase difference shifts maxima and minima continuously, destroying the pattern.
Q10Interference of Light
Two waves are represented by y₁ = a sin(ωt) and y₂ = a sin(ωt − φ), where φ is a constant. Which statement correctly describes these waves?
Not quite — the answer is B.
Both waves share the same angular frequency ω and have a fixed phase difference φ — satisfying both coherence conditions. Option D is the trap: the phase constant φ does not imply different frequencies; ω is identical in both expressions.
Q11Polarisation of Light
An unpolarised light beam strikes a glass surface at Brewster's angle. Which statement is correct?
Not quite — the answer is A.
At Brewster's angle, reflected light is completely plane-polarised perpendicular to the plane of incidence. Refracted light is only partially polarised; both planes of vibration are present but the in-plane component dominates. Option C is wrong because the refracted beam retains both components.
Q12Polarisation of Light
The Brewster angle for an interface between air and a transparent medium must lie in which range for refractive index of the medium greater than 1?
Not quite — the answer is B.
Brewster's law gives tanθB = μ. For μ > 1, tanθB > 1, so θB > 45°. Since μ is always finite, θB never reaches 90°. Therefore Brewster angle lies strictly between 45° and 90°.
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Get RankUp on Google PlayQ13Young's Double Slit Experiment (YDSE)
Two slits in Young's double slit experiment are 1.5 mm apart and the screen is placed at a distance of 1 m from the slits. If the wavelength of light used is 600 × 10⁻⁹ m, the fringe separation is:
Not quite — the answer is D.
Fringe width β = λD/d = (600×10⁻⁹×1)/(1.5×10⁻³) = 4×10⁻⁴ m. Common trap: forgetting to convert 1.5 mm to 1.5×10⁻³ m, giving a power-of-ten error.
Q14Young's Double Slit Experiment (YDSE)
In a Young's double slit experiment, a student observes 8 fringes in a certain region of the screen using monochromatic light of wavelength 600 nm. If the wavelength is changed to 400 nm, the number of fringes observed in the same region will be:
Not quite — the answer is C.
Since β ∝ λ, fringe width decreases with shorter wavelength, so more fringes fit. N₂ = N₁ × (λ₁/λ₂) = 8 × (600/400) = 12. Option A (6) is the trap for students who apply the ratio inverted.
Q15Diffraction of Light
In single-slit diffraction, the first minimum occurs when the path difference between waves from the extreme ends of the slit is:
Not quite — the answer is A.
For the first minimum, the total path difference across the full slit width is λ, giving condition a sinθ = λ. Option B (λ/2) is the most common trap — it is the path difference used in pairing wavelets for the cancellation argument, not the total across the full slit.
Q16Diffraction of Light
In a single-slit diffraction experiment, the angular width of the central maximum is θ₀ for wavelength 6000 Å. The angular width decreases by 30% when wavelength is changed. The new wavelength is:
Not quite — the answer is C.
Angular width 2θ = 2λ/a ∝ λ for fixed a. A 30% decrease means new width is 70% of original, so λ₂ = 0.70 × 6000 = 4200 Å. Option D (5400 Å) results from applying a 10% decrease instead of 30%.
Q17Huygens Principle and Wavefront
Consider the following statements about Huygens' principle: I. It can determine the new position of a wavefront. II. It can derive Snell's law of refraction. III. It explains the particle nature of light. IV. It can determine the absolute speed of light in vacuum from first principles. How many of the above statements are correct?
Not quite — the answer is B.
Huygens' principle locates new wavefronts via secondary wavelets (I: true) and geometrically derives Snell's law (II: true). It is a wave construction and cannot explain the particle nature of light (III: false) nor independently yield the absolute speed of light (IV: false). Only I and II are correct.
Q18Huygens Principle and Wavefront
A wavefront is best defined as:
Not quite — the answer is A.
A wavefront is the locus of all points at the same phase at a given instant. Amplitude and intensity can vary across a wavefront. Wavefronts are perpendicular, not parallel, to the direction of propagation — option D is a common directional trap.
Q19Huygens Principle and Wavefront
A plane wavefront propagates in a homogeneous isotropic medium. Which of the following correctly describes the relationship between rays and the wavefront?
Not quite — the answer is A.
In a homogeneous isotropic medium, energy flows along rays, which are always normal (perpendicular) to the wavefront. Option B is a common confusion between tangent and normal. Options C and D have no physical basis.
Q20Huygens Principle and Wavefront
A plane wavefront is incident normally on a perfectly homogeneous isotropic medium. As it propagates, the wavefront will:
Not quite — the answer is B.
In a homogeneous isotropic medium, every point of a plane wavefront has the same speed. Secondary wavelets after time Δt all have radius vΔt; their forward envelope is a plane parallel to the original. The wavefront does not curve or change shape.
Q21Huygens Principle and Wavefront
Huygens' construction predicts that when light travels from a rarer medium (refractive index n₁) into a denser medium (refractive index n₂ > n₁), the speed of light in the denser medium is:
Not quite — the answer is C.
Snell's law derived via Huygens gives v₂ = v₁(n₁/n₂). Since n₂ > n₁, v₂ < v₁: light slows down in the denser medium. This directly contradicted Newton's corpuscular prediction that light speeds up in a denser medium, making it a critical result of the wave theory.
Q22Huygens Principle and Wavefront
Which instrument is correctly associated with Huygens in the context of wave optics?
Not quite — the answer is B.
Huygens used telescopic observations of Saturn's rings and developed the wave theory of light to explain his observations, including the nature of wavefronts. The diffraction grating was developed much later (Fraunhofer). Newton used a telescope for optics but proposed the corpuscular theory, not the wave theory.
Q23Huygens Principle and Wavefront
A monochromatic source produces a spherical wavefront in an isotropic medium. The wavefront radius increases from 3.0 m to 3.6 m. If the wave speed is 3.0 × 10⁸ m/s, the time interval required is:
Not quite — the answer is D.
Δr = 3.6 − 3.0 = 0.6 m; Δt = Δr/v = 0.6/(3.0 × 10⁸) = 2.0 × 10⁻⁹ s. Option C results from the common error of dividing 0.3 m (half of Δr) by v; option B from misplacing the power of ten by one.
Q24Huygens Principle and Wavefront
In a homogeneous isotropic medium, a point source emits monochromatic light. Which correctly explains why the spherical wavefront remains spherical as it propagates?
Not quite — the answer is A.
In an isotropic medium, wave speed v is the same in all directions. Every surface point generates a wavelet of equal radius vΔt. The forward envelope of equal-radius wavelets from a spherical surface is a larger concentric sphere. Options C and D introduce amplitude and wavelength incorrectly — neither determines wavefront shape.
ELITE question · AIR under 50 level
This chapter has 173 ELITE questions for students aiming at the very top. They are only in the app.
Unlock ELITE questions in the appKey Wave Optics Concepts
Quick revision: most questions in this chapter test these facts.
| Concept | Key Formula |
|---|---|
| YDSE fringe width | β = λD/d; D = screen distance, d = slit separation |
| Bright fringe | Path diff = nλ (n = 0, 1, 2...); central maximum at n = 0 |
| Dark fringe | Path diff = (2n−1)λ/2; first dark fringe closest to centre |
| Single slit diffraction | First minimum at a sinθ = λ; central maximum width = 2λD/a |
| Brewster's angle | tan θ_B = n₂/n₁; reflected light completely polarised |
| Malus's law | I = I₀ cos²θ; intensity after analyser at angle θ to polariser |
What the app covers in this chapter
468 questions in total, each with a detailed explanation.
| PYQ Mixed Concepts | 96 |
| Grand Test | 78 |
| Mixed Revision | 76 |
| Reflection and Refraction of Wavefront | 40 |
| Interference of Light | 40 |
| Polarisation of Light | 40 |
| Young's Double Slit Experiment (YDSE) | 39 |
| Diffraction of Light | 39 |
| Huygens Principle and Wavefront | 20 |
Questions students ask
Is Wave Optics important for NEET?
Yes — YDSE fringe width, interference conditions, single-slit diffraction and polarisation (Brewster's law, Malus's law) are tested regularly.
Which topics should I revise first?
Focus on YDSE fringe width formula, conditions for bright and dark fringes, single-slit diffraction minima, Brewster's angle, and Malus's law for polarisation intensity.
How many questions from this chapter are on RankUp?
The RankUp app has 468 questions on Wave Optics, including 173 ELITE questions. Every question has a detailed explanation.
