Wave Optics – CBSE NCERT Study Resources

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Wave Optics

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Wave Optics - Grade 12 Physics (CBSE)

Chapter Overview

Wave Optics is a fundamental chapter in Class 12 Physics that explores the wave nature of light. It covers phenomena such as interference, diffraction, and polarization, which cannot be explained by ray optics alone. The chapter establishes Huygens' principle and derives mathematical expressions for various wave optics phenomena.

Wave Optics: The branch of optics that studies light as a wave, explaining phenomena like interference, diffraction, and polarization.

Huygens' Principle

Huygens' principle states that every point on a wavefront acts as a source of secondary wavelets. The new wavefront is the tangential surface to all these secondary wavelets.

Wavefront: A surface where all points have the same phase of vibration.

Reflection and Refraction of Plane Waves

Using Huygens' principle, the laws of reflection and refraction can be derived. The principle explains how wavefronts behave at the interface between two media.

Coherent and Incoherent Sources

Coherent sources have a constant phase difference and produce sustained interference patterns, while incoherent sources have random phase differences.

Coherent Sources: Sources emitting light waves with a constant phase relationship.

Interference of Light

When two coherent light waves superimpose, they produce interference patterns of bright and dark fringes. Young's double-slit experiment demonstrates this phenomenon.

Diffraction of Light

Diffraction is the bending of light around obstacles or through slits. Single-slit diffraction produces a central bright fringe with alternating dark and bright fringes.

Diffraction: The spreading of light waves when they pass through an aperture or around an obstacle.

Polarization

Polarization is the phenomenon where light waves oscillate in a single plane. It demonstrates the transverse wave nature of light.

Polarized Light: Light in which the electric field vector vibrates in a single plane.

Applications of Wave Optics

Wave optics principles are applied in various technologies such as optical instruments, holography, and fiber optics communication.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
Define wavefront in wave optics.
Answer:

A wavefront is the locus of all points in a wave that are in the same phase of vibration.

Question 2:
What is the principle behind Huygens' principle?
Answer:

Every point on a wavefront acts as a secondary source of wavelets spreading in all directions.

Question 3:
State the condition for constructive interference in Young's double-slit experiment.
Answer:

Path difference must be an integral multiple of the wavelength (nλ).

Question 4:
What is the effect on fringe width if the distance between slits is increased in Young's experiment?
Answer:

Fringe width decreases as slit separation increases.

Question 5:
Why does light appear to bend around obstacles?
Answer:

Due to diffraction, light waves spread out when passing edges.

Question 6:
What is the condition for destructive interference in thin films?
Answer:

Path difference must be an odd multiple of half-wavelength ((2n+1)λ/2).

Question 7:
How does the intensity of light vary in diffraction pattern?
Answer:

Intensity decreases as we move away from the central maximum.

Question 8:
What is the shape of wavefront for a point source?
Answer:

Spherical wavefront.

Question 9:
Why are coherent sources necessary for sustained interference?
Answer:

Coherent sources maintain a constant phase difference for observable interference.

Question 10:
What happens to fringe pattern if white light is used in Young's experiment?
Answer:

Central fringe is white, colored fringes appear on both sides.

Question 11:
Define polarization of light waves.
Answer:

Restriction of light wave vibrations to one plane.

Question 12:
What is Brewster's angle?
Answer:

Angle of incidence where reflected light is completely polarized.

Question 13:
Why can't we observe interference pattern from two independent sources?
Answer:

Due to rapidly changing phase difference between independent sources.

Question 14:
What is the effect of increasing wavelength on diffraction pattern?
Answer:

Diffraction pattern becomes wider with increasing wavelength.

Question 15:
State Huygens' principle.
Answer:

Huygens' principle states that every point on a wavefront acts as a secondary source of wavelets, and the new wavefront is the envelope of these secondary wavelets.

Question 16:
What is the condition for constructive interference in Young's double-slit experiment?
Answer:

Constructive interference occurs when the path difference between two waves is an integral multiple of the wavelength (Δx = nλ, where n = 0, 1, 2...).

Question 17:
Name the phenomenon responsible for the bending of light around obstacles.
Answer:

The phenomenon is called diffraction, where light waves bend around edges or pass through narrow slits, causing spreading.

Question 18:
What is the coherent source in interference experiments?
Answer:

A coherent source emits waves with a constant phase difference and the same frequency, essential for sustained interference patterns.

Question 19:
Why is a diffraction pattern more pronounced with a narrower slit?
Answer:

Narrower slits cause greater diffraction because the wavefront spreads more, making the bending effect more visible.

Question 20:
Define polarization of light.
Answer:

Polarization is the restriction of light wave oscillations to a single plane, perpendicular to the direction of propagation.

Question 21:
Why does light not show interference patterns in daily life?
Answer:

Ordinary light sources are incoherent (random phase differences) and have short wave trains, preventing sustained interference.

Question 22:
What is the shape of the wavefront from a point source at a large distance?
Answer:

At a large distance, the wavefront becomes nearly plane due to the curvature reducing over distance.

Question 23:
How does the intensity of light vary in Young's interference pattern?
Answer:

Intensity varies sinusoidally, with maxima at constructive interference and minima at destructive interference points.

Question 24:
What is the significance of Brewster's angle in polarization?
Answer:

At Brewster's angle, reflected light is completely polarized, with its electric field perpendicular to the plane of incidence.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
What is the condition for constructive interference in Young's double-slit experiment?
Answer:

Constructive interference occurs when the path difference between the two waves is an integral multiple of the wavelength (λ).
Mathematically: Δx = nλ, where n is an integer (0, 1, 2, ...).

Question 2:
Why is a monochromatic light source used in interference experiments?
Answer:

A monochromatic light source emits light of a single wavelength, ensuring a consistent phase relationship between waves. This clarity is crucial for observing distinct interference fringes.

Question 3:
State Huygens' principle and its significance in wave optics.
Answer:

Huygens' principle states that every point on a wavefront acts as a secondary source of wavelets spreading in all directions.
Significance: It explains wave propagation, reflection, refraction, and diffraction.

Question 4:
What is the effect on fringe width if the distance between the slits is increased in Young's experiment?
Answer:

The fringe width (β) decreases as the slit separation (d) increases.
Formula: β = λD/d, where D is the screen distance and λ is the wavelength.

Question 5:
Differentiate between interference and diffraction.
Answer:
  • Interference: Superposition of waves from two or more coherent sources.
  • Diffraction: Bending of waves around obstacles or through slits, producing fringes from a single source.
Question 6:
Explain why diffraction effects are more pronounced with sound waves than light waves.
Answer:

Sound waves have longer wavelengths (cm to m) compared to light (nm), making them more likely to bend around obstacles, as diffraction is significant when the wavelength is comparable to the obstacle size.

Question 7:
What is the shape of the wavefront when light diverges from a point source?
Answer:

The wavefront is spherical, as all points equidistant from the source lie on a sphere.

Question 8:
How does the intensity of light vary in Young's double-slit interference pattern?
Answer:

Intensity varies between maximum (constructive interference) and zero (destructive interference).
Formula: I = 4I0cos²(πΔx/λ), where I0 is intensity from one slit.

Question 9:
Define resolving power of an optical instrument.
Answer:

Resolving power is the ability to distinguish two closely spaced objects. For a telescope, it is inversely proportional to the smallest angular separation (θ) it can resolve: R = 1/θ.

Question 10:
Why does the sky appear blue? Explain using wave optics.
Answer:

Blue light has a shorter wavelength and scatters more (Rayleigh scattering) by atmospheric particles than other colors, making the sky appear blue during the day.

Question 11:
Define coherent sources in the context of wave optics.
Answer:

Coherent sources are sources of light that emit waves with a constant phase difference and the same frequency. These sources are essential for producing sustained interference patterns.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Explain the phenomenon of interference in wave optics with an example.
Answer:

Interference is the phenomenon where two or more coherent waves superimpose to form a resultant wave of greater, lower, or the same amplitude.

Example: When light from a single source splits into two beams (e.g., via Young's double-slit experiment), they recombine to produce bright (constructive interference) and dark (destructive interference) fringes.

Key condition: The waves must be coherent (same frequency and constant phase difference).

Question 2:
What is the significance of Huygens' principle in understanding wave propagation?
Answer:

Huygens' principle states that every point on a wavefront acts as a secondary source of wavelets, which spread out in all directions with the same speed as the original wave.

Significance:

  • Explains reflection, refraction, and diffraction of waves.
  • Helps derive laws like Snell's law of refraction.
  • Predicts wavefront shapes (e.g., spherical for point sources).

Application: Used to analyze light behavior in lenses and mirrors.

Question 3:
Derive the expression for fringe width in Young's double-slit experiment.
Answer:

Fringe width (β) is the distance between two consecutive bright or dark fringes.

Derivation:
Path difference for bright fringe: Δx = nλ = (d/D)y
For adjacent fringes (n=1), y = β
Thus, β = λD/d

Where:

  • λ = wavelength of light
  • D = distance between slits and screen
  • d = slit separation

Question 4:
How does the diffraction pattern of a single slit differ from that of a double slit?
Answer:

Single-slit diffraction:

  • Central maximum is brightest and widest.
  • Intensity decreases for higher-order fringes.
  • Dark fringes occur when a sinθ = nλ.

Double-slit interference:
  • Equally spaced bright and dark fringes.
  • Uniform intensity distribution (if slits are narrow).
  • Fringe width depends on slit separation (d).

Key difference: Single-slit involves diffraction + interference, while double-slit primarily shows interference.

Question 5:
Why is a diffraction grating more effective than a double slit for measuring wavelengths?
Answer:

Diffraction grating (with thousands of slits) is more effective because:

  • Produces sharper and brighter maxima due to multiple slits.
  • Higher angular dispersion, allowing precise wavelength measurement.
  • Resolves closely spaced wavelengths (better resolving power).

Limitation of double slit: Broad fringes make accurate measurements difficult.

Question 6:
Describe how polarization proves the transverse nature of light waves.
Answer:

Polarization is the alignment of light's electric field vector in a specific direction.

Proof of transversality:

  • Only transverse waves can be polarized (longitudinal waves cannot).
  • Light passing through a polarizer vibrates in a single plane, confirming its transverse nature.

Example: Polaroid sunglasses block horizontally polarized glare, demonstrating this property.

Question 7:
Explain the phenomenon of interference in wave optics and state the necessary conditions for sustained interference.
Answer:

The phenomenon of interference occurs when two or more waves superpose to form a resultant wave of greater, lower, or the same amplitude.

Conditions for sustained interference:

  • The waves must be coherent (constant phase difference).
  • They must have the same frequency.
  • The waves should have comparable amplitudes for clear fringes.
  • The medium should be isotropic and homogeneous.

Interference is observed in phenomena like Young's double-slit experiment, producing bright and dark fringes due to constructive and destructive interference.

Question 8:
What is diffraction of light? How does the diffraction pattern due to a single slit differ from that of a double slit?
Answer:

Diffraction is the bending of light around obstacles or through apertures, causing wavefronts to spread out.

Single-slit diffraction: Produces a central bright fringe with alternating dark and less intense bright fringes on either side. The intensity decreases gradually.

Double-slit diffraction: Shows interference fringes (due to two slits) superimposed on the single-slit diffraction pattern. The fringes are equally spaced and of uniform intensity within the diffraction envelope.

Question 9:
Derive the expression for fringe width in Young's double-slit experiment.
Answer:

In Young's double-slit experiment, the fringe width (β) is the distance between two consecutive bright or dark fringes.

Derivation:
Path difference for constructive interference: Δx = nλ = d sinθ ≈ dθ (for small angles).
Since θ = y/D (where y is the fringe position and D is the slit-to-screen distance),
Δx = d(y/D) = nλ.
For consecutive fringes (n and n+1), fringe width β = yn+1 - yn = (Dλ)/d.

Thus, β = Dλ/d, showing fringe width is directly proportional to wavelength and screen distance, and inversely proportional to slit separation.

Question 10:
Why is light with a longer wavelength diffracted more than light with a shorter wavelength? Explain with an example.
Answer:

Diffraction depends on the ratio of wavelength (λ) to the obstacle/aperture size (a). For significant diffraction, λ ≈ a.

Example: Red light (λ ≈ 700 nm) diffracts more than violet light (λ ≈ 400 nm) through the same slit because its wavelength is closer to the slit width.

Mathematically, angular width of diffraction (θ) ≈ λ/a. Thus, longer wavelengths spread out more, making diffraction patterns broader.

Question 11:
Describe how a diffraction grating works and state its principle.
Answer:

A diffraction grating consists of many equally spaced slits (or rulings) that disperse light into its component wavelengths.

Principle: It works on constructive interference of light waves from multiple slits. The condition for maxima is: d sinθ = nλ, where d is grating spacing, θ is diffraction angle, and n is order of spectrum.

Applications include spectroscopy and precise wavelength measurement due to high resolution.

Question 12:
What is polarization of light? How does a polaroid filter work to produce plane-polarized light?
Answer:

Polarization is the restriction of light vibrations to a single plane perpendicular to the direction of propagation.

Polaroid filter: It contains long-chain molecules aligned in one direction. When unpolarized light passes through, these molecules absorb vibrations perpendicular to their alignment, transmitting only the parallel component.

Thus, the emergent light is plane-polarized, with vibrations confined to one plane.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Explain Huygens' Principle and its application in deriving the laws of reflection and refraction. How does it support the wave theory of light?
Answer:
Theoretical Framework

Huygens' Principle states that every point on a wavefront acts as a secondary source of wavelets, and the new wavefront is the tangent to these wavelets. We studied this in our textbook as the foundation of wave optics.

Evidence Analysis
  • For reflection, the principle shows incident and reflected wavefronts forming equal angles with the surface.
  • For refraction, it derives Snell's Law using wavelet propagation speeds in different media.
Critical Evaluation

This principle validates wave theory by explaining phenomena like diffraction, which particle theory couldn't. However, it doesn't account for light's quantum nature.

Question 2:
Describe Young's Double Slit Experiment with a labeled diagram. How does it demonstrate the wave nature of light?
Answer:
Theoretical Framework

Young's experiment uses two coherent light sources from a single slit to produce interference patterns. Our textbook shows this as definitive proof of light's wave nature.

[Diagram: Monochromatic light through single slit splitting into two coherent sources]Evidence Analysis
  • Alternate bright (constructive interference) and dark (destructive interference) fringes appear on screen.
  • Fringe width (β) = λD/d, where λ is wavelength, proving light's wave properties.
Critical Evaluation

This experiment conclusively disproved Newton's corpuscular theory. Modern applications include measuring microscopic distances.

Question 3:
Compare Fresnel and Fraunhofer diffraction with examples. Why does fringe spacing change with distance in Fresnel diffraction?
Answer:
Theoretical Framework

Fresnel diffraction occurs at finite distances with curved wavefronts, while Fraunhofer uses parallel rays at infinity. We studied both in our wave optics unit.

Evidence Analysis
ParameterFresnelFraunhofer
DistanceFiniteInfinite
WavefrontSphericalPlane
Critical Evaluation

Fresnel fringes vary because secondary wavelets interfere at changing angles. Applications include laser beam profiling (Fraunhofer) and shadow patterns (Fresnel).

Question 4:
Analyze how polarization proves light's transverse wave nature. Include Malus' Law with two real-world applications.
Answer:
Theoretical Framework

Polarization shows light's electric field oscillates perpendicular to propagation direction, confirming transverse waves. Our textbook demonstrates this using polaroid filters.

Evidence Analysis
  • Malus' Law: I = I₀cos²θ, where θ is angle between polarizer axes.
  • Examples: 1) Polarized sunglasses reduce glare 2) LCD screens use polarization to control pixels.
Critical Evaluation

This phenomenon couldn't be explained by longitudinal waves. Modern fiber optics rely on polarization-maintaining cables.

Question 5:
Derive the expression for resolving power of a telescope. How does it explain why radio telescopes need larger apertures than optical ones?
Answer:
Theoretical Framework

Resolving power (R) = λ/Δλ = D/1.22λ, where D is aperture diameter. We studied this in the context of Rayleigh's criterion.

Evidence Analysis
  • Radio waves (λ≈1m) require larger D to match optical telescope resolution (λ≈500nm).
  • Example: The Arecibo telescope (305m) achieves resolution comparable to small optical telescopes.
Critical Evaluation

This limitation drives modern interferometry techniques. Future telescopes like SKA will use arrays to simulate giant apertures.

Question 6:
Explain the Huygens' Principle and its application in deriving the laws of reflection and refraction. How does it support the wave theory of light?
Answer:
Theoretical Framework

Huygens' Principle states that every point on a wavefront acts as a secondary source of wavelets, and the new wavefront is the tangent to these wavelets. We studied how this principle helps derive the laws of reflection and refraction geometrically.

Evidence Analysis
  • For reflection, the angle of incidence equals the angle of reflection, proven using Huygens' construction.
  • For refraction, Snell's law (n1sinθ1 = n2sinθ2) is derived by comparing wavelet speeds in different media.
Critical Evaluation

This principle validates wave optics by explaining light's behavior without assuming particles. However, it doesn't account for polarization or quantum effects.

Question 7:
Describe Young's Double Slit Experiment and derive the expression for fringe width. What factors affect the interference pattern?
Answer:
Theoretical Framework

Young's experiment demonstrates light interference using two coherent sources. Our textbook shows the fringe width (β) formula: β = λD/d, where λ is wavelength, D is screen distance, and d is slit separation.

Evidence Analysis
  • Fringe width increases with longer λ (e.g., red light vs. violet).
  • Increasing D or decreasing d widens fringes, as per the formula.
Critical Evaluation

This experiment confirmed light's wave nature. However, monochromatic light is essential; white light creates overlapping colored fringes.

Question 8:
What is diffraction? Compare the diffraction patterns of a single slit and a circular aperture. How does slit width affect the pattern?
Answer:
Theoretical Framework

Diffraction is light bending around obstacles. A single slit produces alternating dark/bright fringes, while a circular aperture creates concentric rings (Airy pattern).

Evidence Analysis
  • Single-slit intensity follows I = I0(sinβ/β)2, where β = πasinθ/λ.
  • Narrower slits cause wider diffraction patterns due to increased path differences.
Critical Evaluation

Diffraction limits optical resolution (Rayleigh criterion). For example, telescopes struggle to distinguish close stars due to overlapping Airy patterns.

Question 9:
Explain polarization and how it demonstrates light's transverse wave nature. Describe Malus' Law with an example.
Answer:
Theoretical Framework

Polarization is the alignment of light's electric field oscillations. Transverse waves can be polarized; longitudinal cannot, proving light's wave nature.

Evidence Analysis
  • Malus' Law: I = I0cos2θ, where θ is the angle between polarizer axes.
  • Example: Polaroid sunglasses reduce glare by blocking horizontally polarized light reflected from surfaces.
Critical Evaluation

Polarization has applications in LCD screens and 3D movies. However, unpolarized light sources (e.g., sunlight) require polarizers for such effects.

Question 10:
Discuss resolving power of optical instruments. Derive the Rayleigh criterion for a circular aperture. How does it affect telescope design?
Answer:
Theoretical Framework

Resolving power is the ability to distinguish close objects. Rayleigh criterion states two sources are resolvable when the central maximum of one coincides with the first minimum of the other.

Evidence Analysis
  • For a circular aperture: θ = 1.22λ/D, where D is aperture diameter.
  • Larger telescopes (e.g., Hubble Space Telescope) have better resolution due to bigger D.
Critical Evaluation

Atmospheric turbulence limits ground-based telescopes, necessitating adaptive optics. Future telescopes like JWST use infrared to reduce diffraction (longer λ but larger D compensates).

Question 11:
Explain Huygens' Principle and its application in deriving the laws of reflection and refraction.
Answer:
Theoretical Framework

Huygens' Principle states that every point on a wavefront acts as a secondary source of wavelets, and the new wavefront is the tangent to these wavelets. We studied this in our textbook as the foundation for understanding wave propagation.

Evidence Analysis
  • For reflection: Using Huygens' construction, we showed that the angle of incidence equals the angle of reflection.
  • For refraction: The principle explains Snell's Law by considering the change in wave speed between media.
Critical Evaluation

This principle simplifies complex wave behaviors but assumes isotropic media, which may not hold in anisotropic materials.

Question 12:
Describe Young's Double Slit Experiment and derive the expression for fringe width.
Answer:
Theoretical Framework

Young's experiment demonstrates interference by splitting light into two coherent sources. Our textbook shows how this produces alternating bright and dark fringes.

Evidence Analysis
  • Path difference Δx = dsinθ ≈ d(y/D) for small angles.
  • Fringe width β = λD/d, derived from constructive interference conditions.
Critical Evaluation

While revolutionary, the experiment requires monochromatic light and precise slit separation, limiting practical applications with white light.

Question 13:
Compare Fraunhofer and Fresnel diffraction with suitable diagrams.
Answer:
Theoretical Framework

Diffraction occurs when light encounters obstacles. We studied two types: Fraunhofer (far-field) and Fresnel (near-field).

Evidence Analysis
  • Fraunhofer: Parallel rays, observed at infinity (or lens focus). [Diagram: Single slit with lens]
  • Fresnel: Spherical wavefronts, observed at finite distances. [Diagram: Circular obstacle]
Critical Evaluation

Fraunhofer is mathematically simpler but Fresnel describes real-world scenarios better. Modern lasers make Fraunhofer conditions easier to achieve.

Question 14:
Explain how polarization proves the transverse nature of light waves.
Answer:
Theoretical Framework

Polarization is the restriction of light vibrations to one plane. Our textbook shows this only occurs for transverse waves.

Evidence Analysis
  • Experiment: Light intensity varies when rotated through a polarizer.
  • Longitudinal waves (like sound) cannot be polarized as they vibrate parallel to propagation.
Critical Evaluation

This was crucial evidence against Newton's corpuscular theory. However, quantum mechanics later showed light has both wave and particle nature.

Question 15:
Derive the expression for resolving power of a telescope using Rayleigh's criterion.
Answer:
Theoretical Framework

Resolving power is the ability to distinguish close objects. Rayleigh's criterion states two points are resolvable when the central maximum of one coincides with the first minimum of another.

Evidence Analysis
  • For circular aperture: θ = 1.22λ/D
  • Resolving power R = 1/θ = D/1.22λ
Critical Evaluation

This limits telescope performance. Modern adaptive optics can surpass this limit by compensating for atmospheric distortions.

Question 16:
Describe Brewster's Law and its applications in polarizing sunglasses.
Answer:
Theoretical Framework

Brewster's Law states that when light strikes a surface at the polarizing angle, the reflected light becomes completely polarized.

Evidence Analysis
  • Mathematically: tan ip = μ (refractive index)
  • Sunglasses use this principle with vertical polarizers to block horizontally polarized glare from roads/water.
Critical Evaluation

While effective, polarized lenses can interfere with LCD screens. Newer anti-glare technologies combine multiple principles.

Question 17:
Explain Doppler effect in light and its significance in astronomy.
Answer:
Theoretical Framework

The Doppler effect causes frequency shifts when source and observer move relative to each other. For light, this produces redshift or blueshift.

Evidence Analysis
  • Receding sources show redshift (λ increases)
  • Approaching sources show blueshift (λ decreases)
Critical Evaluation

This is crucial for measuring galactic motions and proving universe expansion. However, cosmological redshift also includes gravitational effects.

Question 18:
Analyze the intensity distribution in single slit diffraction pattern.
Answer:
Theoretical Framework

Single slit diffraction produces a central bright fringe with diminishing intensity side fringes. Our textbook derives this using Huygens' principle.

Evidence Analysis
  • Intensity I = I0(sinβ/β)2, where β = πa sinθ/λ
  • Minima occur when a sinθ = nλ
Critical Evaluation

The pattern depends on slit width. Extremely narrow slits produce wide patterns, affecting optical instrument design.

Question 19:
Discuss coherent sources and why they're essential for interference experiments.
Answer:
Theoretical Framework

Coherent sources maintain constant phase difference. We studied this as fundamental for sustained interference patterns.

Evidence Analysis
  • Examples: Laser light, Young's double slit
  • Non-coherent sources (like bulbs) produce rapidly changing patterns that average out.
Critical Evaluation

While lasers provide perfect coherence, traditional methods like sodium lamps are still used due to cost. Quantum optics now explores entanglement as coherence source.

Question 20:
Describe Michelson interferometer and its use in determining wavelength of light.
Answer:
Theoretical Framework

The Michelson interferometer splits light into two perpendicular beams that recombine to form interference fringes.

Evidence Analysis
  • Moving one mirror by λ/2 changes path difference by λ, shifting fringe pattern.
  • By counting fringes N for mirror displacement d: λ = 2d/N
Critical Evaluation

This precise method helped define the meter. Modern applications include gravitational wave detection (LIGO).

Question 21:
Explain the phenomenon of interference in wave optics with the help of Young's double-slit experiment. Derive the expression for the fringe width and discuss the conditions for obtaining sustained interference pattern.
Answer:

Interference is a phenomenon in wave optics where two or more coherent waves superimpose to form a resultant wave of greater, lower, or the same amplitude. Young's double-slit experiment beautifully demonstrates this phenomenon.

Procedure:
1. A monochromatic light source is passed through a single slit to obtain coherent light.
2. This light then falls on two closely spaced slits, S1 and S2, acting as coherent sources.
3. The light waves from S1 and S2 superimpose on a screen placed at a distance, creating alternating bright and dark fringes.

Derivation of Fringe Width (β):
Let the distance between the slits be d, and the screen distance be D.
For constructive interference (bright fringe), the path difference must be an integral multiple of wavelength (λ):
Δx = nλ = d sinθ ≈ d (y/D)
Thus, the position of the nth bright fringe (yn) is:
yn = nλD/d
The fringe width (β) is the distance between two consecutive bright or dark fringes:
β = yn+1 - yn = λD/d

Conditions for Sustained Interference:

  • The sources must be coherent (constant phase difference).
  • The light must be monochromatic (single wavelength).
  • The amplitudes of the interfering waves should be nearly equal.
  • The distance between the slits should be small compared to the screen distance.

Application: This principle is used in interferometers to measure small distances and wavelengths accurately.

Question 22:
Explain the phenomenon of interference in wave optics with the help of Young's double-slit experiment. Derive the expression for the fringe width and discuss the conditions for obtaining sustained interference patterns.
Answer:

Interference is a phenomenon in which two or more waves superimpose to form a resultant wave of greater, lower, or the same amplitude. In wave optics, this is demonstrated using Young's double-slit experiment.

Young's Double-Slit Experiment:
A monochromatic light source is passed through a single slit to produce coherent light. This light then falls on two closely spaced slits, S1 and S2, which act as secondary coherent sources. The light waves from these slits interfere on a screen placed at a distance D from the slits.

Derivation of Fringe Width (β):
The path difference between the waves from S1 and S2 at a point P on the screen is given by:
Δx = S2P - S1P ≈ d sinθ
For constructive interference (bright fringe):
Δx = nλ ⇒ d sinθ = nλ
For small angles, sinθ ≈ tanθ = y/D ⇒ y = nλD/d
Fringe width (β) is the distance between two consecutive bright or dark fringes:
β = λD/d

Conditions for Sustained Interference:

  • The two sources must be coherent (constant phase difference).
  • The sources must be monochromatic (single wavelength).
  • The amplitudes of the interfering waves should be nearly equal.
  • The distance between the slits (d) should be small compared to the screen distance (D).

Application: Interference is used in technologies like anti-reflective coatings and interferometers.

Question 23:
Explain the phenomenon of interference in wave optics with the help of Young's double-slit experiment. Derive the expression for the fringe width and discuss the conditions for obtaining sustained interference pattern.
Answer:

The phenomenon of interference in wave optics refers to the superposition of two or more waves resulting in a new wave pattern. Young's double-slit experiment is a classic demonstration of interference where light waves from two coherent sources overlap to produce bright and dark fringes on a screen.

Derivation of Fringe Width (β):
1. Let the distance between the slits (S1 and S2) be d, and the distance between the slits and the screen be D.
2. The path difference (Δx) between the waves reaching a point P on the screen is given by: Δx = S2P - S1P ≈ d·sinθ ≈ d·(y/D), where y is the distance of P from the central maximum.
3. For constructive interference (bright fringe), Δx = nλ ⇒ yn = nλD/d.
4. For destructive interference (dark fringe), Δx = (2n+1)λ/2 ⇒ yn = (2n+1)λD/2d.
5. The fringe width (β) is the distance between two consecutive bright or dark fringes: β = λD/d.

Conditions for Sustained Interference:

  • The two sources must be coherent (constant phase difference).
  • The sources should be monochromatic (single wavelength).
  • The amplitudes of the waves should be nearly equal for clear contrast.
  • The distance between the slits should be small compared to the screen distance (d << D).

This experiment confirms the wave nature of light and is fundamental in understanding interference patterns.

Question 24:
Explain the phenomenon of interference in thin films and derive the condition for constructive interference in reflected light. Also, discuss any one practical application of this phenomenon.
Answer:

The phenomenon of interference in thin films occurs due to the superposition of light waves reflected from the upper and lower surfaces of the film. When light is incident on a thin film, part of it is reflected from the top surface, and the rest is transmitted and reflected from the bottom surface. These two reflected waves interfere with each other, leading to constructive or destructive interference depending on the path difference.


Condition for Constructive Interference:
For constructive interference in reflected light, the path difference between the two waves must be an integral multiple of the wavelength (λ). The path difference (Δ) is given by:
Δ = 2μt cos(r) + λ/2
where μ is the refractive index of the film, t is the thickness, and r is the angle of refraction. The additional λ/2 arises due to the phase change of π when light reflects from a denser medium.
For constructive interference:
2μt cos(r) + λ/2 = nλ
Simplifying, we get:
2μt cos(r) = (n - 1/2)λ


Practical Application:
One practical application is in anti-reflective coatings on lenses. By carefully choosing the thickness of the coating, destructive interference is achieved for specific wavelengths, reducing glare and improving light transmission. This enhances the clarity of optical instruments like cameras and eyeglasses.

Question 25:
Describe the working principle of a diffraction grating and derive the expression for the angular positions of the principal maxima. How does increasing the number of slits affect the diffraction pattern?
Answer:

A diffraction grating consists of a large number of equally spaced parallel slits. When light passes through the grating, it diffracts and interferes, producing a pattern of bright and dark fringes. The working principle is based on the constructive interference of light waves from multiple slits.


Expression for Principal Maxima:
The condition for constructive interference (principal maxima) is given by the grating equation:
d sinθ = nλ
where:
- d is the grating spacing (distance between adjacent slits),
- θ is the angle of diffraction,
- n is the order of the maximum (0, ±1, ±2, ...),
- λ is the wavelength of light.
This equation shows that the angular position of the maxima depends on the wavelength and the grating spacing.


Effect of Increasing Slits:
Increasing the number of slits in the grating:

  • Sharpens the principal maxima, making them narrower and more intense.
  • Reduces the width of the dark regions between maxima.
  • Improves the resolution of the grating, allowing better separation of closely spaced wavelengths.

This is why diffraction gratings with more slits are used in high-precision spectrometers to analyze light spectra.

Question 26:
Explain the phenomenon of interference in wave optics with the help of Young's double-slit experiment. Derive the expression for fringe width and discuss the conditions for obtaining sustained interference pattern.
Answer:

Interference is a phenomenon in which two or more waves superimpose to form a resultant wave of greater, lower, or the same amplitude. In wave optics, this is demonstrated using Young's double-slit experiment.

Young's Double-Slit Experiment:
1. A monochromatic light source passes through a single slit to produce coherent light.
2. This light then passes through two closely spaced slits, creating two coherent sources.
3. The waves from these slits interfere on a screen, forming alternate bright and dark fringes.

Derivation of Fringe Width (β):
Let the distance between the slits be d, and the distance to the screen be D.
The path difference between the two waves at a point on the screen is given by:
Δx = (d sinθ) ≈ (d y/D) (for small angles)
For constructive interference (bright fringe):
Δx = nλ ⇒ y = nλD/d
For destructive interference (dark fringe):
Δx = (2n+1)λ/2 ⇒ y = (2n+1)λD/2d
The fringe width (β) is the distance between two consecutive bright or dark fringes:
β = λD/d

Conditions for Sustained Interference:

  • The sources must be coherent (constant phase difference).
  • The waves must have the same frequency and wavelength.
  • The amplitudes should be nearly equal for clear contrast.
  • The distance between the slits should be small compared to the screen distance.

Question 27:
Describe the working principle of a diffraction grating. Derive the expression for the angular positions of the principal maxima in the diffraction pattern. How does increasing the number of slits affect the pattern?
Answer:

A diffraction grating consists of a large number of equally spaced parallel slits or rulings. It works on the principle of diffraction and interference to produce a spectrum of light.

Working Principle:
1. When a parallel beam of monochromatic light is incident on the grating, each slit acts as a source of secondary wavelets.
2. These wavelets interfere constructively and destructively, producing a pattern of bright and dark fringes.

Derivation of Angular Positions:
For constructive interference (principal maxima), the path difference between waves from adjacent slits must be an integer multiple of the wavelength (λ).
The condition is given by:
(a + b) sinθ = nλ
where:
- (a + b) = grating element (slit width + spacing)
- θ = angle of diffraction
- n = order of the spectrum (0, ±1, ±2, ...)
Thus, the angular positions of the principal maxima are:
θ = sin⁻¹(nλ / (a + b))

Effect of Increasing the Number of Slits:

  • The principal maxima become sharper and more intense.
  • The secondary maxima become weaker and less noticeable.
  • The resolution of the grating improves, allowing better separation of closely spaced wavelengths.

Question 28:
Explain the phenomenon of interference in thin films and derive the condition for constructive interference in reflected light. How does the color of the film appear when viewed in white light?
Answer:

The phenomenon of interference in thin films occurs due to the superposition of light waves reflected from the upper and lower surfaces of the film. When a beam of light is incident on a thin film, part of it is reflected from the top surface, and the rest is transmitted and reflected from the bottom surface. These two reflected waves interfere with each other.


Condition for constructive interference in reflected light:
For constructive interference, the path difference between the two reflected waves must be an integral multiple of the wavelength (λ). The path difference (Δ) is given by:
Δ = 2μt cos(r) + λ/2
Here, μ is the refractive index of the film, t is the thickness, and r is the angle of refraction. The additional λ/2 arises due to the phase change of π when light reflects from a denser medium.


For constructive interference:
2μt cos(r) + λ/2 = nλ
Simplifying, we get:
2μt cos(r) = (n - 1/2)λ
where n = 1, 2, 3,...


Color appearance in white light:
When viewed in white light, the thin film appears colored because white light consists of multiple wavelengths. Only those wavelengths that satisfy the constructive interference condition for the given thickness and angle are enhanced, while others are diminished. This results in the film displaying vibrant colors due to the selective reinforcement of certain wavelengths.

Question 29:
Describe the working principle of a diffraction grating and derive the expression for the angular position of the nth order maxima. How does increasing the number of slits per unit width affect the diffraction pattern?
Answer:

A diffraction grating consists of a large number of equally spaced parallel slits. When light passes through these slits, it diffracts and interferes, producing a pattern of bright and dark fringes on a screen.


Working Principle:
Each slit acts as a source of secondary wavelets, and the light waves from these slits interfere constructively or destructively depending on their path difference. Constructive interference results in bright fringes (maxima), while destructive interference results in dark fringes (minima).


Derivation for nth order maxima:
For constructive interference, the path difference between waves from adjacent slits must be an integral multiple of the wavelength (λ). If 'd' is the grating element (distance between two adjacent slits), and θ is the angle of diffraction, the condition is:
d sinθ = nλ
where n = 0, 1, 2,... is the order of the maxima.


Thus, the angular position of the nth order maxima is given by:
θ = sin⁻¹(nλ/d)


Effect of increasing slits per unit width:
Increasing the number of slits per unit width decreases the grating element 'd'. This leads to:

  • Sharper and more intense maxima due to increased constructive interference.
  • Better resolution as the angular separation between adjacent maxima increases.
  • More distinct and well-defined diffraction patterns.

Question 30:
Explain the phenomenon of interference in thin films and derive the condition for constructive interference in reflected light. Also, mention any two practical applications of this phenomenon.
Answer:

The phenomenon of interference in thin films occurs due to the superposition of light waves reflected from the upper and lower surfaces of the film. When light is incident on a thin film, part of it is reflected from the top surface, and the rest is transmitted and then reflected from the bottom surface. These two reflected waves interfere with each other, producing constructive or destructive interference based on their path difference.


Condition for constructive interference in reflected light:
For constructive interference, the path difference between the two reflected waves must be an integral multiple of the wavelength (λ). The path difference (Δ) is given by:
Δ = 2μt cos(r) + λ/2
Here, μ is the refractive index of the film, t is the thickness, and r is the angle of refraction. The additional λ/2 term accounts for the phase change of π due to reflection at the denser medium.
For constructive interference:
2μt cos(r) + λ/2 = nλ (where n = 0, 1, 2, ...)
Simplifying, we get:
2μt cos(r) = (2n - 1)λ/2


Practical applications:

  • Anti-reflective coatings: Thin films are used on lenses to reduce glare by causing destructive interference of reflected light.
  • Newton's rings: This interference pattern is used to measure the wavelength of light and the thickness of thin films.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
In a double-slit experiment, monochromatic light of wavelength 600 nm is used. The slits are 1 mm apart, and the screen is placed 2 m away. Calculate the fringe width. Also, explain how the pattern changes if the wavelength is increased to 800 nm.
Answer:
Case Deconstruction

Given: λ = 600 nm, d = 1 mm, D = 2 m. Fringe width (β) = λD/d.

Theoretical Application

β = (600×10⁻⁹ × 2)/10⁻³ = 1.2 mm. If λ increases to 800 nm, β becomes (800×10⁻⁹ × 2)/10⁻³ = 1.6 mm.

Critical Evaluation

Our textbook shows that β ∝ λ, so increasing λ widens fringes. This aligns with wave theory, where longer wavelengths diffract more.

Question 2:
A diffraction grating has 5000 lines/cm. Calculate the angular separation between the first-order maxima for light of wavelengths 450 nm and 650 nm. Discuss why gratings provide sharper patterns than double slits.
Answer:
Case Deconstruction

Grating element (d) = 1/5000 cm = 2×10⁻⁶ m. For λ₁ = 450 nm, sinθ₁ = λ₁/d = 0.225 → θ₁ ≈ 13°. For λ₂ = 650 nm, θ₂ ≈ 19.5°.

Theoretical Application

Δθ = θ₂ - θ₁ = 6.5°. Gratings have multiple slits, causing constructive interference to dominate, sharpening maxima.

Critical Evaluation

As per NCERT, more slits reduce intensity of secondary maxima, enhancing resolution, crucial for spectroscopy.

Question 3:
In Young’s experiment, white light replaces monochromatic light. Predict the observed pattern and justify why central fringe is white but others are colored.
Answer:
Case Deconstruction

White light contains all visible wavelengths (400–700 nm). Central fringe (n=0) has zero path difference, so all colors constructively interfere.

Theoretical Application

For other fringes, β ∝ λ, causing violet (λ≈400 nm) to appear closer to center than red (λ≈700 nm), creating spectra.

Critical Evaluation

Our lab observations confirm this dispersion effect, validating wave theory’s dependence on λ for fringe spacing.

Question 4:
A soap film (μ=1.33) appears dark when viewed under reflected light of λ=589 nm. Determine the minimum thickness for destructive interference. How does thickness affect color perception?
Answer:
Case Deconstruction

For darkness, 2μt = nλ (n=1). t = λ/(2μ) = 589/(2×1.33) ≈ 221 nm.

Theoretical Application

Thinner films cancel shorter λ (e.g., violet), while thicker films cancel longer λ (e.g., red), altering perceived color via thin-film interference.

Critical Evaluation

NCERT highlights this in oil slick examples, showing how t and λ interplay creates iridescence.

Question 5:
Compare Fresnel and Fraunhofer diffraction using two distinct experimental conditions. Which is more suitable for lab demonstrations and why?
Answer:
Case Deconstruction

Fresnel: Near-field, curved wavefronts (e.g., razor blade shadow). Fraunhofer: Far-field, plane wavefronts (e.g., laser through slit).

Theoretical Application

Fraunhofer is simpler in labs as it requires lenses to parallelize light, avoiding complex near-field calculations.

Critical Evaluation

We studied that Fraunhofer’s math is linear (sinθ ≈ θ), making it preferable for quantitative analysis in controlled setups.

Question 6:
In a double-slit experiment, monochromatic light of wavelength 600 nm is used. The slits are 1 mm apart, and the screen is 2 m away. Calculate the fringe width. Also, explain how the pattern changes if the wavelength is increased to 800 nm.
Answer:
Case Deconstruction

Given: λ = 600 nm, d = 1 mm, D = 2 m. Fringe width (β) = λD/d = (600×10⁻⁹×2)/10⁻³ = 1.2 mm.


Theoretical Application

If λ increases to 800 nm, β becomes (800×10⁻⁹×2)/10⁻³ = 1.6 mm. Fringes widen since β ∝ λ.


Critical Evaluation

Our textbook shows that increasing λ reduces angular separation but increases linear fringe width due to the direct proportionality in Young's formula.

Question 7:
A diffraction grating has 5000 lines/cm. Calculate the angular separation between the first-order maxima for light of wavelengths 450 nm and 550 nm. Discuss the resolution of the grating.
Answer:
Case Deconstruction

Grating element (d) = 1/5000 cm = 2×10⁻⁶ m. For λ₁=450 nm: sinθ₁ = λ₁/d = 0.225 → θ₁≈13°. For λ₂=550 nm: θ₂≈16°.


Theoretical Application

Angular separation = θ₂ - θ₁ ≈ 3°. Resolution (R) = Nm, where N=total lines. For 1 cm, R=5000×1=5000.


Critical Evaluation

We studied that higher N improves resolution, as seen here. Example: A 10,000-line grating would double R.

Question 8:
Explain polarization by reflection and derive Brewster's angle for water (refractive index=1.33). How does this prove light's transverse wave nature?
Answer:
Case Deconstruction

Brewster's law: tanθₚ = n = 1.33 → θₚ≈53.1°. At this angle, reflected light is fully polarized.


Theoretical Application

Polarization occurs because transverse waves vibrate perpendicular to propagation. Only transverse waves can be polarized, as longitudinal waves (e.g., sound) cannot.


Critical Evaluation

Our textbook shows polarization as key evidence for light's transverse nature, unlike sound. Example: Polaroid sunglasses use this principle.

Question 9:
In a Michelson interferometer, moving a mirror by 0.1 mm causes 300 fringe shifts. Determine the wavelength of light used. Analyze the effect of using a white light source.
Answer:
Case Deconstruction

Path difference Δx = 2d = 0.2 mm = Nλ → λ = 0.2×10⁻³/300 ≈ 666.7 nm.


Theoretical Application

White light causes overlapping fringes due to multiple λ. Only central fringe (Δx=0) is clear, as studied in interference.


Critical Evaluation

Monochromatic light is preferred for distinct fringes. Example: Sodium lamp (589 nm) is commonly used in such experiments.

Question 10:
A single slit of width 0.2 mm is illuminated by 500 nm light. Calculate the angular width of the central maximum. Contrast this with the double-slit intensity pattern.
Answer:
Case Deconstruction

Angular width = 2θ ≈ 2λ/a = 2×500×10⁻⁹/0.2×10⁻³ = 0.005 rad ≈ 0.29°.


Theoretical Application

Double-slit shows equally spaced bright/dark fringes with uniform intensity, while single-slit has a broad central peak with diminishing side lobes.


Critical Evaluation

We studied that single-slit diffraction modulates double-slit interference, as seen in combined patterns. Example: Young's experiment with finite slit width.

Question 11:
A student performs Young's double-slit experiment using monochromatic light of wavelength 600 nm. The distance between the slits is 0.5 mm, and the screen is placed 1.5 m away. Calculate the fringe width observed on the screen. Also, explain how the fringe width would change if the experiment is conducted underwater (refractive index = 1.33).
Answer:

To calculate the fringe width (β) in Young's double-slit experiment, we use the formula:
β = λD/d
where:
λ = wavelength of light (600 nm = 600 × 10-9 m)
D = distance between slits and screen (1.5 m)
d = distance between slits (0.5 mm = 0.5 × 10-3 m)

Substituting the values:
β = (600 × 10-9 × 1.5) / (0.5 × 10-3)
β = 1.8 × 10-3 m or 1.8 mm

When the experiment is conducted underwater, the wavelength of light changes due to the refractive index (n) of water. The new wavelength (λ') is given by:
λ' = λ/n = 600 nm / 1.33 ≈ 451 nm

The fringe width underwater (β') becomes:
β' = λ'D/d = (451 × 10-9 × 1.5) / (0.5 × 10-3) ≈ 1.35 × 10-3 m or 1.35 mm

Thus, the fringe width decreases underwater because the wavelength of light reduces in a denser medium.

Question 12:
In a diffraction experiment, a single slit of width 0.2 mm is illuminated by light of wavelength 500 nm. The diffraction pattern is observed on a screen placed 2 m away. Determine the angular width of the central maximum. Also, compare the intensity of the central maximum with that of the first secondary maximum.
Answer:

The angular width (2θ) of the central maximum in a single-slit diffraction pattern is given by:
2θ = 2λ/a
where:
λ = wavelength of light (500 nm = 500 × 10-9 m)
a = slit width (0.2 mm = 0.2 × 10-3 m)

Substituting the values:
2θ = 2 × (500 × 10-9) / (0.2 × 10-3)
2θ = 5 × 10-3 radians

To compare intensities:
The intensity (I) of the central maximum (I0) is the brightest. The intensity of the first secondary maximum (I1) is much weaker.
For the first secondary maximum (occurring at β ≈ 3π/2), the intensity is:
I1 = I0 / (3π/2)2 ≈ I0/22

Thus, the central maximum is 22 times brighter than the first secondary maximum. This shows how diffraction causes most of the light to concentrate at the center.

Question 13:
A student performs Young's double-slit experiment using monochromatic light of wavelength 600 nm. The distance between the slits is 0.5 mm, and the screen is placed 1.5 m away.

(i) Calculate the fringe width.
(ii) If the entire setup is immersed in a liquid of refractive index 1.33, how does the fringe width change? Justify your answer.

Answer:

(i) Fringe width calculation:
Given: λ = 600 nm = 600 × 10-9 m, d = 0.5 mm = 0.5 × 10-3 m, D = 1.5 m
Fringe width (β) = λD/d
= (600 × 10-9 × 1.5) / (0.5 × 10-3)
= 1.8 × 10-3 m or 1.8 mm

(ii) Change in fringe width when immersed in liquid:
When immersed in liquid, wavelength decreases as λ' = λ/n, where n = 1.33.
New fringe width (β') = λ'D/d = (λ/n) × D/d = β/n
= 1.8 mm / 1.33 ≈ 1.35 mm
Thus, fringe width decreases because the wavelength of light reduces in the medium.

Question 14:
In a diffraction experiment, a parallel beam of light of wavelength 500 nm is incident normally on a slit of width 0.2 mm. The diffraction pattern is observed on a screen placed 1 m away.

(i) Calculate the angular width of the central maximum.
(ii) How would the diffraction pattern change if the slit width is reduced to 0.1 mm? Explain.

Answer:

(i) Angular width of central maximum:
Given: λ = 500 nm = 500 × 10-9 m, a = 0.2 mm = 0.2 × 10-3 m
Angular width = 2θ ≈ 2(λ/a) (for small angles)
= 2 × (500 × 10-9 / 0.2 × 10-3)
= 5 × 10-3 radians

(ii) Effect of reducing slit width:
If slit width (a) is halved (0.1 mm), angular width doubles as θ ∝ 1/a.
New angular width = 2 × 5 × 10-3 = 10 × 10-3 radians.
The central maximum becomes wider, and intensity decreases because less light passes through the narrower slit. The diffraction pattern spreads out more.

Question 15:
A student performs Young's double-slit experiment using monochromatic light of wavelength 600 nm. The distance between the slits is 0.5 mm, and the screen is placed 1.5 m away.

(i) Calculate the fringe width.
(ii) If the entire setup is immersed in a liquid of refractive index 1.33, how does the fringe width change? Justify your answer.
Answer:

(i) Fringe width (β) in Young's double-slit experiment is given by the formula:
β = λD/d
where λ = wavelength of light (600 nm = 600 × 10⁻⁹ m),
D = distance between slits and screen (1.5 m),
d = distance between slits (0.5 mm = 0.5 × 10⁻³ m).

Substituting the values:
β = (600 × 10⁻⁹ × 1.5) / (0.5 × 10⁻³)
β = 1.8 × 10⁻³ m or 1.8 mm.

(ii) When the setup is immersed in a liquid of refractive index (μ = 1.33), the wavelength of light changes to λ' = λ/μ.
New fringe width (β') = λ'D/d = (λ/μ)D/d = β/μ.
Since μ > 1, β' decreases.

Numerically:
β' = 1.8 mm / 1.33 ≈ 1.35 mm.
Thus, the fringe width reduces due to the decrease in wavelength in the medium.

Question 16:
In a diffraction experiment, a parallel beam of light of wavelength 500 nm is incident normally on a single slit of width 0.2 mm. The diffraction pattern is observed on a screen placed 2 m away.

(i) Calculate the angular width of the central maximum.
(ii) How does the angular width change if the slit width is halved? Explain.
Answer:

(i) Angular width (2θ) of the central maximum in single-slit diffraction is given by:
2θ = 2λ/a
where λ = wavelength (500 nm = 500 × 10⁻⁹ m),
a = slit width (0.2 mm = 0.2 × 10⁻³ m).

Substituting the values:
2θ = 2 × (500 × 10⁻⁹) / (0.2 × 10⁻³)
2θ = 5 × 10⁻³ radians.

(ii) If the slit width (a) is halved (a' = a/2), the angular width becomes:
2θ' = 2λ/a' = 2λ/(a/2) = 4λ/a = 2 × (2θ).

Thus, the angular width doubles when the slit width is halved.
This is because diffraction effects become more pronounced as the slit becomes narrower, causing the central maximum to spread out.

Question 17:
A student observes the interference pattern formed by a double-slit setup using monochromatic light of wavelength 600 nm. The distance between the slits is 0.2 mm, and the screen is placed 1 m away.

(a) Calculate the fringe width.
(b) If the entire setup is immersed in water (refractive index = 4/3), how does the fringe width change? Justify your answer.
Answer:

(a) The fringe width (β) in Young's double-slit experiment is given by the formula:
β = λD/d
Here, λ = 600 nm = 600 × 10-9 m, D = 1 m, and d = 0.2 mm = 0.2 × 10-3 m.
Substituting the values:
β = (600 × 10-9 × 1) / (0.2 × 10-3)
β = 3 × 10-3 m = 3 mm.

(b) When the setup is immersed in water, the wavelength of light changes due to the refractive index (n) of water. The new wavelength (λ') is:
λ' = λ/n = 600 nm / (4/3) = 450 nm.
The new fringe width (β') becomes:
β' = λ'D/d = (450 × 10-9 × 1) / (0.2 × 10-3)
β' = 2.25 × 10-3 m = 2.25 mm.
Thus, the fringe width decreases because the wavelength of light reduces in water.

Question 18:
In a diffraction experiment, a parallel beam of light of wavelength 500 nm is incident normally on a single slit of width 0.2 mm. The diffraction pattern is observed on a screen placed 1 m away.

(a) Calculate the angular width of the central maximum.
(b) How does the width of the central maximum change if the slit width is halved? Explain.
Answer:

(a) The angular width () of the central maximum in single-slit diffraction is given by:
2θ = 2λ/a
Here, λ = 500 nm = 500 × 10-9 m and a = 0.2 mm = 0.2 × 10-3 m.
Substituting the values:
2θ = 2 × (500 × 10-9) / (0.2 × 10-3)
2θ = 5 × 10-3 radians.

(b) If the slit width (a) is halved (i.e., a' = a/2), the angular width becomes:
2θ' = 2λ/a' = 2λ/(a/2) = 4λ/a.
Thus, the new angular width is 2θ' = 4 × (500 × 10-9) / (0.2 × 10-3) = 10 × 10-3 radians.
The width of the central maximum doubles because it is inversely proportional to the slit width.

Question 19:
A student performs Young's double-slit experiment using monochromatic light of wavelength λ. The distance between the slits is d, and the screen is placed at a distance D from the slits.

(a) Derive the expression for the fringe width β.

(b) If the student replaces the monochromatic light with white light, what changes will be observed in the interference pattern? Justify your answer.

Answer:

(a) Derivation of fringe width β:


For constructive interference in Young's double-slit experiment, the path difference must be an integral multiple of the wavelength:

Path difference = = d sinθ
For small angles, sinθ ≈ tanθ = y/D, where y is the distance from the central maximum.

Thus, y = nλD/d
The fringe width β is the distance between two consecutive bright or dark fringes:

β = yn+1 - yn = (n+1)λD/d - nλD/d
β = λD/d

(b) Observation with white light:


When white light is used instead of monochromatic light:
  • The central fringe will be white because all wavelengths constructively interfere at the center (n=0).
  • The fringes on either side will appear colored because different wavelengths (λ) produce maxima at different positions (y = nλD/d).
  • The fringe pattern will blur out after a few fringes due to overlapping of different colored fringes.
Question 20:
In a diffraction experiment, a parallel beam of light of wavelength 600 nm is incident normally on a slit of width 0.2 mm. The diffraction pattern is observed on a screen placed 1 m away from the slit.

(a) Calculate the angular width of the central maximum.

(b) How will the diffraction pattern change if the slit width is reduced to 0.1 mm? Explain.

Answer:

(a) Angular width of the central maximum:


The angular width () of the central maximum in single-slit diffraction is given by:

2θ = 2λ/a
where λ = 600 nm = 600 × 10-9 m, and a = 0.2 mm = 0.2 × 10-3 m.

2θ = 2 × (600 × 10-9)/(0.2 × 10-3)
2θ = 6 × 10-3 radians

(b) Effect of reducing slit width:


If the slit width is reduced to 0.1 mm:
  • The angular width of the central maximum will increase because 2θ ∝ 1/a.
  • The intensity of the central maximum will decrease as less light passes through the narrower slit.
  • The secondary maxima will become less pronounced and more spread out.
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