Wave Optics — Class 12 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Physics chapter "Wave Optics" — 7 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Wave Optics" — 7 important questions with detailed answers for CBSE board exam prepara…
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NCERT Solutions for Class 12 Physics — Wave Optics. Step-by-step answers to all 7 textbook questions from this chapter, written for CBSE board preparation and free to use.
Wave Optics — All 7 Questions Solved
Q1. Explain Huygens' principle and show how it accounts for reflection and refraction.
Huygens' Principle: Every point on a wavefront acts as a source of secondary wavelets, and the new wavefront is the envelope of these secondary wavelets.
Reflection using Huygens' Principle:
- Incident plane waves AB fall on plane surface MM'
- Each point on AB generates secondary wavelets
- At time t, point A has traveled distance d₁, point B has traveled distance d₂
- Secondary wavelets form reflected wave A'B'
- Angle of incidence = Angle of reflection (proven from geometry)
- Reflected ray appears to come from mirror image
Refraction using Huygens' Principle:
- Incident plane waves AB pass from medium 1 to medium 2
- Speed changes from v₁ to v₂
- At interface, secondary wavelets have different speeds
- In time t: distance traveled in medium 1 = v₁t, in medium 2 = v₂t
- Refracted wavefront A'B' at angle θ₂
- From geometry: sin θ₁/v₁ = sin θ₂/v₂
- Using n = c/v: n₁ sin θ₁ = n₂ sin θ₂ (Snell's law)
Key Points:
- Explains wave behavior without assuming particles
- Accounts for diffraction (bending around obstacles)
- Explains polarization (transverse nature)
- Shows why light bends toward normal when entering denser medium
Q2. What is interference? Derive the condition for constructive and destructive interference in Young's double slit experiment.
Interference: Superposition of two coherent waves resulting in maximum or minimum intensity.
Young's Double Slit Experiment:
Two slits S₁ and S₂ separated by distance d
Distance to screen: D >> d
Wavelength: λ
Path difference at point P on screen:
Δ = S₂P - S₁P ≈ d sin θ ≈ d × y/D
Where y = distance from center on screen
Phase difference: δ = 2π/λ × Δ = 2πdy/(λD)
Constructive Interference (Bright fringes):
Path difference = nλ (n = 0, 1, 2, ...)
Δ = nλ
dy/D = nλ
y_n = nλD/d
Phase difference = 2nπ (even multiple of π)
Resultant amplitude: A_R = A₁ + A₂ = 2A
Intensity: I_max = (2A)² = 4I₀ (if both have intensity I₀)
Destructive Interference (Dark fringes):
Path difference = (2n + 1)λ/2 (n = 0, 1, 2, ...)
Δ = (2n + 1)λ/2
dy/D = (2n + 1)λ/2
y'_n = (2n + 1)λD/(2d)
Phase difference = (2n + 1)π (odd multiple of π)
Resultant amplitude: A_R = A₁ - A₂ = 0
Intensity: I_min = 0
Fringe width: β = λD/d
Q3. In Young's double slit experiment, the slits are 1 mm apart and screen is 1 m away. Light of wavelength 500 nm is used. Find fringe width and position of 3rd bright fringe.
Given:
Slit separation: d = 1 mm = 10⁻³ m
Distance to screen: D = 1 m
Wavelength: λ = 500 nm = 500 × 10⁻⁹ m = 5 × 10⁻⁷ m
Fringe width:
β = λD/d
β = (5 × 10⁻⁷ × 1)/(10⁻³)
β = 5 × 10⁻⁴ m = 0.5 mm
Position of 3rd bright fringe (n = 3, counting from center at n = 0):
For bright fringes: y_n = nλD/d
For 3rd bright fringe: y₃ = 3λD/d
y₃ = 3 × (5 × 10⁻⁷) × 1 / (10⁻³)
y₃ = (15 × 10⁻⁷) / (10⁻³)
y₃ = 15 × 10⁻⁴ m = 1.5 mm
Alternatively using fringe width:
y₃ = 3β = 3 × 0.5 = 1.5 mm
Answers:
- Fringe width: 0.5 mm or 500 μm
- Position of 3rd bright fringe: 1.5 mm from center
Q4. Define diffraction and explain single slit diffraction pattern. Derive the condition for minima.
Diffraction: Bending of light around obstacles or edges, caused by wave nature of light. Produces pattern of bright and dark bands.
Single Slit Diffraction:
Slit of width 'a' acts as source of secondary wavelets
Screen at large distance D from slit
Condition for Minima (Dark bands):
Divide slit into two halves of width a/2
Path difference between rays from top and bottom = (a/2) sin θ
For first minimum:
Path difference = λ/2
(a/2) sin θ₁ = λ/2
a sin θ₁ = λ
For nth minimum:
Path difference = nλ/2
a sin θ_n = nλ (n = ±1, ±2, ±3, ...)
On screen at distance y from center:
y_n = nλD/a
Condition for Maxima (Bright bands):
a sin θ = (2n + 1)λ/2 (n = 0, 1, 2, ...)
Central maximum is brightest
Other maxima are progressively fainter
Intensity Pattern:
I = I₀[sin(β)/β]²
Where β = (πa sin θ)/λ
Characteristics:
- Central maximum width = 2λD/a
- Side maxima narrower and weaker
- Pattern symmetric about center
- Width inversely proportional to slit width
Q5. What is diffraction grating? Derive the grating equation and explain its applications.
Diffraction Grating: Optical component with large number of closely-spaced slits (or rulings) producing multiple beams of light.
Grating Equation Derivation:
Consider N slits separated by distance d (grating spacing)
Light of wavelength λ incident normally on grating
Path difference between adjacent slits = d sin θ
For constructive interference:
Path difference = nλ
d sin θ = nλ (n = 0, 1, 2, ...)
Grating Equation: d sin θ = nλ
Where:
d = distance between adjacent slits (grating spacing)
θ = angle of diffraction
n = order of maximum (0 for central, ±1, ±2, ...)
λ = wavelength
Grating Constant:
N = total number of slits per unit width = 1/d
Order of Maximum:
Highest order: n_max = d/λ (when sin θ = 1)
Angular Dispersion:
Dθ/Dλ = n/d × sec θ ≈ n/d (for small angles)
Applications:
1. Determination of wavelength: λ = d sin θ/n
2. Spectroscopy: separates light into spectrum
3. Measuring refractive index
4. Quality control in manufacturing
5. Astronomical observations
6. X-ray diffraction (crystal gratings)
Q6. A diffraction grating has 4000 lines per cm. Find angles of first and second order maxima for light of wavelength 550 nm.
Given:
Number of lines per cm = 4000 lines/cm
Wavelength λ = 550 nm = 550 × 10⁻⁹ m
Grating spacing:
d = 1/(4000 × 10² lines/m) = 1/(4 × 10⁵ lines/m)
d = 2.5 × 10⁻⁶ m = 2500 nm
Using grating equation: d sin θ = nλ
sin θ = nλ/d
For first order (n = 1):
sin θ₁ = (1 × 550 × 10⁻⁹)/(2.5 × 10⁻⁶)
sin θ₁ = 550/(2.5 × 10³) = 550/2500 = 0.22
θ₁ = sin⁻¹(0.22) ≈ 12.7°
For second order (n = 2):
sin θ₂ = (2 × 550 × 10⁻⁹)/(2.5 × 10⁻⁶)
sin θ₂ = 1100/2500 = 0.44
θ₂ = sin⁻¹(0.44) ≈ 26.1°
Verification of maximum order:
n_max = d/λ = (2.5 × 10⁻⁶)/(550 × 10⁻⁹) ≈ 4.5
So maximum order is 4
Answers:
- First order maximum: θ₁ ≈ 12.7°
- Second order maximum: θ₂ ≈ 26.1°
- Maximum possible order: 4th order
Q7. Explain polarization by reflection. Why is polarized light produced when light reflects from a dielectric surface?
Polarization by Reflection: When unpolarized light reflects from a dielectric surface, reflected light becomes partially or completely polarized.
Mechanism:
1. Incident unpolarized light has E components both parallel and perpendicular to plane of incidence
2. At interface, induced charges oscillate due to incident wave
3. Component parallel to surface (⊥ to plane of incidence) radiates efficiently → reflected in all proportions
4. Component perpendicular to surface (parallel to plane of incidence) is absorbed and does not radiate at certain angles
5. At Brewster's angle, parallel component is completely absorbed
Brewster's Angle Explanation:
At Brewster's angle θ_B:
- Reflected ray ⊥ refracted ray
- Component of E parallel to plane of incidence in refracted ray is maximum
- This component cannot radiate back at 90° angle
- Therefore, reflected light contains ONLY ⊥ component → completely polarized
Mathematical Basis:
Reflectivity for parallel component: r_∥ = tan(θ₁ - θ₂)/tan(θ₁ + θ₂)
At Brewster's angle: θ₁ + θ₂ = 90°, so r_∥ = 0
Only perpendicular component reflects: r_⊥ ≠ 0
Degree of Polarization:
At other angles, both components reflect but in different proportions:
P = (I_⊥ - I_∥)/(I_⊥ + I_∥)
At Brewster's angle: P = 1 (complete polarization)
At normal incidence: P = 0 (no polarization)
Wave Optics — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. According to Huygens' Principle, every point on a primary wavefront acts as a source of secondary wavelets. What do these secondary wavelets form when they propagate?
- A. A new primary wavefront
- B. A spherical wave
- C. A secondary wavefront
- D. A point source
Answer: C. A secondary wavefront — Huygens' Principle states that every point on a given wavefront can be considered as a source of secondary wavelets, which spread out in all directions. The envelope of these secondary wavelets at a later instant gives the position of the new wavefront, called the secondary wavefront.
MCQ 2. Which of the following phenomena cannot be explained by the wave theory of light based on Huygens' Principle?
- A. Reflection
- B. Refraction
- C. Diffraction
- D. Photoelectric effect
Answer: D. Photoelectric effect — The photoelectric effect is the emission of electrons when light shines on a material. This phenomenon demonstrates the particle nature of light (photons) and cannot be explained by the classical wave theory of light, which assumes continuous energy transfer.
MCQ 3. For a sustained and observable interference pattern, which of the following conditions is essential for the light sources?
- A. They must be independent
- B. They must be coherent
- C. They must have different amplitudes
- D. They must be widely separated
Answer: B. They must be coherent — Coherent sources maintain a constant phase difference over time and emit waves of the same frequency. This constant phase relationship is crucial for producing stable and observable interference patterns.
MCQ 4. In Young's Double Slit Experiment (YDSE), if the separation between the slits is doubled, how does the fringe width change?
- A. It doubles
- B. It halves
- C. It remains unchanged
- D. It quadruples
Answer: B. It halves — The fringe width (β) in YDSE is given by the formula β = λD/d, where λ is the wavelength, D is the distance to the screen, and d is the slit separation. If d is doubled, β becomes half of its original value.
MCQ 5. What is the path difference for destructive interference (dark fringe) at a point in Young's Double Slit Experiment?
- A. nλ
- B. (n+1)λ
- C. (n + 1/2)λ
- D. nλ/2
Answer: C. (n + 1/2)λ — For destructive interference, the waves arriving at a point must be out of phase by an odd multiple of π radians, which corresponds to a path difference of (n + 1/2)λ, where n = 0, 1, 2, ... .
MCQ 6. In YDSE, if the intensity of light from each slit is I₀, what is the maximum intensity observed in the interference pattern?
- A. I₀
- B. 2I₀
- C. 4I₀
- D. √2 I₀
Answer: C. 4I₀ — If the amplitudes of waves from each slit are A, then their intensities are proportional to A². So, I₀ is the intensity due to one slit. For constructive interference, the amplitudes add up to 2A, and the maximum intensity is proportional to (2A)² = 4A², hence 4I₀.
MCQ 7. A Young's Double Slit experiment is performed in air and then the entire apparatus is immersed in water (refractive index μ > 1). How does the fringe width change?
- A. Increases
- B. Decreases
- C. Remains unchanged
- D. Becomes zero
Answer: B. Decreases — The fringe width (β) is directly proportional to the wavelength (λ) of light. When light passes from air to water, its wavelength decreases (λ_water = λ_air / μ). Therefore, the fringe width will decrease.
MCQ 8. The width of the central maximum in a single-slit diffraction pattern is given by:
- A. λD/a
- B. 2λD/a
- C. λD/2a
- D. D/a
Answer: B. 2λD/a — For single-slit diffraction, the angular width of the central maximum is 2θ (where sinθ ≈ θ = λ/a for the first minimum). The linear width on the screen is 2Dθ = 2λD/a.
MCQ 9. In single-slit diffraction, if the width of the slit is increased, what happens to the width of the central maximum?
- A. Increases
- B. Decreases
- C. Remains unchanged
- D. Becomes brighter but width is same
Answer: B. Decreases — The width of the central maximum in single-slit diffraction is inversely proportional to the slit width (width = 2λD/a). Therefore, if the slit width (a) increases, the width of the central maximum decreases.
MCQ 10. Which phenomenon best demonstrates the transverse nature of light waves?
- A. Interference
- B. Diffraction
- C. Refraction
- D. Polarization
Answer: D. Polarization — Polarization occurs only for transverse waves because the oscillations can be restricted to a single plane. Longitudinal waves, whose oscillations are along the direction of propagation, cannot be polarized.
MCQ 11. According to Malus's Law, if unpolarized light passes through a polarizer and then an analyzer whose pass-axis makes an angle θ with the polarizer's pass-axis, what is the intensity of the transmitted light?
- A. I₀ cosθ
- B. I₀ cos²θ
- C. I₀/2 cos²θ
- D. I₀/2
Answer: C. I₀/2 cos²θ — When unpolarized light passes through the first polarizer, its intensity becomes I₀/2. Then, according to Malus's Law, when this plane-polarized light passes through the analyzer, the transmitted intensity is (I₀/2)cos²θ.
MCQ 12. Brewster's angle (i_p) for an interface between two media is the angle of incidence at which the reflected light is completely plane-polarized. What is the relation between i_p and the refractive index (n) of the second medium with respect to the first?
- A. sin i_p = n
- B. cos i_p = n
- C. tan i_p = n
- D. cot i_p = n
Answer: C. tan i_p = n — Brewster's Law states that tan i_p = n, where i_p is the polarizing angle and n is the refractive index of the denser medium relative to the rarer medium. At this angle, the reflected light is completely polarized, and the reflected and refracted rays are perpendicular to each other.
MCQ 13. In Brewster's Law, when light is incident at the polarizing angle, what is the angle between the reflected and refracted rays?
- A. 0°
- B. 45°
- C. 90°
- D. 180°
Answer: C. 90° — At Brewster's angle, the reflected light is completely plane-polarized, and a key condition is that the reflected ray and the refracted ray are perpendicular to each other, forming an angle of 90°.
MCQ 14. If the amplitudes of two interfering waves are in the ratio 3:1, what is the ratio of maximum to minimum intensity in the interference pattern?
- A. 9:1
- B. 4:1
- C. 16:1
- D. 2:1
Answer: C. 16:1 — Let the amplitudes be 3A and A. Maximum intensity I_max is proportional to (3A+A)² = (4A)² = 16A². Minimum intensity I_min is proportional to (3A-A)² = (2A)² = 4A². The ratio I_max : I_min is 16A² : 4A² = 4:1.
MCQ 15. What is the condition for observing diffraction?
- A. The wavelength of light must be much smaller than the obstacle/aperture size.
- B. The wavelength of light must be comparable to or larger than the obstacle/aperture size.
- C. Light sources must be incoherent.
- D. Light must be monochromatic and unpolarized.
Answer: B. The wavelength of light must be comparable to or larger than the obstacle/aperture size. — Diffraction is the bending of waves around obstacles or through apertures. It is significant when the wavelength of the wave is comparable to or larger than the size of the obstacle or aperture. If the wavelength is much smaller, light travels in straight lines (geometrical optics).
MCQ 16. Which of the following statements is true regarding the intensity of successive maxima in single-slit diffraction?
- A. All maxima have the same intensity.
- B. Intensity increases with increasing order of maxima.
- C. Intensity decreases rapidly with increasing order of maxima.
- D. Only the central maximum is visible, others are too dim.
Answer: C. Intensity decreases rapidly with increasing order of maxima. — In single-slit diffraction, the central maximum is the brightest. The intensity of subsequent secondary maxima decreases rapidly as the order of the maximum increases, unlike interference where bright fringes have nearly equal intensity.
MCQ 17. To increase the resolving power of a telescope, which of the following changes would be most effective?
- A. Increase the focal length of the objective lens.
- B. Decrease the diameter of the objective lens.
- C. Increase the wavelength of light used.
- D. Increase the diameter of the objective lens.
Answer: D. Increase the diameter of the objective lens. — The resolving power of a telescope is given by D/(1.22λ), where D is the diameter of the objective lens and λ is the wavelength of light. To increase resolving power, one should increase D and/or decrease λ.
MCQ 18. The resolving power of a microscope depends on which of these factors?
- A. Only the wavelength of light used.
- B. Only the numerical aperture of the objective.
- C. Both the wavelength of light and the numerical aperture of the objective.
- D. The magnification of the eyepiece.
Answer: C. Both the wavelength of light and the numerical aperture of the objective. — The resolving power of a microscope is given by 2n sinθ / (1.22λ), where n sinθ is the numerical aperture and λ is the wavelength of light. Thus, it depends on both the wavelength and the numerical aperture.
MCQ 19. What happens to the central fringe in Young's Double Slit Experiment if one of the slits is covered with a thin, transparent film of refractive index μ and thickness t?
- A. It becomes darker.
- B. It shifts towards the covered slit.
- C. It shifts away from the covered slit.
- D. Its width increases.
Answer: B. It shifts towards the covered slit. — The thin film introduces an additional optical path length (μ-1)t for the light passing through that slit. To maintain the zero path difference condition for the central bright fringe, the fringe shifts towards the slit covered by the film.
MCQ 20. When white light is used in Young's Double Slit Experiment, what is observed at the center of the screen?
- A. A dark fringe.
- B. A white fringe.
- C. A red fringe.
- D. A violet fringe.
Answer: B. A white fringe. — At the center of the screen, the path difference for all wavelengths is zero. Since path difference is zero, all wavelengths interfere constructively, resulting in a bright white fringe.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.