Important Questions Class 12 Physics Chapter 10: Wave Optics

Wave optics studies light as a wave that can bend, overlap, interfere, diffract, and polarise.
The wave model explains why light forms fringes, spreads through narrow slits, and shows transverse behaviour.

Light behaves like a wave when its wavelength becomes relevant in experiments involving narrow slits, coherent sources, and polaroids. Important Questions Class 12 Physics Chapter 10 help students practise Huygens principle, reflection, refraction, interference, Young’s double-slit experiment, diffraction, and polarisation. The CBSE 2026 chapter uses wavefronts, phase difference, path difference, fringe width, single-slit diffraction, and Malus law from NCERT Wave Optics.

Key Takeaways

• Wavefront: A wavefront is a surface of constant phase.
• Young’s Experiment: Two coherent sources produce equally spaced bright and dark fringes.
• Single Slit Diffraction: Minima occur at a sin θ = nλ for n = 1, 2, 3.
• Malus Law: Transmitted intensity through two polaroids follows I = I0 cos²θ.

Important Questions Class 12 Physics Chapter 10 Structure 2026

Important Questions Class 12 Physics Chapter 10 with Answers

Wave optics begins when ray optics cannot explain the observed pattern.
Students should connect every fringe, dark band, and intensity change with phase difference.
These wave optics class 12 important questions follow the NCERT 2026 sequence.

1. What does Important Questions Class 12 Physics Chapter 10 mainly test?

Important Questions Class 12 Physics Chapter 10 mainly test Huygens principle, wavefronts, interference, diffraction, and polarisation. The chapter links wave theory with measurable light patterns.

1. Wavefront Skill: Define and construct wavefronts.
2. Interference Skill: Use path difference and phase difference.
3. Diffraction Skill: Apply single-slit minima conditions.
4. Polarisation Skill: Use transverse nature and Malus law.
5. Final Result: The chapter tests wave behaviour of light.

2. Why did wave optics become necessary?

Wave optics became necessary because light shows interference, diffraction, and polarisation. Ray optics cannot explain these effects.

1. Ray Optics Limit: It assumes wavelength tends to zero.
2. Observed Effect: Light forms bright and dark fringes.
3. Wave Explanation: Superposition gives intensity variation.
4. Final Result: Wave optics explains fringe and diffraction patterns.

3. Why did Young’s experiment support the wave theory of light?

Young’s experiment supported wave theory because it showed interference fringes. Such fringes need coherent wave superposition.

1. Experiment: Light passed through two close pinholes or slits.
2. Observation: Bright and dark bands appeared.
3. Meaning: Light amplitudes added constructively and destructively.
4. Final Result: Young’s experiment proved wave behaviour of light.

Huygens Principle Class 12 Physics Questions

Huygens principle gives a geometrical method for building a new wavefront.
It treats every point on a known wavefront as a source of secondary wavelets.
These Huygens principle Class 12 Physics questions cover wavefront construction and laws of optics.

4. What is Huygens principle in Class 12 Physics?

Huygens principle states that every point on a wavefront acts as a source of secondary wavelets. Their common forward envelope forms the new wavefront.

1. Initial Wavefront: Known at a given time.
2. Secondary Wavelets: Spread with wave speed.
3. New Wavefront: Common tangent gives later wavefront.
4. Final Result: Huygens principle constructs wavefronts geometrically.

5. What is a wavefront in Wave Optics?

A wavefront is a surface of constant phase. All points on it vibrate in the same phase.

1. Phase Condition: Same phase at all points.
2. Point Source: Produces spherical wavefronts.
3. Distant Source: Produces nearly plane wavefronts.
4. Final Result: A wavefront is a constant-phase surface.

6. What is the direction of energy propagation relative to a wavefront?

Energy travels perpendicular to the wavefront. Rays are normal to wavefronts in wave optics.

1. Wavefront: Surface of constant phase.
2. Ray Direction: Normal to wavefront.
3. Plane Wave: Rays remain parallel.
4. Final Result: Energy travels normal to the wavefront.

7. What is a spherical wavefront?

A spherical wavefront forms when a point source emits light uniformly in all directions. The wavefronts are spheres.

1. Source Type: Point source.
2. Shape: Spherical surface.
3. Centre: Source position.
4. Final Result: A point source produces spherical wavefronts.

8. What is a plane wavefront?

A plane wavefront is a small part of a spherical wavefront far from the source. Its surface appears flat.

1. Source Distance: Very large distance.
2. Wavefront Curvature: Negligible over small region.
3. Ray Direction: Parallel rays.
4. Final Result: A distant source gives a plane wavefront.

Wavefront Class 12 Physics Questions on Reflection and Refraction

Reflection and refraction laws can be derived from Huygens construction.
The derivation also explains why speed and wavelength change across media.
These wavefront Class 12 Physics questions cover Snell’s law and reflected wavefronts.

9. How does Huygens principle prove the law of reflection?

Huygens principle proves that angle of incidence equals angle of reflection. The reflected wavefront construction gives congruent triangles.

1. Incident Wavefront: Reaches the reflecting surface.
2. Secondary Wavelets: Start from the first contact point.
3. Tangent Plane: Gives reflected wavefront.
4. Final Result: i = r for reflection.

10. How does Huygens principle derive Snell’s law?

Huygens principle derives Snell’s law by comparing wave speeds in two media. The result is n1 sin i = n2 sin r.

1. Medium 1 Speed: v1.
2. Medium 2 Speed: v2.
3. Speed Relation: sin i/sin r = v1/v2.
4. Refractive Form: n1 sin i = n2 sin r.
5. Final Result: Huygens construction gives Snell’s law.

11. What changes when light enters a denser medium?

Speed and wavelength decrease when light enters a denser medium. Frequency remains unchanged.

1. Speed: Decreases in denser medium.
2. Wavelength: Decreases with speed.
3. Frequency: Source fixes frequency.
4. Final Result: Frequency remains constant during refraction.

12. Why does light bend towards the normal in a denser medium?

Light bends towards the normal because its speed decreases in the denser medium. The refracted angle becomes smaller than the incident angle.

1. Speed Relation: v2 < v1.
2. Angle Relation: r < i.
3. Snell’s Law: n1 sin i = n2 sin r.
4. Final Result: Lower speed bends light towards the normal.

13. What is critical angle in wave optics?

Critical angle is the angle of incidence in denser medium for which refraction angle becomes 90°. It satisfies sin ic = n2/n1.

1. Condition: Light travels from denser to rarer medium.
2. Refraction Angle: r = 90°.
3. Formula Used: sin ic = n2/n1.
4. Final Result: Critical angle marks the start of total internal reflection.

Coherent Sources Class 12 Questions

Stable interference needs sources with the same frequency and fixed phase difference.
Two independent lamps do not keep a constant phase relation.
These coherent sources Class 12 questions explain why Young used one source to create two sources.

14. What are coherent sources in Wave Optics?

Coherent sources emit waves with the same frequency and constant phase difference. They produce a stable interference pattern.

1. Frequency: Same for both sources.
2. Phase Difference: Constant with time.
3. Result: Fixed bright and dark fringes.
4. Final Result: Coherent sources give stable interference.

15. Why do two independent sodium lamps not produce interference fringes?

Two independent sodium lamps do not produce fringes because their phase difference changes rapidly. The intensity averages out.

1. Independent Sources: Emit light with random phase changes.
2. Phase Relation: Not fixed.
3. Observed Result: Intensities add.
4. Final Result: Independent lamps act as incoherent sources.

16. How did Young create coherent sources?

Young created coherent sources by illuminating two close pinholes from the same source. Both pinholes inherited the same phase changes.

1. Primary Source: Single source S.
2. Secondary Sources: S1 and S2.
3. Phase Lock: Same original wave feeds both.
4. Final Result: Young’s two slits behave as coherent sources.

17. What happens when two coherent waves meet in phase?

Constructive interference occurs when two coherent waves meet in phase. The resultant intensity becomes maximum.

1. Path Difference: Δx = nλ.
2. Phase Difference: Δφ = 2nπ.
3. Equal Source Intensity: Resultant intensity = 4I0.
4. Final Result: In-phase waves produce bright fringes.

18. What happens when two coherent waves meet out of phase?

Destructive interference occurs when coherent waves meet with phase difference π. The resultant intensity becomes minimum.

1. Path Difference: Δx = (n + 1/2)λ.
2. Phase Difference: Δφ = (2n + 1)π.
3. Equal Source Intensity: Resultant intensity = 0.
4. Final Result: Out-of-phase waves produce dark fringes.

Interference of Light Class 12 Questions

Interference comes from superposition of light waves with a stable phase difference.
It redistributes light energy into bright and dark regions on a screen.
These interference of light Class 12 questions cover intensity, path difference, and phase relation.

19. What is interference of light?

Interference of light is redistribution of intensity due to superposition of coherent waves. It creates alternate bright and dark regions.

1. Cause: Superposition of waves.
2. Bright Region: Constructive interference.
3. Dark Region: Destructive interference.
4. Final Result: Interference produces intensity redistribution.

20. What is the condition for constructive interference?

Constructive interference occurs when path difference equals an integral multiple of wavelength. The condition is Δx = nλ.

1. Path Difference: Δx = nλ.
2. Here: n = 0, 1, 2, 3.
3. Phase Difference: Δφ = 2nπ.
4. Final Result: Constructive interference gives bright fringes.

21. What is the condition for destructive interference?

Destructive interference occurs when path difference equals an odd half multiple of wavelength. The condition is Δx = (n + 1/2)λ.

1. Path Difference: Δx = (n + 1/2)λ.
2. Here: n = 0, 1, 2, 3.
3. Phase Difference: Δφ = (2n + 1)π.
4. Final Result: Destructive interference gives dark fringes.

22. What is resultant intensity when phase difference is φ?

The resultant intensity for equal source intensities is I = 4I0 cos²(φ/2). It depends on phase difference.

1. Source Intensity: I0 from each source.
2. Phase Difference: φ.
3. Formula Used: I = 4I0 cos²(φ/2).
4. Final Result: Intensity varies with cos²(φ/2).

23. If path difference is λ/3 and intensity at path difference λ is K, find intensity.

The intensity is K/4. At path difference λ, intensity is maximum K.

1. Given: Maximum intensity K = 4I0.
2. Path Difference: Δx = λ/3.
3. Phase Difference: φ = 2π/3.
4. Intensity:
   I = 4I0 cos²(π/3)
   I = 4I0 × 1/4
   I = I0 = K/4
5. Final Result: Intensity = K/4.

Young Double Slit Experiment Class 12 Questions

Young’s double-slit experiment measures wavelength using equally spaced fringes.
The slit separation must be small compared with the screen distance.
These Young double slit experiment Class 12 questions focus on fringe position and fringe width.

24. What is Young’s double-slit experiment?

Young’s double-slit experiment shows interference of light from two coherent sources. It produces alternate bright and dark fringes on a screen.

1. Source: Single light source illuminates two close slits.
2. Slits: Act as coherent sources.
3. Screen: Shows interference fringes.
4. Final Result: YDSE proves wave nature of light.

25. What is fringe width in Young’s double-slit experiment?

Fringe width is the distance between two consecutive bright or dark fringes. Its formula is β = λD/d.

1. Wavelength: λ.
2. Screen Distance: D.
3. Slit Separation: d.
4. Formula Used: β = λD/d.
5. Final Result: Fringe width equals λD/d.

26. What is position of nth bright fringe in YDSE?

The position of nth bright fringe is xn = nλD/d. The central bright fringe corresponds to n = 0.

1. Bright Condition: Δx = nλ.
2. Small Angle Result: path difference = xd/D.
3. Formula Used: xn = nλD/d.
4. Final Result: Bright fringes are equally spaced.

27. What is position of nth dark fringe in YDSE?

The position of nth dark fringe is x = (n + 1/2)λD/d. It lies midway between two bright fringes.

1. Dark Condition: Δx = (n + 1/2)λ.
2. Small Angle Result: path difference = xd/D.
3. Formula Used: x = (n + 1/2)λD/d.
4. Final Result: Dark fringes lie between bright fringes.

28. In YDSE, d = 0.28 mm, D = 1.4 m, and fourth bright fringe is 1.2 cm from centre. Find λ.

The wavelength is 600 nm. Use x = nλD/d.

1. Given Data:
   d = 0.28 mm = 2.8 × 10^-4 m
   D = 1.4 m
   n = 4
   x = 1.2 cm = 1.2 × 10^-2 m
2. Formula Used: x = nλD/d.
3. Calculation:
   λ = xd/(nD)
   λ = (1.2 × 10^-2 × 2.8 × 10^-4)/(4 × 1.4)
   λ = 6.0 × 10^-7 m
4. Final Result: λ = 600 nm.

29. Two wavelengths 650 nm and 520 nm form YDSE fringes. Where does third bright fringe of 650 nm occur?

The third bright fringe occurs at 3λD/d for 650 nm light. Its exact value needs D and d.

1. Bright Fringe Formula: x = nλD/d.
2. For Third Bright: n = 3.
3. Wavelength: λ = 650 nm.
4. Final Result: x = 3 × 650 nm × D/d.

30. When do bright fringes from two wavelengths coincide?

Bright fringes coincide when n1λ1 = n2λ2. For 650 nm and 520 nm, the least common match is 4λ1 = 5λ2.

1. Condition: n1λ1 = n2λ2.
2. Wavelengths: 650 nm and 520 nm.
3. Ratio: 650/520 = 5/4.
4. Least Match: 4 × 650 = 5 × 520 = 2600 nm.
5. Final Result: Fourth 650 nm bright fringe matches fifth 520 nm bright fringe.

Diffraction Class 12 Physics Questions

Diffraction appears when light spreads around edges or passes through narrow openings.
Its effect becomes noticeable when aperture size becomes comparable to wavelength.
These diffraction Class 12 Physics questions explain ray optics limits and fringe formation.

31. What is diffraction of light?

Diffraction is the bending and spreading of light near an aperture or obstacle edge. It proves the wave nature of light.

1. Opening: Narrow slit or small aperture.
2. Effect: Light enters geometrical shadow region.
3. Pattern: Bright and dark regions appear.
4. Final Result: Diffraction is wave spreading of light.

32. Why do we not see diffraction of light easily in daily life?

We do not see diffraction easily because visible light has very small wavelength. Most everyday objects are much larger than wavelength.

1. Visible Wavelength: Around 10^-7 m.
2. Common Objects: Much larger than wavelength.
3. Effect: Diffraction remains weak.
4. Final Result: Small wavelength hides everyday diffraction.

33. How does diffraction limit optical instruments?

Diffraction limits optical instruments by spreading point images into finite patterns. Very close objects cannot be resolved beyond a limit.

1. Lens Aperture: Acts like an opening.
2. Diffraction Pattern: Central maximum has finite width.
3. Resolution Limit: Nearby images overlap.
4. Final Result: Diffraction sets resolution limits.

34. What is the difference between interference and diffraction?

Interference usually involves superposition from a few coherent sources. Diffraction involves superposition from many parts of the same wavefront.

1. Interference: Often two coherent sources.
2. Diffraction: Many secondary sources across an aperture.
3. Common Principle: Both use superposition.
4. Final Result: Both are wave superposition effects.

Single Slit Diffraction Class 12 Questions

A single slit produces a broad central maximum and weaker side maxima.
The central maximum is wider and brighter than the secondary maxima.
These single slit diffraction Class 12 questions cover angular minima and intensity pattern.

35. What is single-slit diffraction?

Single-slit diffraction is the spreading of light after passing through one narrow slit. It creates a central bright band and side bands.

1. Slit Width: a.
2. Central Pattern: Broad central maximum.
3. Side Pattern: Alternating dark and weak bright bands.
4. Final Result: Single slit gives a diffraction pattern.

36. What is the condition for minima in single-slit diffraction?

The minima condition is a sin θ = nλ. Here n = 1, 2, 3 for dark bands.

1. Slit Width: a.
2. Wavelength: λ.
3. Order: n = 1, 2, 3.
4. Final Result: Dark bands occur at a sin θ = nλ.

37. Where does the central maximum occur in single-slit diffraction?

The central maximum occurs at θ = 0. All parts of the slit contribute in phase at the centre.

1. Central Direction: θ = 0.
2. Path Difference: Zero across the slit.
3. Intensity: Maximum.
4. Final Result: Central maximum lies along the slit normal.

38. Why does central maximum become narrower when slit width increases?

The central maximum becomes narrower because angular width is proportional to λ/a. Larger slit width reduces diffraction spread.

1. First Minimum: sin θ = λ/a.
2. Small Angle: θ ≈ λ/a.
3. Width Dependence: Central width decreases when a increases.
4. Final Result: Larger slit gives narrower central maximum.

39. What is the angular position of first minimum for slit width a?

The first minimum occurs at sin θ = λ/a. For small angles, θ ≈ λ/a.

1. Minima Condition: a sin θ = nλ.
2. First Minimum: n = 1.
3. Formula: sin θ = λ/a.
4. Final Result: First dark band occurs at sin θ = λ/a.

Polarisation Class 12 Physics Questions

Polarisation proves that light waves are transverse electromagnetic waves.
A polaroid transmits only the component along its pass-axis.
These polarisation Class 12 Physics questions cover unpolarised light, plane polarisation, and polaroids.

40. What is polarisation of light?

Polarisation is the restriction of electric field vibrations to one plane. It occurs only for transverse waves.

1. Unpolarised Light: Electric vector has random transverse directions.
2. Plane Polarised Light: Electric vector vibrates in one plane.
3. Device: Polaroid produces plane polarised light.
4. Final Result: Polarisation proves transverse nature of light.

41. Why can only transverse waves be polarised?

Only transverse waves can be polarised because their vibrations occur perpendicular to propagation direction. Longitudinal waves cannot show such restriction.

1. Transverse Wave: Vibration has multiple transverse directions.
2. Longitudinal Wave: Vibration stays along propagation direction.
3. Polarisation: Selects one transverse direction.
4. Final Result: Polarisation is possible only for transverse waves.

42. What happens when unpolarised light passes through one polaroid?

Its intensity becomes half of the incident intensity. The transmitted light becomes plane polarised.

1. Incident Light: Unpolarised.
2. Polaroid Action: Passes one electric field component.
3. Intensity: I = Iincident/2.
4. Final Result: One polaroid halves unpolarised light intensity.

43. Why does rotating one polaroid alone not change transmitted intensity?

Rotating one polaroid alone does not change intensity because unpolarised light has all transverse directions equally. Every orientation passes half intensity.

1. Incident Light: Unpolarised.
2. Electric Vectors: Random transverse directions.
3. Average Transmission: Half intensity for any axis.
4. Final Result: Single polaroid rotation keeps intensity constant.

44. What happens when two polaroids are crossed?

Crossed polaroids transmit nearly zero light. Their pass-axes are at 90°.

1. First Polaroid: Produces plane polarised light.
2. Second Polaroid: Has perpendicular pass-axis.
3. Component Along Axis: Zero.
4. Final Result: Crossed polaroids block transmitted light.

Malus Law Class 12 Questions

Malus law gives the transmitted intensity through an analyser.
The intensity depends on the angle between polarised light and analyser axis.
These Malus law Class 12 questions cover intensity control using two polaroids.

45. State Malus law.

Malus law states that transmitted intensity equals I0 cos²θ. Here θ is the angle between polarisation direction and analyser axis.

1. Incident Polarised Intensity: I0.
2. Angle: θ.
3. Formula Used: I = I0 cos²θ.
4. Final Result: Malus law gives transmitted polarised intensity.

46. What is transmitted intensity when θ = 60°?

The transmitted intensity is I0/4. Use I = I0 cos²θ.

1. Given Data: θ = 60°.
2. Formula Used: I = I0 cos²θ.
3. Calculation:
   I = I0 cos²60°
   I = I0 × (1/2)²
   I = I0/4
4. Final Result: Transmitted intensity = I0/4.

47. What is transmitted intensity when two polaroids are at 90°?

The transmitted intensity is zero for ideal polaroids. Use I = I0 cos²90°.

1. Given Data: θ = 90°.
2. Formula Used: I = I0 cos²θ.
3. Calculation:
   I = I0 cos²90°
   I = 0
4. Final Result: Crossed ideal polaroids transmit zero light.

48. When is intensity maximum through two polaroids?

Intensity is maximum when their pass-axes are parallel. The angle θ equals 0°.

1. Given Data: θ = 0°.
2. Formula Used: I = I0 cos²θ.
3. Calculation: I = I0 cos²0° = I0.
4. Final Result: Maximum transmitted intensity is I0.

Wave Optics Numericals Class 12

Numericals in this chapter often need unit conversion from nm, mm, and cm.
Students should identify whether the question uses refraction, interference, diffraction, or polarisation.
These wave optics numericals Class 12 follow the NCERT 2026 exercise style.

49. Monochromatic light of 589 nm enters water of refractive index 1.33. Find refracted wavelength.

The refracted wavelength is 443 nm. Frequency remains unchanged during refraction.

1. Given Data:
   λair = 589 nm
   nwater = 1.33
2. Formula Used: λwater = λair/n.
3. Calculation:
   λwater = 589/1.33
   λwater = 443 nm
4. Final Result: Wavelength in water = 443 nm.

50. What is speed of light in water with refractive index 1.33?

The speed is 2.26 × 10^8 m/s. Use v = c/n.

1. Given Data:
   c = 3.0 × 10^8 m/s
   n = 1.33
2. Formula Used: v = c/n.
3. Calculation:
   v = 3.0 × 10^8/1.33
   v = 2.26 × 10^8 m/s
4. Final Result: Speed in water = 2.26 × 10^8 m/s.

51. What is the shape of wavefront from a distant star reaching Earth?

The intercepted wavefront is approximately plane. Earth receives only a tiny part of a huge spherical wavefront.

1. Source: Distant star.
2. Wavefront Type: Spherical near the star.
3. At Earth: Small portion appears plane.
4. Final Result: Distant starlight gives a plane wavefront.

52. What is the shape of wavefront after a convex lens when source is at focus?

The emerging wavefront is plane. Rays become parallel after passing through the convex lens.

1. Source Position: At focus.
2. Lens Action: Converts diverging wavefront to plane wavefront.
3. Ray Behaviour: Emergent rays are parallel.
4. Final Result: The wavefront becomes plane.

NCERT Class 12 Physics Chapter 10 Questions

NCERT questions from Wave Optics test exact formula choice more than lengthy derivations.
The common traps involve frequency constancy, phase difference, and rms-free intensity ratios.
These NCERT Class 12 Physics Chapter 10 questions follow the 2026 textbook pattern.

53. Does reflected light have the same frequency as incident light?

Yes, reflected light has the same frequency as incident light. Atoms oscillate at the incident frequency and re-emit light at that frequency.

1. Incident Light: Drives atomic oscillators.
2. Oscillator Frequency: Same as incident frequency.
3. Reflected Light: Same frequency.
4. Final Result: Reflection does not change frequency.

54. Does refracted light have the same frequency as incident light?

Yes, refracted light has the same frequency as incident light. Speed and wavelength change, but frequency stays fixed.

1. Frequency: Set by the source.
2. Speed: Changes in a new medium.
3. Wavelength: Changes with speed.
4. Final Result: Refraction does not change frequency.

55. Does lower wave speed mean lower light energy?

No, lower wave speed does not mean lower light energy. In wave optics, energy depends on amplitude.

1. Speed: Depends on medium.
2. Energy: Depends on amplitude in wave picture.
3. Photon Picture: Intensity depends on photon number per area per time.
4. Final Result: Speed reduction does not imply energy reduction.

56. Why does diffraction not violate conservation of energy?

Diffraction does not violate conservation of energy because light energy redistributes. Dark regions lose intensity, and bright regions gain intensity.

1. Dark Band: Lower intensity.
2. Bright Band: Higher intensity.
3. Total Energy: Conserved across pattern.
4. Final Result: Diffraction redistributes energy.

57. Why does a convex lens make a converging spherical wavefront?

A convex lens delays the central part of a plane wavefront more than the edges. The emerging wavefront becomes spherical and convergent.

1. Lens Thickness: Maximum at centre.
2. Speed in Glass: Less than air.
3. Delay: Central part delays most.
4. Final Result: The wavefront converges to the focus.

Class 12 Physics Chapter 10 Derivations

Derivations in Wave Optics need clear geometry and path difference logic.
Students should write the wavefront condition before using trigonometric relations.
These Class 12 Physics Chapter 10 derivations cover Snell’s law, YDSE, and Malus law.

58. How do you derive Snell’s law using Huygens principle?

Snell’s law follows by comparing wavefront travel distances in two media. The final relation is n1 sin i = n2 sin r.

1. Incident Medium Speed: v1.
2. Refracted Medium Speed: v2.
3. Time Taken: t.
4. Distances:
   BC = v1t
   AE = v2t
5. Triangle Relations:
   sin i = v1t/AC
   sin r = v2t/AC
6. Ratio: sin i/sin r = v1/v2.
7. Using n = c/v: n1 sin i = n2 sin r.
8. Final Result: Snell’s law is n1 sin i = n2 sin r.

59. How do you derive fringe width in Young’s double-slit experiment?

Fringe width comes from the separation between two consecutive bright fringes. The result is β = λD/d.

1. Bright Fringe Position: xn = nλD/d.
2. Next Bright Fringe: xn+1 = (n + 1)λD/d.
3. Fringe Width:
   β = xn+1 − xn
   β = λD/d
4. Final Result: Fringe width = λD/d.

60. How do you derive intensity formula for two coherent waves?

The intensity formula comes from adding two equal-amplitude waves with phase difference φ. The result is I = 4I0 cos²(φ/2).

1. Wave 1: y1 = a cos ωt.
2. Wave 2: y2 = a cos(ωt + φ).
3. Resultant:
   y = 2a cos(φ/2) cos(ωt + φ/2)
4. Amplitude: 2a cos(φ/2).
5. Intensity: I = 4I0 cos²(φ/2).
6. Final Result: Resultant intensity = 4I0 cos²(φ/2).

61. How do you derive Malus law?

Malus law follows by resolving the electric field along the analyser axis. The intensity varies as the square of amplitude.

1. Incident Polarised Field: E0.
2. Analyser Component: E = E0 cos θ.
3. Intensity Relation: I ∝ E².
4. Final Formula: I = I0 cos²θ.
5. Final Result: Malus law is I = I0 cos²θ.

Class 12 Physics Chapter-Wise Important Questions