Wave optics

NEET

Wave Optics

1. Huygens' Principle

Every point on a wavefront acts as a source of secondary wavelets, spreading out in all directions with the speed of light. The new wavefront is the tangent envelope to these secondary wavelets.

Wavefront Shapes:
- Point Source → Spherical Wavefront.
- Line Source → Cylindrical Wavefront.
- Source at Infinity → Plane Wavefront.

2. Interference (YDSE)

Superposition of waves from two coherent sources resulting in redistribution of energy (Bright and Dark fringes).

Path Difference (Δx): d sinθ ≈ xd/D.
Constructive (Bright): Δx = nλ (n=0,1,2...).
Destructive (Dark): Δx = (2n-1)λ/2 (n=1,2...).
Fringe Width (β): β = λD/d.
Young's Double Slit Experiment

(Asked in NEET 2017, 2019, 2022)

3. Diffraction (Single Slit)

Bending of light around corners of an obstacle.

Condition for Minima (Dark): a sinθ = nλ (n=1,2...).
Condition for Maxima (Bright): a sinθ = (2n+1)λ/2.
Width of Central Maximum: 2λD/a (Double the width of other fringes).
Single Slit Diffraction Pattern

(Asked in NEET 2016, 2020)

4. Polarisation

Restriction of vibrations of light wave to a single plane.

Malus' Law: I = I0 cos2θ.
Brewster's Law: Reflected light is completely polarised when reflected at Brewster angle ip.
tan ip = μ.
Polarisation and Malus Law

(Asked in NEET 2018, 2021)

Numericals: Wave Optics

Q1. In YDSE, slits are 2 mm apart and screen is 1.5 m away. Wavelength is 600 nm. Find fringe width.
Solution:
d = 2×10-3 m, D = 1.5 m, λ = 600×10-9 m.
β = λD / d.
β = (600×10-9 × 1.5) / (2×10-3).
β = (900 × 10-9) / (2 × 10-3) = 450 × 10-6 m.
β = 0.45 mm.
Q2. Ratio of intensities of two waves is 9:1. Find ratio of maximum to minimum intensity.
Solution:
I1/I2 = 9/1 → a1/a2 = 3/1.
Imax/Imin = (a1 + a2)2 / (a1 - a2)2.
Ratio = (3+1)2 / (3-1)2 = 42 / 22 = 16/4 = 4:1.
Q3. Screen is 2m from single slit of width 1mm. Light of 500nm used. Find width of central maximum.
Solution:
Width = 2λD / a.
Width = 2 × (500×10-9) × 2 / (1×10-3).
Width = 2000 × 10-6 m = 2 mm.
Q4. Two polaroids are oriented with axes at 30°. Unpolarized light of intensity I0 falls on first. Find final intensity.
Solution:
After 1st polaroid: I1 = I0/2.
After 2nd polaroid: I2 = I1 cos2(30°).
I2 = (I0/2) × (√3/2)2 = (I0/2) × (3/4).
I2 = 3I0/8.
Q5. Distance of 3rd dark fringe from central maximum in YDSE? (λ=500nm, D=1m, d=1mm).
Solution:
For dark fringe: xn = (2n-1)λD / 2d.
For 3rd dark fringe, n=3. (2(3)-1) = 5.
x3 = 5λD / 2d.
x3 = 5 × (500×10-9) × 1 / (2 × 10-3).
x3 = 1250 × 10-6 m = 1.25 mm.
Q6. Light is reflected from glass plate (μ=1.732) completely polarised. Find angle of refraction.
Solution:
tan ip = μ = 1.732 = √3 → ip = 60°.
At polarising angle, reflected and refracted rays are perpendicular.
ip + r = 90°.
r = 90 - 60 = 30°.
Q7. Angular fringe width in air is 0.2°. Apparatus immersed in water (μ=4/3). New angular width?
Solution:
θwater = θair / μ.
θwater = 0.2 / (4/3) = 0.6 / 4.
θwater = 0.15°.
Q8. Find path difference at a point where phase difference is π/3.
Solution:
Phase difference φ = (2π/λ) Δx.
π/3 = (2π/λ) Δx.
1/3 = 2/λ Δx → Δx = λ/6.
Path difference = λ/6.
Q9. Mica sheet (μ=1.5, t=6μm) placed in path of one wave in YDSE. λ=600nm. How many fringes shift?
Solution:
Shift Δx = (μ-1)t.
Number of fringes N = Δx / λ = (μ-1)t / λ.
N = (1.5 - 1)(6 × 10-6) / (600 × 10-9).
N = (0.5 × 6 × 1000) / 600 = 3000 / 600 = 5 fringes.
Q10. Width ratio of two slits is 4:1. Ratio of amplitudes?
Solution:
Width w ∝ Intensity I ∝ Amplitude2 (a2).
w1/w2 = 4/1 → I1/I2 = 4/1.
a1/a2 = √(I1/I2) = √4/1 = 2:1.

Important Formulae

1. Interference (YDSE)

Path Difference:

Δx = xd/D

Fringe Width:

β = λD/d

Resultant Intensity:

I = I1 + I2 + 2√(I1I2) cosφ
2. Diffraction (Single Slit)

Minima Angles:

sin θn = nλ/a

Linear Width (Central Max):

Width = 2λD/a
3. Polarisation

Malus' Law:

I = I0 cos2θ

Brewster's Law:

μ = tan ip

20 NEET Golden Facts

  • 1. Wave Nature: Light exhibits both wave and particle nature (Dual nature). Interference/Diffraction prove wave nature.
  • 2. Coherent Sources: Must have constant phase difference. Independent sources can never be coherent.
  • 3. YDSE in Water: If whole apparatus is immersed in water (μ), fringe width decreases (β' = β/μ).
  • 4. White Light in YDSE: Central fringe is white. Other fringes are coloured (Violet nearest to center).
  • 5. Doppler Effect: Red shift (Source moving away, λ increases). Blue shift (Source approaching, λ decreases).
  • 6. Intensity: I ∝ A2. For equal amplitudes a, Imax = 4I0, Imin = 0.
  • 7. Energy Conservation: Interference redistributes energy. Average intensity remains <I1 + I2>.
  • 8. Diffraction limits: Resolving power of optical instruments is limited by diffraction.
  • 9. Resolving Power (Microscope): RP = 2μ sinθ / 1.22λ. (Increase μ with oil immersion to increase RP).
  • 10. Resolving Power (Telescope): RP = D / 1.22λ. (Larger aperture D → Better resolution).
  • 11. Fresnel Distance: Zf = a2/λ. For Z < Zf, Ray optics is valid.
  • 12. Phase Difference: φ = (2π/λ) Δx.
  • 13. Missing Fringes: If slit width a is such that 1st diffraction minimum coincides with nth interference maximum, that order is missing.
  • 14. Thin Film Interference: Soap bubbles appear colored due to interference.
  • 15. Crossed Polaroids: If angle between pass axes is 90°, Intensity I = 0.
  • 16. Unpolarised Light: Intensity becomes I0/2 after passing through first polaroid.
  • 17. Transverse Nature: Polarisation proves light is a transverse wave. (Sound waves cannot be polarised).
  • 18. Shift of Fringes: Introducing a slab of thickness t shifts fringes by Δx = (μ-1)t D/d.
  • 19. Sustained Interference: Needs coherent sources, same frequency, same direction, nearly same amplitude.
  • 20. Angular Width: θ = λ/d (Interference), θ = λ/a (Diffraction Half Width).
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