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#light propagation

5 public questions tagged with this topic.

What allows light to propagate through a vacuum without requiring a traditional medium?

**Polarization requires transverse waves** because only transverse can have orientation perpendicular to propagation, longitudinal cannot be polarized, wave theory requires light transverse to explain polarization, polaroids transmit only component along pass-axis, unpolarized has random transverse orientations, after polaroid polarized. Light consists of oscillating electric and magnetic fields that sustain each other, enabling propagation without a material medium, as explained by electromagnetic theory. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n

Ref: NCERT > Physics Book > Wave Optics > Optical Phenomena and Applications

Why does the wave theory suggest that light’s behavior approximates straight-line propagation in large-scale scenarios?

**Single-slit pattern** intensity I = I₀ (sinα/α)², α=π a sinθ/λ, central maximum at α=0, minima at α=nπ, so a sinθ=nλ, width increases with λ and D decreases with a, for a=15 μm λ=750 nm first minimum sinθ=750/15000=0.05 θ≈2.87°, angular width of central maximum 2θ≈5.74°. Light’s wavelength is extremely small compared to everyday objects, minimizing wave effects like diffraction, making straight-line propagation dominant. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation gives High speed, illustrating interference, diffra

Ref: NCERT > Physics Book > Wave Optics > Diffraction - Single-Slit and Central Maximum

What is the wavelength of light in a medium with refractive index 1.6 if its wavelength in vacuum is \( 640 \, \text{nm}

**Refractive index** n = c/v, c=3×10⁸ m/s vacuum, v speed in medium, for v=2.25×10⁸ m/s n=3/2.25=1.33, for v=1.8×10⁸ n=1.67, for n=1.3 v=3×10⁸/1.3=2.31×10⁸ m/s, for n=1.6 v=1.875×10⁸ m/s, for n=1.2 λ_medium = λ_vacuum/n =600/1.2=500 nm, for n=1.75 λ=700/1.75=400 nm, for n=1.5 λ=750/1.5=500 nm, wavelength in medium λ' = λ/n. Wavelength in a medium λ_m = (λvₐcuuₘ/n) . Given λvₐcuuₘ = 640 nm , n = 1.6 , λ_m = (640/1.6) = 400 nm . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a

Ref: NCERT > Physics Book > Wave Optics > Refraction, Refractive Index and Critical Angle

What is the direction of rays relative to the wavefront in a light wave?

**Huygens principle** predicts shape of wavefront after propagation, for point source close spherical, far plane, after reflection from plane mirror spherical wave becomes spherical with centre mirrored, plane wave remains plane but direction changes angle of incidence equals reflection, after passing through thin prism plane wavefront tilts due to different path. Rays are perpendicular to the wavefront, representing the direction of energy propagation. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation gives Perpendicular,

Ref: NCERT > Physics Book > Wave Optics > Wavefront and Huygens Principle

What is the key factor that determines whether light undergoes total internal reflection at an interface?

**Lens maker's formula** 1/f = (n-1)(1/R₁ - 1/R₂), n refractive index, R₁,R₂ radii of curvature (m), sign convention R positive if surface convex towards incident light. For double convex R₁=12 cm, R₂=-12 cm, n=1.5, 1/f=(0.5)(1/12 -1/(-12))=(0.5)(2/12)=1/12, f=12 cm, converging. Total internal reflection occurs when light travels from a denser to a rarer medium and the angle of incidence exceeds the critical angle. The critical angle depends on the refractive indices of the two media, making the angle of incidence the decisive factor for the phenomenon. Substituting values gives Angle of incid

Ref: NCERT > Physics Book > Ray Optics > Refraction at Spherical Surfaces and Lens Maker's Formula