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#wavelength

133 public questions tagged with this topic.

Optical Phenomena and Applications

What is the angular position of the first minimum in a single-slit diffraction pattern if the slit width is \( 15.0 \, \

**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. First minimum occurs at sin θ = (λ/a) . λ = 7.5 × 10⁻⁷ m , a = 1.5 × 10⁻⁵ m . sin θ = (7.5 × 10⁻⁷/1.5 × 10⁻⁵) = 0.05 , θ = sin⁻¹(0.05) ≈ 2.9° . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ'

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Optical Phenomena and Applications

What is the wavelength of light in a medium with refractive index 1.5 if its wavelength in vacuum is \( 750 \, \text{nm}

**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. Wavelength in a medium λ_m = (λvₐcuuₘ/n) . Given λvₐcuuₘ = 750 nm , n = 1.5 , λ_m = (750/1.5) = 500 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 cos(φ/2), calculation gives 500 nm, illustrating interf

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Optical Phenomena and Applications

In a double-slit experiment, if \( \lambda = 400 \, \text{nm} \), \( d = 0.4 \, \text{mm} \), and \( D = 2.0 \, \text{m}

**Incoherent sources** intensity adds I = I₁+I₂, no interference pattern because phase random, two independent sources cannot produce stable interference because phase difference fluctuates rapidly, coherent sources required with constant phase, laser coherent, visibility of fringes requires coherence, degree of coherence determines contrast. Bright fringe position x_n = (n λ D/d) . For the third bright fringe, n = 3 . λ = 4.0 × 10⁻⁷ m , d = 4.0 × 10⁻⁴ m , D = 2.0 m . x₃ = (3 × 4.0 × 10⁻⁷ × 2.0/4.0 × 10⁻⁴) = 6.0 × 10⁻³ m = 6.0 mm . Using Δ = d sinθ,

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Optical Phenomena and Applications

What is the distance of the fourth bright fringe from the central maximum in a double-slit experiment if \( \lambda = 56

**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. Bright fringe position x_n = (n λ D/d) . For the fourth bright fringe, n = 4 . λ = 5.6 × 10⁻⁷ m , d = 3.5 × 10⁻⁴ m , D = 1.4 m . x₄ = (4 × 5.6 × 10⁻⁷ × 1.4/3.5 × 10⁻⁴) = 8.96 × 10⁻³ m = 8.96 mm . Using Δ = d sinθ, y = n λ

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Wave Properties, Frequency and Energy Conservation

What is the frequency of light with a wavelength of \( 510 \, \text{nm} \) in air, given the speed of light in air is \(

**Intensity not depend on speed** when enters denser medium because intensity I ∝ n E₀²? Actually Poynting vector S = E×H, energy density u =½ ε E², for same amplitude E₀ intensity proportional to n, but amplitude changes at interface due to reflection, total energy conserved incident = reflected + transmitted, interference does not destroy energy, it redistributes. Frequency nu = (c/λ) . λ = 5.1 × 10⁻⁷ m , c = 3.0 × 10⁸ m/s . nu = (3.0 × 10⁸/5.1 × 10⁻⁷) ≈ 5.88 × 10¹⁴ Hz . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ,

Ref: NCERT > Physics Book > Wave Optics > Wave Properties, Frequency and Energy Conservation

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Wave Properties, Frequency and Energy Conservation

What is the wavelength of light in a medium with refractive index 1.25 if its wavelength in air is \( 625 \, \text{nm} \

**Frequency of light** remains unchanged when refracts from air into water because frequency determined by source, energy E= h f conserved, speed decreases v= c/n, wavelength decreases λ'=v/f= λ/n, energy of wave proportional to amplitude² not speed, intensity I =½ c ε₀ E₀², energy not depend on speed directly, when speed decreases amplitude may change but energy conserved, interference redistributes energy, total energy same, bright regions gain from dark. Wavelength in a medium λ_m = (λₐir/n) . Given λₐir = 625 nm , n = 1.25 , λ_m = (625/1.25) = 500 nm . Using Δ = d sinθ, y = n λ D/d, a

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Polarization and Malus Law

What is the fringe width in a double-slit experiment if \( \lambda = 590 \, \text{nm} \), \( d = 0.25 \, \text{mm} \), a

**Polarization** transverse wave property, light after polaroid polarized along pass-axis, Malus law I = I₀ cos²θ, θ angle between pass-axes, initial unpolarized intensity I₀ after first polaroid I₁ = I₀/2, after second at 45° I₂ = I₁ cos²45°= I₀/2×0.5= I₀/4, after two perpendicular 90° I=0 because cos90°=0, for 60° I= I₀/2×cos²60°= I₀/2×0.25= I₀/8. Fringe width β = (λ D/d) . λ = 5.9 × 10⁻⁷ m , d = 2.5 × 10⁻⁴ m , D = 1.2 m . β = (5.9 × 10⁻⁷ × 1.2/2.5 × 10⁻⁴) = 2.832 × 10⁻³ m = 2.83 mm . Using Δ = d sinθ, y = n

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Wave Properties, Frequency and Energy Conservation

What is the frequency of light with a wavelength of \( 460 \, \text{nm} \) in air, given the speed of light in air is \(

**Wave model predicts** light bends away from normal when entering rarer medium because speed increases, Snell's law n₁ sinθ₁ = n₂ sinθ₂, n₁>n₂ so sinθ₂>sinθ₁ θ₂>θ₁ away from normal, towards normal when denser, wavefront slows in denser, Huygens construction shows bending. Frequency nu = (c/λ) . λ = 4.6 × 10⁻⁷ m , c = 3.0 × 10⁸ m/s . nu = (3.0 × 10⁸/4.6 × 10⁻⁷) ≈ 6.52 × 10¹⁴ Hz . 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

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Wavefront and Huygens Principle

In a double-slit experiment, if \( \lambda = 530 \, \text{nm} \), \( d = 0.15 \, \text{mm} \), and \( D = 1.8 \, \text{m

**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. Fringe width β = (λ D/d) . λ = 5.3 × 10⁻⁷ m , d = 1.5 × 10⁻⁴ m , D = 1.8 m . β = (5.3 × 10⁻⁷ × 1.8/1.5 × 10⁻⁴) = 6.36 × 10⁻³ m = 6.36 mm . Using Δ = d sinθ, y = n λ D/d,

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Superposition, Resultant Amplitude and Intensity

What is the phase difference corresponding to a path difference of \( 2\lambda \) in a double-slit experiment?

**Intensity at point** in double-slit I = I_max cos²(φ/2), φ = (2π/λ)Δ, for Δ=λ/4 φ=π/2 I= I_max cos²(π/4)= I_max/2 =2I₀, for Δ=λ/3 φ=2π/3 I= I_max cos²(π/3)= I_max×0.25= I₀, for Δ=5λ/8 φ=5π/4? Actually φ=2π×5/8=5π/4, cos²(5π/8)=?, path difference for destructive φ=(2n+1)π, constructive 2nπ. Phase difference Φ = (2π/λ) Δ . For Δ = 2λ , Φ = (2π/λ) · 2λ = 4π . 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 4π, illustrating interference, diffraction and polarization principles.

Ref: NCERT > Physics Book > Wave Optics > Superposition, Resultant Amplitude and Intensity

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Wavefront and Huygens Principle

What is the angular width of the central maximum in a single-slit diffraction pattern if the slit width is \( 5.0 \, \mu

**Wavefront** is locus of points in same phase, spherical from point source, plane at large distance because radius large, Huygens principle every point on wavefront acts as secondary source of wavelets, new wavefront envelope of secondary wavelets, allows prediction of new wavefront shape from known wavefront, explains reflection and refraction. Angular width 2θ = (2λ/a) . λ = 6.5 × 10⁻⁷ m , a = 5.0 × 10⁻⁶ m . sin θ = (λ/a) = (6.5 × 10⁻⁷/5.0 × 10⁻⁶) = 0.13 , θ = sin⁻¹(0.13) ≈ 7.5° , 2θ ≈ 15° . Using Δ = d sinθ, y = n λ D/d, a sinθ

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Wavefront and Huygens Principle

What is the angular position of the third minimum in a single-slit diffraction pattern if the slit width is \( 10.0 \, \

**Wavefront types** point source spherical, distant point source plane, after convex lens plane wave focuses to point because lens adds phase delay proportional to thickness, converging spherical wavefront, after concave mirror plane wave becomes spherical converging to focus, illustrating Huygens construction. Minima occur at sin θ = (nλ/a) . For the third minimum, n = 3 . λ = 6.0 × 10⁻⁷ m , a = 1.0 × 10⁻⁵ m . sin θ = (3 × 6.0 × 10⁻⁷/1.0 × 10⁻⁵) = 0.18 , θ = sin⁻¹(0.18) ≈ 10.4° . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ,

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

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