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#fringe width

17 public questions tagged with this topic.

Wave Properties, Frequency and Energy Conservation

What is the fringe width in a double-slit experiment if \( \lambda = 660 \, \text{nm} \), \( d = 0.3 \, \text{mm} \), an

**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. Fringe width β = (λ D/d) . λ = 6.6 × 10⁻⁷ m , d = 3.0 × 10⁻⁴ m , D = 1.5 m . β = (6.6 × 10⁻⁷ × 1.5/3.0 × 10⁻⁴) =

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

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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

Ref: NCERT > Physics Book > Wave Optics > Polarization and Malus Law

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Polarization and Malus Law

What happens to the fringe width in a double-slit experiment if the slit separation is halved?

**Polaroid rotation** intensity varies sinusoidally with angle due to Malus law I = I₀ cos²θ, when polaroid rotated 90° from initial, intensity goes from max to zero, for unpolarized light rotating polaroid does not change intensity after first polaroid because average, but second polaroid intensity depends on relative angle, explains why intensity transmitted through two polaroids drops to zero when perpendicular. Fringe width β = (λ D/d) . If d is halved, β doubles. 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 =

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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,

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

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Diffraction - Single-Slit and Central Maximum

In a double-slit experiment, if \( \lambda = 450 \, \text{nm} \), \( d = 0.15 \, \text{mm} \), and \( D = 1.5 \, \text{m

**Single-slit diffraction** central maximum width W =2λ D/a, a slit width, D distance, angular width θ =2λ/a, first minimum at a sinθ = λ, fourth minimum a sinθ=4λ, sinθ=4λ/a, for a=5.0 μm λ=500 nm sinθ=4×0.5/5=0.4 θ≈23.6°, central maximum width increases when slit width reduced to half doubles width, when wavelength quadrupled width quadruples, when slit tripled width one-third. Fringe width β = (λ D/d) . λ = 4.5 × 10⁻⁷ m , d = 1.5 × 10⁻⁴ m , D = 1.5 m . β = (4.5 × 10⁻⁷ × 1.5/1.5 × 10⁻⁴) = 4.5 × 10⁻³ m = 4.5 mm . Using Δ =

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

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Interference - Double-Slit and Coherence

In a double-slit experiment, if \( \lambda = 520 \, \text{nm} \), \( d = 0.4 \, \text{mm} \), and \( D = 2.5 \, \text{m}

**Coherence** requires constant phase difference, two independent sources not coherent because phase random, prevents interference pattern, laser coherent, visibility of fringes determined by coherence, intensity and phase difference, coherent sources produce stable interference, incoherent intensity adds I = I₁+I₂ no interference, interference pattern distinguishes from diffraction. Fringe width β = (λ D/d) . λ = 5.2 × 10⁻⁷ m , d = 4.0 × 10⁻⁴ m , D = 2.5 m . β = (5.2 × 10⁻⁷ × 2.5/4.0 × 10⁻⁴) = 3.25 × 10⁻³ m = 3.25 mm . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ,

Ref: NCERT > Physics Book > Wave Optics > Interference - Double-Slit and Coherence

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Interference - Double-Slit and Coherence

In a double-slit experiment, if \( \lambda = 580 \, \text{nm} \), \( d = 0.25 \, \text{mm} \), and \( D = 2.5 \, \text{m

**Interference** occurs when two coherent waves superpose, path difference Δ = d sinθ, bright fringe when Δ = n λ, n integer, seventh bright Δ=7λ, dark fringe Δ = (2n-1)λ/2, second dark Δ=3λ/2, third dark 5λ/2, sixth dark 11λ/2, distance of bright fringe from central y = n λ D/d, D screen distance, d slit separation, for λ=650 nm d=0.5 mm D=1.0 m fifth dark? Actually dark y=(2n-1)λ D/(2d). Fringe width β = (λ D/d) . λ = 5.8 × 10⁻⁷ m , d = 2.5 × 10⁻⁴ m , D = 2.5 m . β = (5.8 × 10⁻⁷ × 2.5/2.5 × 10⁻⁴) =

Ref: NCERT > Physics Book > Wave Optics > Interference - Double-Slit and Coherence

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Diffraction - Single-Slit and Central Maximum

In a double-slit experiment, if the wavelength is tripled, what happens to the fringe width?

**Diffraction bending** property of light waves causes bending around corners, width of central maximum inversely proportional to slit width, intensity of secondary maxima decreases with order because less constructive interference, angular position of minima θ_n = n λ/a, n=±1,±2..., second minimum n=2, third n=3, condition for third secondary maximum approx a sinθ = (2n+1)λ/2. Fringe width β = (λ D/d) . If λ is tripled, β triples. 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 Triples, illustrating interference,

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

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Interference - Double-Slit and Coherence

What is the fringe width in a double-slit experiment if \( \lambda = 620 \, \text{nm} \), \( d = 0.5 \, \text{mm} \), an

**Interference** occurs when two coherent waves superpose, path difference Δ = d sinθ, bright fringe when Δ = n λ, n integer, seventh bright Δ=7λ, dark fringe Δ = (2n-1)λ/2, second dark Δ=3λ/2, third dark 5λ/2, sixth dark 11λ/2, distance of bright fringe from central y = n λ D/d, D screen distance, d slit separation, for λ=650 nm d=0.5 mm D=1.0 m fifth dark? Actually dark y=(2n-1)λ D/(2d). Fringe width β = (λ D/d) . λ = 6.2 × 10⁻⁷ m , d = 5.0 × 10⁻⁴ m , D = 2.0 m . β = (6.2 × 10⁻⁷ × 2.0/5.0 × 10⁻⁴) =

Ref: NCERT > Physics Book > Wave Optics > Interference - Double-Slit and Coherence

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Diffraction - Single-Slit and Central Maximum

What happens to the fringe width in a double-slit experiment if the wavelength of light is halved?

**Diffraction bending** property of light waves causes bending around corners, width of central maximum inversely proportional to slit width, intensity of secondary maxima decreases with order because less constructive interference, angular position of minima θ_n = n λ/a, n=±1,±2..., second minimum n=2, third n=3, condition for third secondary maximum approx a sinθ = (2n+1)λ/2. Fringe width β = (λ D/d) . If λ is halved, β is halved. 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 Halves, illustrating interference,

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

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Interference - Double-Slit and Coherence

In a double-slit experiment, if the screen distance is doubled, what happens to the fringe width?

**Coherence** requires constant phase difference, two independent sources not coherent because phase random, prevents interference pattern, laser coherent, visibility of fringes determined by coherence, intensity and phase difference, coherent sources produce stable interference, incoherent intensity adds I = I₁+I₂ no interference, interference pattern distinguishes from diffraction. Fringe width β = (λ D/d) . If D is doubled, β doubles. 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 Doubles, illustrating interfer

Ref: NCERT > Physics Book > Wave Optics > Interference - Double-Slit and Coherence

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Diffraction - Single-Slit and Central Maximum

In Young’s double-slit experiment, if the slit separation is \( 0.25 \, \text{mm} \), the screen is \( 1.0 \, \text{m} \

**Diffraction bending** property of light waves causes bending around corners, width of central maximum inversely proportional to slit width, intensity of secondary maxima decreases with order because less constructive interference, angular position of minima θ_n = n λ/a, n=±1,±2..., second minimum n=2, third n=3, condition for third secondary maximum approx a sinθ = (2n+1)λ/2. Fringe width β = (λ D/d) . Given β = 2.4 mm = 2.4 × 10⁻³ m , D = 1.0 m , d = 0.25 mm = 2.5 × 10⁻⁴ m . λ = (β d/D) = (2.4 × 10⁻³ × 2.5 × 10⁻⁴/1.0) = 6.0 × 10⁻⁷ m =

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

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