Skip to content
Login Register

#slit separation

9 public questions tagged with this topic.

Wave Properties, Frequency and Energy Conservation

Why does the interference pattern from two slits disappear if the slits are too far apart?

**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. Large slit separation reduces the overlap of wavefronts, disrupting the consistent path difference needed for stable interference. Using Δ

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

View Details
Superposition, Resultant Amplitude and Intensity

In a double-slit experiment, if \( \lambda = 460 \, \text{nm} \), \( d = 0.2 \, \text{mm} \), and \( D = 2.0 \, \text{m}

**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. Bright fringe position x_n = (n λ D/d) . For the third bright fringe, n = 3 . λ = 4.6 × 10⁻⁷ m , d = 2.0 × 10⁻⁴ m , D = 2.0 m . x₃ = (3 × 4.6 × 10⁻⁷ × 2.0/2.0 × 10⁻⁴) = 6.9 × 10⁻³ m = 6.9 mm . Using Δ = d sinθ, y =

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

View Details
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 =

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

View Details
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

View Details
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

View Details
Interference - Double-Slit and Coherence

What is the distance of the second bright fringe from the central maximum in a double-slit experiment if \( \lambda = 67

**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. Bright fringe position x_n = (n λ D/d) . For the second bright fringe, n = 2 . λ = 6.7 × 10⁻⁷ m , d = 6.0 × 10⁻⁴ m , D = 1.8 m . x₂ = (2 × 6.7 × 10⁻⁷ × 1.8/6.0 × 10⁻⁴) = 4.02 × 10⁻³ m = 4.02 mm . Using Δ

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

View Details
Interference - Double-Slit and Coherence

What happens to the fringe width in a double-slit experiment if both the slit separation and screen distance are doubled

**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 and d are both doubled, β = (λ (2D)/2d) = (λ D/d) , so it remains the same. 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 > Interference - Double-Slit and Coherence

View Details
Interference - Double-Slit and Coherence

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

**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) . If d is reduced to (d/3) , β increases to 3β , i.e., triples. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ,

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

View Details
Interference - Double-Slit and Coherence

In a double-slit experiment, if the slit separation 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 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 interfe

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

View Details