Diffraction of Light — Important Questions

6 × 10⁻⁴ rad

Step 1: Find wavelength: λ = c/f = (3 × 10⁸) / (5 × 10¹⁴) = 6 × 10⁻⁷ m. Step 2: Half angular spread: sinθ = λ/e = (6 × 10⁻⁷) / (2 × 10⁻³) = 3 × 10⁻⁴. Step 3: Since sinθ is very small, θ ≈ 3 × 10⁻⁴ rad (half spread). Step 4: Total angular spread = 2θ = 6 × 10⁻⁴ rad. Note: The correct answer is 6 × 10⁻⁴ rad. The mistake students make is forgetting to double the half-angular spread to get the total width.

1/22

Step 1: The intensity formula for single-slit diffraction is I = I₀(sinβ/β)², where β = (π/λ) × path difference. Step 2: For the 1st secondary maximum, path difference = 3λ/2, so β = (π/λ)(3λ/2) = 3π/2. Step 3: I = I₀(sin(3π/2) / (3π/2))² = I₀(-1 / (3π/2))² = I₀ × (1/(3π/2))² = I₀ × 4/(9π²). Step 4: 4/(9π²) = 4/(9 × 9.87) = 4/88.8 ≈ 0.045 ≈ 1/22. Step 5: 1/9 is incorrect (ignores the π² factor). 4/π² is the unapproximated value, and 1/4 is far too high and represents no real physical value in this context.

Violet, because it has the shortest wavelength

Step 1: The angular width of the central maximum = 2sin⁻¹(λ/e) ≈ 2λ/e for small angles. Step 2: This shows that angular width ∝ λ (directly proportional to wavelength). Step 3: Violet light has the shortest wavelength (~400 nm) among visible colours, so it produces the narrowest central maximum. Step 4: Red light (~700 nm) has the longest wavelength and therefore diffracts the most, giving the widest central maximum. Step 5: The option 'all colours same width' is wrong because width directly depends on λ. 'Green in middle' is a misconception—position in spectrum does not mean average diffracti

Width doubles; intensity becomes one-quarter

Step 1: Linear width of central maximum = 2λD/e. Since width ∝ 1/e, halving e doubles the width. Step 2: Intensity of central maximum ∝ (amplitude)² ∝ (slit width)² = e². Step 3: If e is halved, intensity ∝ (e/2)² = e²/4, so intensity becomes one-quarter of original. Step 4: Physical reason: A narrower slit passes less total light (fewer wavelets), so energy is spread over a wider area, reducing peak intensity. Step 5: Option 'width doubles, intensity doubles' is wrong because intensity ∝ e², not e. Option 'width halves' is completely opposite to the correct result.