Skip to content

Question

What is the angular position of the second minimum in a single-slit diffraction pattern if the slit
width is \( 2.0 \, \mu\text{m} \) and the wavelength is \( 400 \, \text{nm} \)?

Options

Choose one · Correct answer highlighted

Explanation

**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. Minima occur at sin θ = (nλ/a) . For the second minimum, n = 2 . λ = 4.0 × 10⁻⁷ m , a = 2.0 × 10⁻⁶ m . sin θ = (2 × 4.0 × 10⁻⁷/2.0 × 10⁻⁶) = 0.4 , θ = sin⁻¹(0.4) ≈ 23.6° . Using Δ

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.