A monochromatic light of wavelength 6000 Å is incident on the single slit of width 0.01 mm. If the diffraction pattern is formed at the focus of the convex lens of focal length 20 cm, the linear width of the central maximum is :

Solution

Physics sir · 4 steps

The central maximum spans from the first minimum on one side to the first minimum on the other, so double it.

1. First minimum. For a single slit of width a, the first dark fringe is at angle θ where a sin θ = λ. The angle is small, so θ ≈ λ/a.
2. On the screen. The lens focuses the pattern at distance f, so the first minimum sits at y = fθ = λ f/a from the centre. The central maximum stretches to the first minimum on both sides, so its width is twice this.

   W = (2λ f)/a = (2 × (6 × 10⁻⁷ m) × (0.2 m))/(1 × 10⁻⁵ m)

3. Put in numbers. Convert first: 6000 Å = 6 × 10⁻⁷ m, 0.01 mm = 10⁻⁵ m, 20 cm = 0.2 m. The numerator is 2.4 × 10⁻⁷ m², so W = 2.4 × 10⁻² m.
4. So. The linear width of the central maximum is 0.024 m = 24 mm, option (b).

The idea to keep. Central maximum width is 2λ f/a (or 2λ D/a for a screen). The factor 2 comes from going minimum to minimum, and every length must be in metres before you divide.

Answer from the NTA final answer key. Solution by PrepWiser.