Orange light of wavelength 6000 × 10 –10 m illuminates a single
slit of width 0.6 × 10 –4 m. The maximum possible number of diffraction minima produced on both sides of the central maximum is ___________.

Solution

Physics sir · 4 steps

Since sin θ can never exceed 1, the ratio a/λ caps the order of the minima.

1. Condition for minima. For a single slit, the nth dark fringe appears where the path difference across the slit is a whole number of wavelengths. Here n = 1, 2, 3, … on each side of the centre.

   a sin θ = nλ ⇒ n = (a sin θ)/λ ≤ a/λ

2. Put in numbers. a/λ = (0.6 × 10⁻⁴ m)/(6000 × 10⁻¹⁰ m) = (6 × 10⁻⁵)/(6 × 10⁻⁷) = 100. So the highest possible order is n = 100, reached at the limit sin θ = 1.
3. Count both sides. Orders 1 to 100 appear above the central maximum and orders 1 to 100 appear below it. That makes 2 × 100 minima in all.
4. So. The maximum possible number of diffraction minima is 200.

The common slip: 100. Counting the minima on only one side of the central maximum gives 100. The pattern repeats on the other side, so double it.

The idea to keep. For a maximum count of fringes, set sin θ = 1 in the fringe condition. The pattern is symmetric, so double the count from one side.

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