In a single slit experiment, a parallel beam of green light of wavelength 550 nm passes through a slit of width 0.20 mm. The transmitted light is collected on a screen 100 cm away. The distance of first order minima from the central maximum will be x × 10⁻⁵ m. The value of x is :

Solution

Physics sir · 3 steps

First minimum sits where a sin θ = λ; small angle turns that into y = Dλ/a.

1. Condition for minimum. For a single slit of width a, the first dark fringe appears where a sin θ = λ. The angle is tiny, so sin θ ≈ tan θ = y/D.

   y₁ = Dλ/a

2. Put in numbers. Use SI units: D = 1 m, λ = 550 × 10⁻⁹ m, a = 0.20 × 10⁻³ m. Then y₁ = (1 × 550 × 10⁻⁹)/(0.20 × 10⁻³) = 2.75 × 10⁻³ m.
3. So. 2.75 × 10⁻³ m = 275 × 10⁻⁵ m, so x = 275.

The common slip: 2.75. Stopping at y = 2.75 mm gives 2.75. The question wants the distance as a multiple of 10⁻⁵ m, and 2.75 × 10⁻³ m = 275 × 10⁻⁵ m.

The idea to keep. In single slit diffraction the nth minimum is at yₙ = nDλ/a. Put every length in metres first, then match the power of ten the question asks for.

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