In a Young's double slit experiment, the slits are separated by 0.3 mm and the screen is 1.5 m away from the plane of slits. Distance between fourth bright fringes on both sides of central bright is 2.4 cm. The frequency of light used is ______________ × 10¹⁴ Hz.

Solution

Physics sir · 4 steps

Fourth bright on the left to fourth bright on the right is 8 fringe widths, so start by finding one.

1. Count fringe widths. The fourth bright fringe sits 4β from the centre, on each side. The gap between them is 8β = 2.4 cm, so β = 0.3 cm = 3 × 10⁻³ m.

   β = (λ D)/d ⇒ λ = (β d)/D

2. Find wavelength. With d = 3 × 10⁻⁴ m and D = 1.5 m: λ = (3 × 10⁻³ × 3 × 10⁻⁴)/1.5 = 6 × 10⁻⁷ m.
3. Convert to frequency. f = c/λ = (3 × 10⁸)/(6 × 10⁻⁷) = 5 × 10¹⁴ Hz.
4. So. The frequency is 5 × 10¹⁴ Hz, so the blank is 5.

The common slip: 2.5. Taking 4β = 2.4 cm forgets that the span covers both sides of the centre. It doubles β and λ and gives 2.5, not 5.

The idea to keep. When a question says fringes on both sides of the central maximum, the distance spans both halves. The nth bright on each side are 2nβ apart, not nβ.

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