StreakPeaked· Practice

ExamsJEE MainPhysics

Two slits 1 mm apart are illuminated by light of wavelength 6.5*10⁻⁷ m. The interference pattern is seen on a screen 1 m from the slits. What is the distance between the third dark fringe and the fifth bright fringe (on the same side of the centre)?

  1. 1.63 mm
  2. 0.65 mm
  3. 3.25 mm
  4. 4.88 mm

Correct answer: 1.63 mm

Solution

Fringe scale lambda*D/d = (6.5e-7 * 1)/(1e-3) = 6.5e-4 m. Fifth bright: 5 * 6.5e-4. Third dark: 2.5 * 6.5e-4. Difference = (5 - 2.5)*6.5e-4 = 2.5*6.5e-4 = 1.625e-3 m ~ 1.63 mm.

Related JEE Main Physics questions

⚔️ Practice JEE Main Physics free + battle 1v1 →