JEE Main202621 January 2026Morning ShiftPhysicsWave OpticsActual
In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness t and refractive index n(=1.5) , the central fringe shifts by 0.2 cm. The value of t is _ _ _ _ cm.
Options
- A5.0 10⁻³
- B6.0 10⁻³
- C8 10⁻⁴
- D5.6 10⁻⁴
Correct answer
C. 8 10⁻⁴
Step-by-step solution
Given: d = 0.1 cm, D = 50 cm, n = 1.5 , Shift = 0.2 cm Optical path difference by sheet: = (n-1)t Shift of central fringe: Shift = (n-1)t D d 0.2 = (1.5-1) t 50 0.1 = 0.5 50 t 0.1 = 250t t = 0.2 250 = 8 10⁻⁴ cm