In a YDSE, the distance between the two slits is 2 mm and the wavelength of incident light is 578.9 nm . Path difference between the waves from lower and upper slits reaching a point P where a bright fringe is formed is 1.36 ~m . What is the distance of point P from the central bright fringe if the distance between the screen and the slits is 250 cm ?

Step-by-step solution

The path difference x at a point P on the screen in a Young's Double Slit Experiment is given by the formula x = d d = d y D , where d is the distance between the slits, y is the distance of point P from the central bright fringe, and D is the distance between the slits and the screen. Given values are: d = 2 mm = 2 10⁻³ m D = 250 cm = 2.5 m x = 1.36 m = 1.36 10⁻⁶ m Rearranging the formula for y : y = x D d Substituting the values: y = 1.36 10⁻⁶ 2.5 2 10⁻³ y = 3.4 10⁻⁶ 2 10⁻³ y = 1.7 10⁻³ m y = 1.7 mm Answer: 1.7 m