In a Young's double slit experiment, the distance of the screen from the slits is 2.0 m , the separation between the slits is 2.0 mm , and the wavelength of the light used is 600 nm . Given that the intensity at the centre of the central maximum is 0.20 W m ⁻² , determine the intensity at a point located 0.5 cm away from this centre along the width of the fringes.

Step-by-step solution

Given D = 2.0 m , d = 2.0 10⁻³ m , = 600 10⁻⁹ m , I₀ = 0.20 W m ⁻² , and y = 0.5 10⁻² m . The path difference at distance y from the central maximum is given by: x = y d D Substituting the values: x = 0.5 10⁻² 2.0 10⁻³ 2.0 = 5 10⁻⁶ m The corresponding phase difference is: = 2 x = 2 600 10⁻⁹ 5 10⁻⁶ = 50 3 The intensity at this point is given by: I = I₀ ^2 ( 2 ) I = 0.20 ^2 ( 25 3 ) I = 0.20 ^2 (8 + 3 ) = 0.20 ^2 ( 3 ) I = 0.20 ( 1 2 )^2 = 0.05 W m ⁻²