In a Young's double slit experiment, the distance between the slits is 1.0 ~mm and the screen is placed at a distance of 1.0 ~m from the slits. If the wavelength of light used is 600 ~nm , the intensity of light at a point on the screen located at a distance of 0.1 ~mm from the central maximum is x % of the maximum intensity. The value of x is ________.

Step-by-step solution

The path difference x at a distance y from the central maximum is given by: x = y d D Given: y = 0.1 ~mm = 10⁻⁴ ~m d = 1.0 ~mm = 10⁻³ ~m D = 1.0 ~m Substituting the values: x = 10⁻⁴ 10⁻³ 1.0 = 10⁻⁷ ~m = 100 ~nm The phase difference is related to the path difference by: = 2 x Given = 600 ~nm : = 2 600 100 = 3 The intensity I at this point is given by: I = I_ ^2 ( 2 ) I = I_ ^2 ( 6 ) = I_ ( 3 2 )^2 = 3 4 I_ = 0.75 I_ Thus, the intensity is 75 % of the maximum intensity. The value of x is 75 . Answer: 75