In a double slit interference experiment, a source emits light of wavelengths 480 nm and 600 nm . The separation between the slits is 0.25 mm , and the interference is observed on a screen placed at 150 cm from the slits. Determine the linear separation between the first maximum (adjacent to the central maximum) corresponding to the two wavelengths.

Step-by-step solution

The position of the n -th maximum in a double slit interference pattern is given by y_n = n D d . For the first maximum adjacent to the central maximum, n = 1 . The position is y = D d . The linear separation between the first maxima corresponding to the two wavelengths ₁ and ₂ is: y = y₂ - y₁ = ( ₂ - ₁) D d Given values are: ₁ = 480 nm = 480 10⁻⁹ m ₂ = 600 nm = 600 10⁻⁹ m D = 150 cm = 1.5 m d = 0.25 mm = 0.25 10⁻³ m Substituting these values into the expression for y : y = (600 - 480) 10⁻⁹ 1.5 0.25 10⁻³ y = 120 10