In a Young's double slit experiment, the separation between the two slits is d = 3 , where is the wavelength of the monochromatic light used. If the intensity of the central maximum is I₀ , the intensity of light at an angular position of 30^ on the screen will be :

Step-by-step solution

The path difference x at an angular position is given by x = d . Given d = 3 and = 30^ : x = ( 3 ) (30^ ) = 3 1 2 = 6 The corresponding phase difference is: = 2 x = 2 6 = 3 The intensity I at this point is given by I = I₀ ^2 ( 2 ) , where I₀ is the intensity of the central maximum. Substituting the value of : I = I₀ ^2 ( 6 ) = I₀ ( 3 2 )^2 = 3I₀ 4 Answer: 3I₀ 4