In Young's double slit experiment the intensities at two points, for the path difference 4 and 3 ( = wavelength of light used) are I₁ and I₂ respectively. If I₀ denotes the intensity produced by each one of the individual slits then I₁+I₂ I₀ is equal to ( 60^ =0.5, 45^ = 1 2 )

Step-by-step solution

Phase difference, = 2 l For first point, ₁= 2 ( 4 ) aligned ₁ & = ^c 2 I ₁ & =2 I ₀ (1+ ₁ ) I ₁ & =2 I ₀ . [ ₁= ( / 2)=0 ] aligned Similarly, for second point, ₂= 2 ^e 3 array ll I ₂ & =2 I ₀ (1+ ₂ ) I ₂ & =2 I ₀ (1- 1 2 ) . . [ ( 2 3 )= -1 2 ] I ₂ & = I ₀ array Hence, I₁+I₂ I₀ = 2 I₀+I₀ I₆ =3