JEE Class main Answered

Intensity I = I₁ + I₂ + 2 (I₁ I₂ )^1/2cosδ

 where I₁ and I₂ are intensities of light coming out of slit-1 and slit-2

If I₁ = I and I  = I  , then by using cosδ=+1, we have maximum intensity I_max = 4I

by using cosδ=-1, we have minimum intensity I_min = 0

If I₁ = I and I  = I/2  , then by using cosδ=+1, we have maximum intensity I_max = I [ (3/2) + (2/√2) ] =  I[ (3/2) + √2 ]

by using cosδ=-1, we have minimum intensity I_min = I [ (3/2) - (2/√2) ] =  I[ (3/2) - √2 ]

Hence if intensity of one slit is halved by introducing a transparent plate, intensity of bright fringe is getting reduced

and intensity of dark fringe is increased