Question

If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern.

Answer 1

We know that intensity is directly proportional to the square of an amplitude

`I prop a^2`

if `I_1 = I/2`

if intensity reduced to 50%, the amplitude will be `a/sqrta2`

then `r = sqrt2`

`I_"max"/I_"min" = (r+1)^2/(r-1)^2 = (sqrt2 + 1)^2/(sqrt2 - 1)^2`

`I_"max"/I_"min" = ((2.414)/(0.414))^2 = (5.83)^2`

`I_"max"/I_"min"` = 33.98 ≅ 34