Question

In a double slit experiment with d=5*a, a=40* $\lambda$ , compare (as a ratio) the intensity of the third order interference maximum with that of the zero order maximum.

Answer 1

the positon of the constructive interference on the screen

y = m*lambda*d/a

a = slit width

d = distance of the sreen from the slit

lambda = wave length

the position of the 3 order maximum = y = (3*lambda*5a)/(a)

y = 15*lambda

the intensity of light on the screen other than the central maxima

I = Io*cos^2(pi*a*y/lambda*a)

Io = intensity of the entral maxima

I / Io = cos^2(pi*a*y/lambda*d)

I/Io = cso^2(pi*a*15*lambda/lambda*5a)

I / Io = cos^2(pi*3) = 0.99