A coherent parallel beam of microwaves of wavelength `lambda = 0.5 mm` falls on aYoung's double- slit apparatus. The separation between the slits is 1.0 mm. The intensity of microwaves is measured on a screen placed parallel to the plane of the slits at a distance of 1.0 m from it as shown in figure

If the incient beam falls normally on the double-slit apparatus, find the order of the interference minima on the screen

Explanation

(a) Given seperation between two slits = 2d 1.0mm, D=1.0m,
Path diffenrece = `2dsin theta`
For minima `2 d sintheta = (2n -1 ) lambda/2`
`therafore 1 xx 10^(-3) = ((2n-1)xx(0.5xx10^(-3)))/2` as `theta=90^@` for the highest possible order of minimum.
or `(2n-1) = (2xx10^(-3))/(0.5xx10^(-3)) implies n=2.5`
Thus, only two minima are possible on either side of central maxima. Thus total number of minima = 2+2 =4
For minima `y= pm ((2n-1)lambdaD)/(4d) = (2n-1)0.25`
`y_(1)=0.25m , y_(2)=0.75m` and `y_(4)= -0.75m`
(b) Initial path difference = `2d sin 30^@`
For the central maxima
`therefore y/D xx2d= 2d sin30^@`
`y=Dsin30^@=0.5m`
For first minima
`y/D xx2d -2dsin 30^@ = om lambda/2`
or `y= pm 0.25 + 0.5`
or `y_(1)0.75m ` and `y_(2) = 0.25 m`.