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 Fig. 2.42.
If the incident beam makes an angle or `30^(@)` with the x-axis (as in the dotted arrow shown in the figure), find the y-coordinates of the first minima on either side of the central maximum.

Answer

(i). `+-0.26m,+-1.13m`
(ii) `0.26m,1.13m`

Explanation

(i). When the incident beam falls normally let path difference `=Deltax`
`thereforeDeltax=S_(2)P-S_(1)P=dsintheta`
For minimum intensity
`impliesdsintheta=(2n-1)(lamda)/(2)impliesintheta=((2n-1)lamda)/(2d)`
`impliessintheta=((2n-1)(0.5))/(2xx1)impliessintheta=(2n-1)/(4)`
Since `sin theta le 1 therefore `n can be either 1 or 2
When n=1, `sintheta_(1)=(1)/(4),tantheta_(1)=(1)/(sqrt(15))`
When `n=2,sintheta_(2)=(3)/(4),tantheta_(2)=(3)/(sqrt(7))`
`thereforey=Dtantheta` or `y=1xxtantheta`
For minima above centre O
`y_(1)=tantheta_(1)=(1)/(sqrt(15))=0.26m`
`y_(2)=tantheta_(2)=(3)/(sqrt(7))=1.13m`
There will be four minima due to interference at positions `+-0.26m,+-1.13m`

When incident beam makes as angle of `30^(@)` width x-axis
For central maxima path difference should be zero
`Deltax_(1)=0` or `Deltax_(2)=0`
`dsintheta=dsinalphaimpliesalpha=theta=30^(@)`
`becausey=Dtanthetathereforey=1xxtan30^(@)=0.58m`
for first minima `dsintheta-dsinalpha=(lamda)/(2)`
`impliesdsintheta=(lamda)/(2)+dsinalphaimpliessintheta=(lamda)/(2d)+sinalpha`
`sintheta=(0.5)/(2xx1)+sin30^(@)impliessintheta=0.75=(3)/(4)`
`thereforey_(2)=Dtantheta=1xx(3)/(sqrt(7))=1.13m`
For first minima on either side
`d sin theta=(lamda)/(2)impliesintheta=(lamda)/(2d)=(0.5)/(2xx0.1)=(1)/(4)`
`thereforetantheta=(1)/(sqrt(15))=0.26m`
`thereforey_(1)=1xxtantheta=0.26m`
Therefore y coordinates of the first minima on either side of the central maximum are `y=0.26`
m and `y_(2)=1.13m`