If the two slits in Young's double slit experiment have width ratio 1:25, then the ratio of the intensity of maxima and minima in the interference pattern will .....

Options

A. `(4)/(9)`
B. `(9)/(4)`
C. `(121)/(49)`
D. `(49)/(121)`

Answer

B

Explanation

Let width of slits `W_(1) and W_(2)`
Now `I prop W`
`:.(I_(1))/(I_(2))=(W_(1))/(W_(2))=(1)/(25)`
`:.(I_(1))/(I_(1))=(25)/(I)`
`:. (sqrt(I_(2)))/(sqrt(I_(1)))=(5)/(I)`
but `I prop A^(2)` (A is amplitude)
`:. (I_(2))/(I_(1))=((A_(2))/(A_(1)))^(2)`
`:. (A_(2))/(A_(1))= sqrt((I_(2))/(I_(1)))`
`:. A_(2)= sqrt(I_(2)) and A_(1)=sqrt(I_(1))`
Now `(I_("max"))/(I_(""min"))=((A_(2)+A_(1))^(2))/((A_(2)-A_(1))^(2))=([sqrt(I_(2))+sqrt(I_(1))]^(2))/([sqrt(I_(2))-sqrt(I_(1))]^(2))`
`=[((sqrt(1_(2)))/(sqrt(I_(2))/(sqrt(I_(1)))+1))/((sqrt(I_(2)))/(sqrt(I_(1)))-1)]^(2)`
`=[((5)/(1)+1)/((5)/(1)-1)]^(2)`
`=((6)/(4))^(2)=((3)/(2))^(2)`
`=(9)/(4)`