In Young's slit experiment, if the distance between two slits is halved and the distance between the slits and the screen is doubled then the width of fringe will be.....

Explanation

Here, situation given in the statement is depicted in above figure. As per the statement,
`x=D=(d)/(2)" "....(1)`
Now, path difference between the waves superposing at point P is,
`r_(2)-r_(1)=S_(2)P-S_(1)P`
`=sqrt((S_(2)T_(2))^(2)+(T_(2)P)^(2))-sqrt((S_(1)T_(1))^(2)+(T_(1)P)^(2))`
`=sqrt(D^(2)+(2D)^(2))-sqrt(D^(2)+(D-D)^(2))`
`=sqrt5D-D` (From fig. and equation (1))
`("":.T_(1)P=OP-OT_(1))`
`=D(sqrt5-1)" "....(2)`
For first order dark fringe, putting n = 1 in the condition `(r_(2)-r_(1))=(2n-1)(lamda)/(2)` for destructive interference.
`r_(2)-r_(1)=(lamda)/(2)`
`:.D(sqrt5-1)=(lamda)/(2)`
`:.D=(lamda)/(2(sqrt5-1))=(lamda)/(2(2.236-1))=(lamda)/(2xx1.236)`
`:.D=(lamda)/(2.472)`
`:.D=0.404lamda`