What will be the diffraction angle of first order maxima in diffraction obtained due to light of wavelength 55nm and width of slit 0.55 mm ?

Explanation

Here given slab is placed in the path of lower ray `AS_(2)` and so entire interference pattern will shift down from point O. Now, suppose new first order bright fringe is obtained at point `P_(0)`, at distance `x_(0)` below point O. Here width of slit `=S_(1)S_(2)=2d`. Now geometrical path difference between the waves superposing at point `P_(0)` is `S_(1)M=2dsintheta_(0)` whereas optical path difference is
`=(mu-1)t`
`=(mu-1)L" (":."here "t=L)`
`=(1.5-1)((d)/(4))("":.L=(d)/(4))`
`=(d)/(8)`
Net path difference `=2dsintheta_(0)+(d)/(8)`

Here for bright fringe, above net path difference should be `nlamda` and so :
`:.2dsintheta_(0)+(d)/(8)=nlamda`
`:.2dsintheta_(0)+(d)/(8)=0`
( `:.` For central bright fringe n=0)
`:.2dsintheta_(0)=-(d)/(8)`
`:.sintheta_(0)=-(1)/(16)`
Here `theta_(0)` is quite small and so
`sintheta_(0)~~tantheta_(0)=(x_(0))/(D)`
`:.(x_(0))/(D)=-(1)/(16)`
`:.x_(1)=-(D)/(16)`
`implies` Now, new central bright fringe will be at distance `(D)/(16)`, below point O.
Now, suppose first order dark fringe is obtained on the screen at point `P._(1)`, at distance `x_(1).`, below point 0. At this point, if corresponding angle of diffraction is `theta_(1).`, then :
Net path difference `=(2n-1)(lamda)/(2)`
`2dsintheta._(1)+(d)/(8)=(2n-1)(lamda)/(2)`
`:.2dsintheta._(1)+(d)/(8)=(lamda)/(2)`
( `:.` For first order dark fringe, n=1)
Here for appreciable amount of diffraction, we should take `lamda` = width of slit = 2d
( `:.` Here width of slit is 2d). Hence,
`2dsintheta._(1)+(d)/(8)=d`
`:.2dsintheta._(1)=d-(d)/(8)=(7d)/(8)`
`:.sintheta._(1)=(7)/(16)`
`:.tantheta._(1)=(7)/(16)`
`:.(x._(1))/(D)=(7)/(16)`
`:.x._(1)=(7D)/(16)`
`implies` New first order dark fringe will be obtained on the screen at distance `(7D)/(16)` below point O.