What is the answer ?

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1 Answer
Aug 26, 2017

Condition of minima at point #O#

#d=(n+1/2)lambda#
where #n# is #0,+-1,+-2,+-3....#

Hence, we have minimum at #O# for
#d=(lamda)/2,+-(3lamda)/2,+-(5lamda)/2,+-(7lamda)/2#

From above #"A" and "D"# are true.

Condition of maxima at point #O#

#d=nlambda#
where #n# is #0,+-1,+-2,+-3....#

Given, and from the figure

#d=lambda=>n=1#

It has single value. Hence there will be only one maximum at #O#
Hence, #"B"# is true.

Consider a point #P# at a distance #r# on the screen from #O#. Assuming distance #D# is measured from mid point#"*"# of sources.

Draw a perpendicular from #S_2# on line #S_1P# as shown. Let #S_1P# make an angle #theta# with the horizontal.
As #D">>"d#, path difference between wave-front reaching at #P# from #S_1and S_2# can be approximated as

#d/2costheta#

Condition for a minimum at point #P# is

#d/2costheta=(n+1/2)lambda#
where #n# is #0,+-1,+-2,+-3....#

Given is #d=4.8lambda#. Inserting in above we get

#(4.8lambda)/2costheta=(n+1/2)lambda#
#=>costheta=1/2.4(n+1/2)#

We know that #|costheta|<=1#

We get values of #n=-2,-1,0,1,# which satisfy the equation. There are #4# minima.
Hence, #"C"# is not true.
-.-.-.-.-.-.-.-.-.-.-.

It is interesting to investigate:
If #D# is measured from #S_2#
we get path difference as
#dcostheta#.
In such a case we get

#=>costheta=1/4.8(n+1/2)#

We know that #|costheta|<=1#

We get values of #n=-5,-4,-3,-2,-1,0,1,2,3,4# which satisfy the equation. There are #10# minima.
In this case #"C"# is true.
........................................

#"*"#In general

Distance of source #S_1# from #P# #=sqrt((D+d/2)^2+r^2)#
Similarly distance of source #S_2##=sqrt((D-d/2)^2+r^2)#
Actual Path difference between the two sources at the point #P# of screen will be difference of these two distances.