Question

What is the answer ?

Answer 1

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.