What is the answer ?

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1 Answer
Aug 26, 2017

Condition of minima at point OO

d=(n+1/2)lambdad=(n+12)λ
where nn 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.