Correct Option: A

Since both are fed with currents of equal phase and magnitude so we get from the synopsis the relation of radiation pattern with separation d, having same phase and magnitude

Enor = cos		βd	cos θ
		2

where,

β =	2π	, d = 2λ
	λ

Enor = cos		2π	2λ	cos θ		= cos (2π cos θ)
		λ	2

for maxima E_nor = ± 1
cos (2 π cos θ) = ± 1
∴    2 π cos π = ± nπ
when n = 0, 1, 2....
or   

2 cos θ = ±	n
	2

θ = cos −1		±	n
			2

Values of θ may be for n = 0, 1, 2

n = 0,	n	= 0
	2

n = 1 ±	n	= ±	1	;
	2		2

θ = 60º, 240º, 120º

n = 1 ±	n	= ± 1;
	2

θ = 0º, 180º
for minima values,we have
cos (2 π cos θ)=0

2 π cos θ = ± (2n + 1)	π
	2

where n = 0, 1, 2, .....

cos θ = ±	(2n + 1)
	4

for   n = 0,cos θ = ±	1	given θ = 75.5°,255.5°
	4

n = 1,cos θ = ±	3	given θ = 41.4°,221.4°
	4

n = 2,cos θ = ±	5	(which is not valid)
	4

so lobes are maxima at
0º, 60º, 180º, 240º, 120º, 310º
minima at
41.4º, 75.5º, 221.4º, 255.5º, 104.5º, 138.5º

so we see from the above that in horizontal plane there are two maxima at 0º and 180º.