Two beams of light having intensities I and 4 I interfere to produce a fringe pattern on a screen. The phase difference between the two beams are π / 2 and π / 3 at points A and B respectively. The difference between the resultant intensities at the two points is xI. The value of x will be

Question

Two beams of light having intensities I and 4 I interfere to produce a fringe pattern on a screen. The phase difference between the two beams are π / 2 and π / 3 at points A and B respectively. The difference between the resultant intensities at the two points is xI. The value of x will be  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

φ A =(π/2) φ B =(π/3) I A = I +4 I +2 √ I √4 I cos ((π/2)) =5 I +4 I (0)=5 I I B = I +4 I +2 √ I √4 I cos (60°) =5 I +4 I × (1/2)=7 I I B - I A =7 I -5 I =2 I ,( x =2)

Q. Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the two beams are π/2 and π/3 at points A and B respectively. The difference between the resultant intensities at the two points is xI. The value of x will be ____

ϕA​=2π​ ϕB​=3π​ IA​=I+4I+2I​4I​cos(2π​) =5I+4I(0)=5I IB​=I+4I+2I​4I​cos(60∘) =5I+4I×21​=7I IB​−IA​=7I−5I=2I,(x=2)

$ϕ_{A} = \frac{π}{2}$
$ϕ_{B} = \frac{π}{3}$
$I_{A} = I + 4 I + 2 I 4 I cos (\frac{π}{2})$
$= 5 I + 4 I (0) = 5 I$
$I_{B} = I + 4 I + 2 I 4 I cos (6 0^{\circ})$
$= 5 I + 4 I \times \frac{1}{2} = 7 I$
$I_{B} - I_{A} = 7 I - 5 I = 2 I, (x = 2)$