In the Young's double slit experiment, the resultant intensity at a point on the screen is 75 % of the maximum intensity of the bright fringe. Then the phase difference between the two interfering rays at that point is

Question

In the Young's double slit experiment, the resultant intensity at a point on the screen is 75 % of the maximum intensity of the bright fringe. Then the phase difference between the two interfering rays at that point is (A) (π/6) (B) (π/4) (C) (π/3) (D) (π/2)

Tardigrade Admin · Accepted Answer

Given, IR=75 % of I max =(3/4) I max =(3/4)(4 a2)=3 a2 ⇒ 4 a2 cos 2 (φ/2)=3 a2 cos 2 (φ/2)=(3/4) or cos (φ/2)=(√3/2) ⇒ (φ/2)=(π/6) or φ=(π/3)

Q. In the Young's double slit experiment, the resultant intensity at a point on the screen is $75%$ of the maximum intensity of the bright fringe. Then the phase difference between the two interfering rays at that point is

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{2}$

Given, $I_{R} = 75%$ of $I_{m a x}$
$= \frac{3}{4} I_{m a x}$
$= \frac{3}{4} (4 a^{2}) = 3 a^{2}$
$\Rightarrow 4 a^{2} cos^{2} \frac{ϕ}{2} = 3 a^{2}$
$cos^{2} \frac{ϕ}{2} = \frac{3}{4}$ or $cos \frac{ϕ}{2} = \frac{3}{2}$
$\Rightarrow \frac{ϕ}{2} = \frac{π}{6}$ or $ϕ = \frac{π}{3}$

Q. In the Young's double slit experiment, the resultant intensity at a point on the screen is 75% of the maximum intensity of the bright fringe. Then the phase difference between the two interfering rays at that point is

6π​

4π​

3π​

2π​