In the YDSE arrangement shown here, the intensity on the screen due to slit-2 is four times that of slit-1. If resultant intensity at the position of central maxima O is I , the resultant intensity at point P, where the phase difference between two waves coming from two slits is ( textcos)-1 ((1/4)) is

Question

In the YDSE arrangement shown here, the intensity on the screen due to slit-2 is four times that of slit-1. If resultant intensity at the position of central maxima O is I , the resultant intensity at point P, where the phase difference between two waves coming from two slits is ( textcos)-1 ((1/4)) is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-pgtbkj36ajw7.png" /> (A) (I/3) (B) (2 I/3) (C) (I/4) (D) (I/2)

Tardigrade Admin · Accepted Answer

I = (√I0 + √4 I0)2 = 9 I0 I text0 text= (I/ textg) I' = I0 + 4 I0 + 2 √I0 text. 4 I0 × (1/4) = 6 I0 = (2/3) I

$\frac{I}{3}$

$\frac{2 I}{3}$

$\frac{I}{4}$

$\frac{I}{2}$

$I = (I_{0} + 4 I_{0})^{2} = 9 I_{0}$
$I_{0} = \frac{I}{g}$
$I^{^{'}} = I_{0} + 4 I_{0} + 2 I_{0} . 4 I_{0} \times \frac{1}{4}$
$= 6 I_{0}$
$= \frac{2}{3} I$

3I​

32I​

4I​

2I​

I=(I0​​+4I0​​)2=9I0​ I0​=gI​ I′=I0​+4I0​+2I0​.4I0​​×41​ =6I0​ =32​I

$\frac{I}{3}$

$\frac{2 I}{3}$

$\frac{I}{4}$

$\frac{I}{2}$

$I = (I_{0} + 4 I_{0})^{2} = 9 I_{0}$
$I_{0} = \frac{I}{g}$
$I^{^{'}} = I_{0} + 4 I_{0} + 2 I_{0} . 4 I_{0} \times \frac{1}{4}$
$= 6 I_{0}$
$= \frac{2}{3} I$