JEE Main 2020 — Wave Optics Question with Solution

Given,

Two sources od same wavelength, $\lambda =1 \mathrm{m}$

Distance between two sources,$S_1{S}_2=1.5 \mathrm{m}$

Distance of listener from the source $S_2$, $S_{2}L=2 \mathrm{m}$

Condition for minimum intensity, at $2 \mathrm{m}$ is

Path difference, $P.D.=(2n+1)\frac{\lambda }{2} ...\left(1\right)$

$\Rightarrow P.D.=\sqrt{2^2+(1.5{)}^2}=(2n+1)\frac{\lambda }{2}$
$\Rightarrow 2.5=(2n+1)\frac{\lambda }{2} ...\left(1\right)$

Now, the path difference between the rays is so that the adjacent maxima is obtained,

$P.D.=d-2=n\lambda ...\left(2\right)$

The path difference between adjacent maxima and minima is $\frac{\lambda }{2}$, form equation (1) and (2) we get,

$(d-2)-2.5=\frac{\lambda }{2} ...\left(3\right)$
$\Rightarrow d-2=2.5+\frac{1}{2}$

$\Rightarrow d=5 \mathrm{m}$