Q.
We have certain modification of YDSE apparatus/original YDSE apparatus. The value of intensity at point $P$ is $n I_{0}$. Find $n$. Given: $d=1\, mm , D=1\, m , \lambda=500\, nm$ for all cases.
Intensity due to either slit on screen $=I_{0} . P$ is centre of screen
Wave Optics
Solution:
$I =4 I_{0} \cos ^{2}\left(\frac{\pi}{\lambda} \times d \times 0.02\right)$
$=4 I_{0} \cos ^{2}\left(\frac{\pi \times 10^{-3} \times 0.02}{500 \times 10^{-9}}\right)$
$=4 I_{0} \cos ^{2}(40 \pi)=4 I_{0}$
