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Q. In a Young's double slit experiment, first maxima is observed at a fixed point $P$ on the screen. Now the screen is continuously moved away from the plane of slits. The ratio of intensity at point $P$ to the intensity at point $O$ (centre of the screen)Physics Question Image

Wave Optics

Solution:

$I_{p}=\frac{I_{\max }}{2}[1+\cos \phi]=\frac{I_{\max }}{2}\left(1+\cos \frac{2 \pi y}{\beta}\right)$
Where $\beta=\frac{D \lambda}{d}$
First maxima is observed at $P$ i.e., $\cos \frac{2 \pi y}{\beta}=1 ;$ as $D$
increases $b$ will increase, the value of $\cos \frac{2 \pi y}{\beta}$ should
be negative hence the ratio $I_{P} / I_{\max }$ starts decreasing but starts increasing again as $\cos \frac{2 \pi y}{\beta}$ again starts becoming positive.