Wave Optics Question 7
Question 7 - 2024 (29 Jan Shift 2)
In Young’s double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is $\frac{7 \lambda}{4}$. The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is :
(1) $1 / 2$
(2) $3 / 4$
(3) $1 / 3$
(4) $1 / 4$
Show Answer
Answer: (1)
Solution:
$\Delta \mathrm{x}=\frac{7 \lambda}{4}$
$\phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=\frac{2 \pi}{\lambda} \times \frac{7 \lambda}{4}=\frac{7 \pi}{2}$
$\mathrm{I}=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\phi}{2}\right)$
$\frac{\mathrm{I}}{\mathrm{I}_{\max }}=\cos ^{2}\left(\frac{\phi}{2}\right)=\cos ^{2}\left(\frac{7 \pi}{2 \times 2}\right)=\cos ^{2}\left(\frac{7 \pi}{4}\right)$
$=\cos ^{2}\left(2 \pi-\frac{\pi}{4}\right)$
$=\cos ^{2} \frac{\pi}{4}$
$=\frac{1}{2}$
