JEE PYQ: Wave Optics Question 37
Question 37 - 2019 (11 Jan Shift 1)
In a Young’s double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is $\frac{1}{8}$th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to:
(1) 0.74
(2) 0.85
(3) 0.94
(4) 0.80
Show Answer
Answer: (2)
Solution
Given, path difference, $\Delta x = \frac{\lambda}{8}$
Phase difference ($\Delta\phi$) is given by
$\Delta\phi = \frac{2\pi}{\lambda}(\Delta x)$
$\Delta\phi = \frac{(2\pi)}{\lambda} \cdot \frac{\lambda}{8} = \frac{\pi}{4}$
For two sources in different phases,
$I = I_0\cos^2\left(\frac{\pi/4}{8}\right)$
$\frac{I}{I_0} = \cos^2\left(\frac{\pi}{8}\right)$
$= \frac{1 + \cos\frac{\pi}{4}}{2} = \frac{1 + \frac{1}{\sqrt{2}}}{2} = 0.85$
BETA
