JEE PYQ: Wave Optics Question 28
Question 28 - 2019 (10 Apr Shift 2)
In a Young’s double slit experiment, the ratio of the slit’s width is 4 : 1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be :
(1) 25 : 9
(2) 9 : 1
(3) 4 : 1
(4) $(\sqrt{3} + 1)^4 : 16$
Show Answer
Answer: (2)
Solution
Let $I_1$, $I_2$ are intensity of light coming from two slits.
As, $I \propto W$
$\therefore \frac{I_1}{I_2} = \frac{4}{1}$
$I_1 = 4I_0$, $I_2 = I_0$
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2} = \frac{(\sqrt{4} + 1)^2}{(\sqrt{4} - 1)^2} = \frac{9}{1}$
BETA
