JEE Main 2019 — Physics solution

JEE Main2019Physics

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 :

A
$(3 +)^{4} : 16$
B
4 : 1
C
25 : 9
D
9 : 1

Solution

$l_{1} = 4 l_{0} l_{2} = l_{0} \frac{1 _{m a x}}{l _{m i n}} = \frac{( l _{1} + l _{2} ) ^{2}}{( l _{1} - l _{2} ) ^{2}} = (\frac{9}{1})$

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