Optics Question 1086
Question: In Young’s double slit experiment, if the widths of the slits are in the ratio 4 : 9, the ratio of the intensity at maxima to the intensity at minima will be
[Manipal MEE 1995]
Options:
A) 169 : 25
B) 81 : 16
C) 25 : 1
D) 9 : 4
Show Answer
Answer:
Correct Answer: C
Solution:
Slit width ratio $ =4:9; $ hence $ I _{1}:I _{2}=4:9 $
$ \therefore \frac{a _{1}^{2}}{a _{2}^{2}}=\frac{4}{9}\Rightarrow \frac{a _{1}}{a _{2}}=\frac{2}{3} $
$ \therefore \frac{{I _{\max }}}{{I _{\min }}}=\frac{{{(a _{1}+a _{2})}^{2}}}{{{(a _{1}-a _{2})}^{2}}}=\frac{25}{1} $
