Question

In Young's double slit experiment, first slit has width four times the width of the second slit. The ratio of the maximum intensity to the minimum intensity in the interference fringe system is:

• A. $2:1$
• B. $4:1$
• C. $9:1$
• D. $8:1$

Answer 1

Answer: (C)

$9:1$

Let intensity from the smaller slit be $I$ so, the intensity from the bigger slit will be $4I$ since it is 4 times of smaller slit.
Now, $I_{max}=4I+I+2\sqrt{I\left(4I\right)}$
$=5I+2\sqrt{4I^2}$
$=5I+2\left(2I\right)$
$=9I$
$Imin=4I+I-2\sqrt{I\left(4I\right)}$
$=5I-2\left(2I\right)$
$=I.$
So, $\frac{I_{max}}{I_{min}}=\frac{9I}{I}=\frac{9}{1}.$