Q. In young’s double slit experiment, the intensity of the maxima is$I$. If the width of each slit is doubled, the intensity if the maxima will be
A) $I/2$
B) $2I$
C) $4I$
D) $I$
Explanation
(b) $I_{\max} = I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}$
When width of each slit is doubled, intensity from each slit becomes twice ie,
$I_{1}^{'} = 2I_{1}\ and\ I_{2}^{'} = 2I_{2}$
$\therefore\ \ I_{\max}^{'} = I^{'} = I_{1}^{'} + I_{2}^{'} + 2\sqrt{I_{1}^{'} \times I_{2}^{'}}$
$= 2I_{1} + 2I_{2} + 2\sqrt{{2I}_{1} \times 2I_{2}}$
$= 2\left( I_{1} + I_{2} + 2\sqrt{I_{1} \times I_{2}} \right) = 2I$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.