Q. The width of one of the two slits in Young's double slit experiment is $d$ while that of the other slit is $x\,d$. If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is $9:4$ then what is the value of $x$ ? (Assume that the field strength varies according to the slit width.)
A) 4
B) 5
C) 3
D) 2
Explanation
[Solution by Dr. Abubacker] The field amplitudes are proportional to the widths: $a_1=d$, $a_2=xd$. $\frac{I_{max}}{I_{min}}=\left(\frac{x+1}{x-1}\right)^2=\frac94\Rightarrow\frac{x+1}{x-1}=\frac32$, so $x=5$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.