Search any question & find its solution
Question:
Answered & Verified by Expert
In Young's double slit experiment if the slit widths are in the ratio $1: 9 .$ The ratio of the intensity at minima to that at maxima will be
Options:
Solution:
1916 Upvotes
Verified Answer
The correct answer is:
1,4
Amplitude of the superimposing waves are
$$
\begin{array}{l}
\frac{a_{1}}{a_{2}}=\left(\frac{1}{9}\right)^{1 / 2}=\frac{1}{3} \\
\frac{I_{\operatorname{minima}}}{I_{\operatorname{maxima}}}=\frac{\left(a_{1}-a_{2}\right)^{2}}{\left(a_{1}+a_{2}\right)^{2}}=\frac{1}{4}
\end{array}
$$
$$
\begin{array}{l}
\frac{a_{1}}{a_{2}}=\left(\frac{1}{9}\right)^{1 / 2}=\frac{1}{3} \\
\frac{I_{\operatorname{minima}}}{I_{\operatorname{maxima}}}=\frac{\left(a_{1}-a_{2}\right)^{2}}{\left(a_{1}+a_{2}\right)^{2}}=\frac{1}{4}
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.