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MHT CET · Physics · Wave Optics

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

  1. A 1
  2. B \(1 / 9\)
  3. C 1,4
  4. D \(1 / 3\)
Verified Solution

Answer & Solution

Correct Answer

(C) 1,4

Step-by-step Solution

Detailed explanation

Amplitude of the superimposing waves are
\(
\begin{array}{l}
\frac{a_{1}}{a_{2}}=\left(\frac{1}{9}\right)^{1 / 2}=\frac{1}{3} \\
\frac{I_{\operatorname{minima}}}{I_{\operatorname{maxima}}}=\frac{\left(a_{1}-a_{2}\right)^{2}}{\left(a_{1}+a_{2}\right)^{2}}=\frac{1}{4}
\end{array}
\)