A beam of light consisting of two wavelengths \(7000\; \mathring A\) and \(5500 \;\mathring A\) is used to obtain interference pattern in Young's double slit experiment. The distance between the slits is \(2.5\,mm\) and the distance between the plane of slits and the screen is \(150\,cm\). The least distance from the central fringe, where the bright fringes due to both the wavelengths coincide, is \(n \times 10^{-5}\,m\). The value of \(n\) is .............

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Answer & Solution

Correct Answer

(C) \(462\)

Detailed explanation

\(d =2.5\,mm , D =150\,cm\)

Fringe width \(\beta=\frac{\lambda D}{d}\)

Let \(n^{\text {th }}\) bright tringle of \(\lambda_1\) match with \(m ^{\text {th }}\) bright triangle of \(\lambda_2\)

\(\Rightarrow n \beta_1=m \beta_2\)

\(\Rightarrow n \lambda_1=m \lambda_2 \Rightarrow \frac{n}{m}=\frac{\lambda_2}{\lambda_1}=\frac{5500}{7000}\)

\(\Rightarrow \frac{n}{m}=\frac{11}{14}\)

Distance where bright fringe will match

\(=n \beta_1=\frac{11 \times 7000 \mathring A \times 150\, cm }{0.25\,cm }\)

\(=462 \times 10^{-5}\)

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