A double slit interference experiment performed with a light of wavelength 600 nm forms an interference fringe pattern on a screen with 10th bright fringe having its centre at a distance of 10 mm from the central maximum. Distance of the center of the same 10^th bright fringe from the central maximum when the source of light is replaced by another source of wavelength 660 nm would be_____mm.

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answer is 11.

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Detailed Solution

In case of YDSE the distance of n^th maxima from central maxima is given by

$\begin{aligned} Y = \frac{n λ D}{d} \end{aligned}$
Here n , D & d are same
$\begin{aligned} So, y \times λ \\ \begin{aligned} \Rightarrow \frac{y_{2}}{y_{1}} = \frac{λ_{2}}{λ_{1}} \Rightarrow \frac{y_{2}}{10 mm} = \frac{660 nm}{600 nm} \\ \Rightarrow y_{2} = 11 mm \end{aligned} \end{aligned}$

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