Sodium light of wavelengths $650 nm$ and $655 nm$ is used to study diffraction at a single slit of aperture $0.5 mm$ . The distance between the slit and the screen is $2.0 m$ . The separation between the positions of the first maxima of diffraction pattern obtained in the two cases is $____\times 10^{- 5} m$ .

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answer is 3.

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Detailed Solution

$\begin{aligned} a \sin θ = \frac{3 λ}{2} \\ \frac{y}{L} = θ = \frac{3 λ}{2 a} \\ y_{1} = \frac{3 λ_{1} L}{2 a} \\ y_{2} = \frac{3 λ_{2} L}{2 a} \end{aligned} \begin{aligned} L = 2 m \\ λ_{2} = 655 nm \\ λ_{1} = 650 nm \\ a = 0.5 nm \end{aligned} \begin{aligned} Δy = y_{2} - y_{1} = \frac{3 (λ_{2} - λ_{1})}{2 a} L \\ = \frac{3 (655 - 650)}{2 \times 0.5 \times 10^{- 3}} \times 2 \times 10^{- 9} \\ = \frac{3 \times 5 \times 2}{1 \times 10^{- 3}} \times 10^{- 9} \\ = 3 \times 10^{- 5} \end{aligned}$