Question

two wavelengths of sodium light 590nmand 596nm are used in diffraction for a aperture of 2× 10^-4m. thedistance between the slit and screen is 1.5m calcuate separation between positions of the first maxima of diffraction patterns in oth cases.

Answer 1

$$\begin{gathered} Differencebetweenthetwomaximaisgivenbythepostionoftheminima. \\ Positionoftheminimaisgivenby: \\ y=\frac{D\lambda }{a} \\ D=differencebetweentheslitandthescreen \\ aistheaperture \\ For\lambda =590nm \\ y=\frac{2\times 590\times {10}^{-9}}{2\times {10}^{-4}} \\ y=590\times {10}^{-5} \\ y=0.0590m \\ for\lambda =596nm \\ y=\frac{2\times 596\times {10}^{-9}}{2\times {10}^{-4}}=0.0596m \end{gathered}$$