Two wavelengths of sodium light 590 nm, 596 nm are used, in turn, to study the diffraction taking place at a single slit of aperture 2 × 10^-6 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the position of first maximum of the diffraction pattern obtained in the two cases.

Solution

Wavelength of first sodium line,

λ_{1}

= 590 nm
Wavelength of second sodium line,

λ_{2}

= 596 nm
Aperture or width of single slit, d = 2 × 10^-6 m
Distance between the slit and the screen , D = 1.5 m

Separation between the first secondary maximum in the two cases is,

x_{2} - x_{1} = \frac{3}{2} \frac{{Dλ}_{2}}{d} - \frac{3}{2} \frac{{Dλ}_{1}}{d}

\Rightarrow

x_{2} - x_{1} = \frac{3 D}{2 d} (λ_{2} - λ_{1})

\Rightarrow

x_{2} - x_{1} = \frac{3 \times 1.5}{2 \times 2 \times 10^{- 6}} (596 \times 10^{- 9} - 590 \times 10^{- 9})

\Rightarrow

x_{2} - x_{1} = \frac{3 \times 1.5 \times 6 \times 10^{- 3}}{4}

= 6.75 \times 10^{- 3} m = 6.75 mm .

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